AP Inter 2nd Year

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 2 Electrostatic Potential and Capacitance Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 2nd Lesson Electrostatic Potential and Capacitance Questions and Answers

I. Multiple Choice Questions

Question 1.
Two points P and Q are maintained at the potential of 10V and -4V respectively. The work done in moving 100 electrons from P to Q is
1) -9.6 × 10^-17J
2) 9.6 × 10^-17J
3) -2.24 × 10^-16J
4) 2.24 × 10^-16J
Answer:
4) 2.24 × 10^-16J
The potential difference between two points P & Q is ∆W = V_P – V_Q = 10 – (-4) = 14 V
No.of electrons n = 100, e = 1.6 × 10¹⁹ C
Total charge q = ne = 100 × (1.6 × 10¹⁹C) = -1.6 × 10^-17C
Work done W = q(∆V) = (-1.6 ) × (14) × 10^-17= 2.24 × 10^-17

Question 2.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface becomes 80V. Then the potential at the centre of the sphere is
1) 80V
2) 800V
3) 8V
4) 0
Answer:
1) 80V
The potential must be a constant value throughout the interior of the sphere.
∴ V_surface = V_centre = 80 V

Question 3.
The electric potential of a short dipole varies with distance r as
1) 1/r²
2) 1/r
3) 1/r
4) 1/r⁴
Answer:
1) 1/r²
For a short dipole V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p} \cos \theta}{\mathrm{r}^2}\)
∴ V ∝ \(\frac{1}{r^2}\)

Question 4.
The examples of polar molecules are
1) HCl
2) H₂O
3) NH₃
4) Both (1) and (2)
Answer:
4) Both (1) and (2)
Polar molecules are formed when atoms share the electrons unequally.

Question 5.
Two conducting spheres of radii 5 cm and 10 cm are given a charge of 15 μC each. After the two spheres are joined by a conducting wire, the charge on the smaller sphere is
1) 5 μC
2) 10 μC
3) 15 μC
4) 20 μC
Answer:
2) 10 μC
Total charge Q = q₁ + q₂ = 15 + 15 = 30 μC
At equilibrium, V₁ = V₂
\(\frac{K q_1}{r_1}=\frac{K q_2}{r_2}\) ⇒ \(\frac{r_2}{r_1}=\frac{q_2}{q_1}\)
⇒ \(\frac{r_2}{r_1}\) + 1 ⇒ \(\frac{q_2}{q_1}\) + 1
⇒ \(\frac{r_2+r_1}{r_1}=\frac{q_2+q_1}{q_1}\)
∴ For smaller sphere, charge q₁ = \(\frac{\left[q_1+q_2\right] r_1}{r_1+r_2}=\frac{Q r_1}{r_1+r_2}=\frac{30 \times 5}{10+5}=\frac{150}{15}\) = 10 = 10 μC

Question 6.
A hollow conducting sphere of radius R has a charge +Q on its surface. What is the electric potential within the sphere at a distance r = \(\frac{\mathrm{R}}{\mathrm{3}}\) from its centre?
1) Zero
2) \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\)
3) \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)
4) \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^2}\)
Answer:
3) \(\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\)
Electric potential remains constant in the entire interior of the sphere.
∴ V_surface = V_inside ⇒ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{R}}\) [∵ E = 0, \(\frac{\mathrm{dV}}{\mathrm{dx}}\) = 0]

Question 7.
An electric dipole consisting of two opposite charges of 2 μC each separated by a distance of 3 cm is placed in an electric field of 2 × 10⁵ N/C . The maximum torque on the dipole in Nm will be
1) 12 × 10^-1
2) 12 × 10^-3
3) 24 × 10^-1
4) 24 × 10^-3
Answer:
2) 12 × 10^-3
Maximum Torque = PE = (q × 2a)E = (2 × 10^-6 × 3 × 10^-2) (2 × 10⁵) = 12 × 10^-3 Nm

Question 8.
Electric potential at an equatorial point of a small dipole with dipole moment P (r, distance from the dipole) is
1) V = Zero
2) V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^2}\)
3) V = \(\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)
4) V = \(\frac{2 \mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}\)
Answer:
1) V = Zero
Potential V = \(\frac{1}{4 \pi \varepsilon_0} \frac{p \cos \theta}{r^2}\), On equatorial line, θ = 90°, So cos θ = 0
∴ Potential V = 0

Question 9.
A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant k of the oil is
1) 0.45
2) 0.55
3) 1.10
4) 2.20
Answer:
4) 2.20
When immersed in oil, new capacity of the capacitor C’ = KC [∵ C = Actual capacity]
∴ K = \(\frac{C^{\prime}}{C}=\frac{110}{50}\) = 2.2

Question 10.
The potentials of the two plates of a capacitor are +10V and -10V . The charge on one of the plates is 40C. The capacitance of the capacitor is
1) 2F
2) 4F
3) 0.5F
4) 0.25F
Answer:
1) 2F
Potential difference ∆V = V₁ – V₂ = 10 – (-10) = 20 V, Charge Q = 40C
∴ C = \(\frac{\mathrm{Q}}{\Delta \mathrm{~V}}=\frac{40}{20}\) = 2

Question 11.
The capacitance between the points A and B in the given circuit (fig) will be

1) 1 μF
2) 2 μF
3) 3 μF
4) 4 μF
Answer:
1) 1 μF
Here 1.5 μF capacitor with 1.5 μF are connected in parallel.
∴ C_eq = C₁ + C₂ = 1.5 + 1.5 = 3 μF
Now 3 μF, 3 μF with 3 μF capacitors are connected in series.

∴ \(\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}\) = 1
∴ C_S = 1 μF

Question 12.
Three capacitors of capacitance 1 μF, 2 μF and 3 μF are connected in series and a potential difference of 11 V is applied across the combination. Then, the potential difference across the plates of 1 μF capacitor is
1) 2 V
2) 4 V
3) 1 V
4) 6 V
Answer:
4) 6 V
In series combination \(\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\) ⇒ C_S = \(\frac{6}{11}\)
∴ Q = C_S V = \(\frac{6}{11}\) × 11 = 6 μC
Potential difference across 1 μF capacitor is V₁ = \(\frac{Q}{C_1}=\frac{6}{1}\) = 6V
∴ V₁ = 6 V

Question 13.
A parallel plate air capacitor is charged and then isolated. When a dielectric material is inserted between the plates of the capacitor completely, then which of the following does not change?
1) Electric field between the plates
2) Potential difference across the plates
3) Charge on the plates
4) Energy stored in the capacitor
Answer:
3) Charge on the plates
The total charge remains constant because the capacitor is isolated, to there is no external path for charge to flow. Charge on plates is constant

Question 14.
The capacity of a capacitor is 48 μp. When it is charged from 0.1C to 0.5C the change in energy stored is
1) 2500 J
2) 2.5 × 10^-3 J
3) 2.5 × 10⁶ J
4) 2.42 × 10^-2 J
Answer:
1) 2500 J
Energy stored in a capacitor E = \(\frac{Q^2}{2 C}\)
Change in energy ∆E = E₂ – E₁ = ∆\(\frac{Q_2^2}{2 C}-\frac{Q_1^2}{2 C}\)
= \(\frac{1}{2 C}\left[Q_2^2-Q_1^2\right]\) = \(\frac{1}{2 \times 48 \times 10^{-6}}\)[(0.5)² – (0.1)²] = 2500 J

Question 15.
A capacitor is charged by a battery and the energy stored is U . The battery is now removed and the separation distance between the plates is doubled. Then the energy stored is now
1) U / 2
2) U
3) 2U
4) 4U
Answer:
3) 2U
Energy stored in a capacitor E = \(\frac{Q^2}{2 C}\) But C = \(\frac{\varepsilon_0 A}{d}\)
∴ E = \(\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \varepsilon_0 \mathrm{~A}}\) ⇒ E ∝ d
∴ \(\frac{E_2}{E_1}=\frac{d_2}{d_1}=\frac{2 d_1}{d_1}\) = 2 ⇒ E₂ = 2E₁

II. Fill in the Blanks

Question 1.
The potential at a point P at a distance 9cm away from a point charge 1 × 10^-7 C is _________
Answer:
10⁴ V.
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}=\frac{9 \times 10^9 \times 10^{-7}}{9 \times 10^{-2}}\) = 10⁴

Question 2.
Work done in moving a charge q from one point to another inside a uniformly charged conducting sphere is always ________
Answer:
zero.
Inside a charged conductor E = 0. So, potential is constant.
Thus potential difference ∆V = 0.
∴ Workdone = Q∆V = 0

Question 3.
The electric dipole moment per unit volume is called ___________.
Answer:
polarisation.
Polarisation P = \(\frac{1}{V}\) × Dipole moment

Question 4.
1 GeV is ___________Joules
Answer:
1.6 × 10^-10 .
1 GeV = 1 × 10⁹ × 1.6 × 10^-19 = 1.6 × 10^-10J

Question 5.
The potential energy of a system consisting of two charges 7 μC and -2 μC placed at (-9cm, 0, 0) and (9cm, 0, 0) is _______
Answer:
-0.7 J.
d = 9 – (-9) = 18 cm
P.E. = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{~d}}=\frac{9 \times 10^9 \times\left(7 \times 10^{-6}\right) \times\left(-2 \times 10^{-6}\right)}{18 \times 10^{-2}}\) = -0.7 J

Question 6.
The potential at infinity due to a point charge is ________
Answer:
zero.
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}\); when r → ∞ we get V = 0

Question 7.
A 25 μF capacitor is charged to a potentiâl of 18V . Then charge stored in the capacitor is ________
Answer:
4.5 × 10^-4 C.
Charge Q = CV = (25 × 10^-6)(18) = 4.5 × 10^-4 C

Question 8.
The negative electrical potential gradient is called ___________
Answer:
Intensity of electric field.
The negative gradient of electric potential iscalled Electric intensity of field.

Question 9.
The dielectric field strength of air is about _______
Answer:
3 × 10⁶ V/m.

Question 10.
The product ε₀ K is called __________
Answer:
permittivity of the medium.
Dielectric constant K = \(\frac{\varepsilon}{\varepsilon_0}\) ⇒ Kε₀ = ε
ε is the permitivity of the medium.

Question 11.
The energy density of the electric field in a region with electric field (E) is ________.
Answer:
\(\frac{1}{2}\) ε₀E²
Energy U = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\left[\frac{\varepsilon_0 A}{d}\right]\) [ED]² = \(\frac{1}{2}\left[\frac{\varepsilon_0 A}{d}\right]\) E²d²
= \(\frac{1}{2}\) ε₀E²(Ad) = 2}[/latex] ε₀E²V [∵ Ad = V]
⇒ \(\frac{U}{V}=\frac{1}{2}\) ε₀E²
Energy per unit volume is energy density.

Question 12.
The potential energy of a dipole of dipole moment p in a uniform field E is ________
Answer:
U = \(-\vec{\mathrm{P}} . \vec{\mathrm{E}}\)
Potential energy of a dipole U = -pEcosθ = – \(\vec{\mathrm{P}} . \vec{\mathrm{E}}\)

Question 13.
The total capacitance when two capacitors of capacitance C₁ and C₂ are connected in parallel is ________
Answer:
C = C₁ + C₂
In parallel combination q = q₁ + q₂ ⇒ CV = C₁ V + C₂ V
⇒ CV = V(C₁ + C₂) [∵ q = CV]
∴ C = C₁ + C₂

III. One Word Answer Questions

Question 1.
Write the expression for potential due to a point charge q at a distance r from it.
Answer:
Potential V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}}\). Here q charge, r distance and ε₀ is the permittivity of free space.

Question 2.
Define the unit of electric potential.
Answer:
One volt (V) is equal to one joule of work performed per unit coulomb of charge
Unit of electric potential is volt (V).

Question 3.
What is the electric potential of a dipole on the equatorial plane?
Answer:
The electrical potential V at any point on the equatorial plane of an electric dipole is ‘zero’.

Question 4.
What is the shape of equipotential surface for a uniform electric field?
Answer:
In a uniform electric field, the equipotential surfaces take the shape of equidistant parallel planes perpendicular to the direction of the electric field lines.

Question 5.
What is the SI unit of electric potential?
Answer:
SI unit of electric potential is volt (V)

Question 6.
What is an equipotential surface?
Answer:
It is the surface at every point of which, the electric potential is the same.

Question 7.
Is the electric potential a scalar or vector quantity?
Answer:
Electric potential is a scalar quantity.
It possesses only magnitude and does not have a specific direction.

Question 8.
What is the potential difference between two points on an equipotentiai surface?
Answer:
The potential difference between any two points on an equipotential surface is zero.
V is constant at every point.
∴ V_A – V_B = 0

Question 9.
Define the capacitance of a capacitor.
Answer:
Capacitance (C) is the ratio of charge stored (Q) on a conductor to the potential difference (V) applied.
Formula: C = \(\frac{\mathrm{Q}}{\mathrm{~V}}\)

Question 10.
What is the SI unit of capacitance?
Answer:
SI unit of capacitance is farad (F).

Question 11.
How does the capacitance of a parallel plate capacitor change when the distance between the plates is halved?
Answer:
The new capacity becomes double to the initial capacity.
Capacitance C = \(\frac{\varepsilon_0 A}{d}\) ⇒ C ∝ \(\frac{1}{d}\)
∴ \(\frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\mathrm{d}_1}{\mathrm{~d}_2}=\frac{\mathrm{d}_1}{\mathrm{~d}_1 / 2}\) = 2 ⇒ \(\frac{\mathrm{C}_2}{\mathrm{C}_1}\) = 2 ⇒ C₂ = 2C₁

Question 12.
Write the expression for capacitance of a parallel plate capacitor with air between plates.
Answer:
Capacitance of a parallel capacitor is C = \(\frac{\varepsilon_0 A}{d}\). Here ε₀ is permittivity of the air,
A is the area of the plates, d is the separation between the plates.

Question 13.
What is the total capacitance when two capacitors of capacitance C, and are connected in series?
Answer:
Total capacitance C = \(\frac{C_1 C_2}{C_1+C_2}\)

In series combination, V \(\frac{1}{\mathrm{C}}\)
V = V₁ + V₂ ⇒ \(\frac{Q}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}\) [∵ V = \(\frac{Q}{C}\)]
∴ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{C_1+C_2}{C_1 C_2}\)
⇒ C = \(\frac{C_1 C_2}{C_1+C_2}\)

IV. Very Short Answer Questions

Question 1.
Can there be electric potential at a point with zero electric intensity? Give an example.
Answer:
Yes.
Ex: Inside a charged metal spherical shell, potential is non-zero but electric intensity is zero .

Question 2.
Can there be electric intensity at a point with zero electric potential ? Give an example.
Answer:
Yes.
Ex: When two charges +q and -q are separated by a distance, electric intensity at the mid point is non-zero. But the potential at the midpoint is zero [∵ + V – V = 0. ]

Question 3.
The potential difference between two points is 1 V.
What is the work done to move 1 C charge between them?
Answer:
Given potential difference V = 1 Volt and charge q = 1C
∴ Work done W = Vq = 1 × 1 = 1 Joule

Question 4.
Why is the electric field always at right angles to the equipotential surface ? Explain.
Answer:
The potential difference between any two points on an equipotential surface dV = 0.
Potential difference dV = –\(\vec{\mathrm{E}} . \mathrm{d} \vec{\mathrm{r}}\) ⇒ 0 = -Edrcosθ
⇒ cosθ =0 ⇒ θ = 90°
∴ Electric field \(\vec{\mathrm{E}}\) is always at right angles to the equipotential surface.

Question 5.
Three capacitors of capacitances 1 μF, 2 μF and 3 μF are connected in parallel.
a) What is the ratio of charges on each capacitor?
b) What is the ratio of potential differences across each capacitor?
Answer:
Given that C₁ = 1 μF, C₂ = 2 μF, C₃ = 3 μF
In the parallel combination of capacitors we have Q ∝ C and V is same,
a) Q₁ : Q₂ : Q₃ = C₁ : C₂ : C₃ = 1 : 2 : 3
b) V₁ = V₂ = V₃ = 1 : 1 : 1 [∵ V is same]

Question 6.
Three capacitors of capacitances 1 μF, 2μF and 3 μF are connected in series,
a) What is the ratio of charges?
b)What is the ratio of potential differences?
Answer:
Given that C₁ = 1 μF, C₂ = 2 μF, C₃ = 3 μF
In the series combination of capacitors we have Q is same and V ∝ \(\frac{1}{\mathrm{C}}\)
a) Q₁ : Q₂ : Q₃ = 1 : 1 : 1 [∵ Q is same]
b) V₁ : V₂ : V₃ = \(\frac{1}{C_1}: \frac{1}{C_2}: \frac{1}{C_3}=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=\frac{6}{1}: \frac{6}{2}: \frac{6}{3}\) = 6 : 3 : 2

Question 7.
What happens to the capacitance of a parallel plate capacitor if the area of its plates is doubled? ‘
Answer:
Capacitance C = \(\frac{\varepsilon_0 A}{d}\) ⇒ C ∝ A Here A₂ = 2A₁
∴ \(\frac{C_2}{C_1}=\frac{A_2}{A_1}=\frac{2 A_1}{A_1}\) = 2 ⇒ C₂ = 2C₁
∴ Capacitance becomes double.

Question 8.
The dielectric strength of air is 3 × 10⁶V m^-1 at certain pressure. A parallel plate capacitor with air in between the plates has a plate separated by I cm. Can you charge the capacitor to 3 × 10⁶ V. Give reason.
Answer:
No.
Since air is the dielectric, the maximum possible potential of the capacitor is V = Ed
V = 3 × 10⁶ × 10^-2 = 3 × 10⁴ V
But required potential is 3 × 10⁶ V. Required potential > Possible potential.
So it is not possible for the capacitor to charge potential 3 × 10⁶ V .

Question 9.
What is electrostatic shielding? (Jive an example.
Answer:
Electrostatic shielding is the process of isolating a specific region from external electric fields by enclosing it within a conducting material.
Ex: Metallic car body protects passengers from thunderstorm as the electric field inside is zero.

Question 10.
What happens to the capacitance of a parallel plate capacitor if the distance between plates is doubled?
Answer:
Capacitance \(\frac{\varepsilon_0 A}{d}\) ⇒ C ∝ \(\frac{1}{d}\)
Here d₂ = 2d₁
∴ \(\frac{C_2}{C_1}=\frac{d_1}{d_2}=\frac{d_1}{2 d_1}=\frac{1}{2}\) ⇒ \(\frac{C_2}{C_1}=\frac{1}{2}\)
⇒ C₂ = \(\frac{C_1}{2}\)
∴ Capacitance becomes half.

Question 11.
In a hydrogen atom, the electron and proton are at a distance of 1Å . Find the dipole moment of the system.
Answer:
Given length of dipole is 2a = 1Å = 1 × 10^-10 m, we know q = 1.6 × 10^-19 C, p = ?
Formula; Dipole moment p = q(2a) = 1.6 × 10^-19 (1 × 10^-10) = 1.6 × 10^-29 C-m

Question 12.
What will be the shapes of equipotential surfaces in case of a uniform electric field and a point charge?
Answer:
In a uniform electric field, the equipotential surfaces take the shape of equidistant parallel planes perpendicular to direction of the electric field lines.
At a point charge, equipotential surfaces are concentric spheres centered on the point charge.

V. Short Answer Questions

Question 1.
Derive an expression for the electric potential due to a point charge.
Answer:
Potential due to a Point Charge:
Consider a positive point charge q at O.
Let P be a point at a distance r from O.
Consider a unit positive test charge (+1 C) at A on the line OP at a distance x from O.

At A, electrostatic force on unit positive charge (+1 C) is F = \(\frac{q \times 1}{4 \pi \varepsilon_0 x^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{x^2}\) ……….. (1)
If dW is the small work done in moving the unit positive charge at A against the force through a small distance (AB) = dx, then dW = – F dx [∵ the two positive charges repel as θ = 1 From (1), dW = –\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{x}^2}\) dx
Total work done(W) is obtained by integrating this between the limits from ∞ to r.
W = –\(\int_{\infty}^r \frac{1}{4 \pi \varepsilon_0} \frac{q}{x^2}=\frac{-q}{4 \pi \varepsilon_0} \int_{\infty}^r \frac{1}{x^2} d x=\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{x}\right]_{\infty}^r=\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{r}\right)\) (∵ \(\int \frac{1}{x^2} d x=\frac{-1}{x}\))
The electric potential due to a point charge is this work done.
∴ V = \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{q}{r}\right)\)

Question 2.
Derive an expression for the electrostatic potential energy of a system of two point charges and find its relation with electric potential of a charge.
Answer:
Potential Energy of System of Two Charges :
Consider a positive charge q₁ located at A.

Let another charge q₂ be brought from infinity to the point B.
Let r be the distance between A and B.
Potential at B due to charge q₁ is V₁ = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r}\)
The work done to bring the charge q₂ from infinity to B is W = V₁(q₂) = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
As this is the total work done to build the system, it gives the potential energy of the system.
Potential energy U = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}\)
The same result is obtained while bringing the charge q₁ first and q₂ later.

Question 3.
Derive an expression for the potential energy of an electric dipole placed in a uniform electric field.
Answer:
Consider an electric dipole AB consisting of two
equal and opposite charges -q & q separated by a distance 2a placed in a uniform electric field E.

The torque on the dipole is \(\vec{\tau}=\vec{\mathrm{p}} \times \vec{\mathrm{E}}\) = pE sin θ
If an external torque 𝜏_ext is applied to rotate the dipole from θ₀ to θ₁, the work done is given by
W = \(\int_{\theta_0}^{\theta_1} \tau_{\text {ext }} \mathrm{d} \theta=\int_{\theta_0}^{\theta_1} \mathrm{pE} \sin \theta \mathrm{~d} \theta=\mathrm{pE}[-\cos \theta]_{\theta_0}^{\theta_1}\) = pE(cosθ₀ – cos θ₁)
This work done is stored as the potential energy in the system.
If θ₀ = π/2 and θ₁ = 0, the potential energy of the dipole at an angle θ, in uniform electric field is
U = pE (cos π/2 – cos θ) = pE(0 – cos θ) = – pE cos 0 = – \(\vec{\mathrm{p}} \cdot \vec{\mathrm{E}}\)
∴ U = – \(\vec{\mathrm{p}} \cdot \vec{\mathrm{E}}\)
This is the expression for potential energy of dipole at an angle θ in a uniform electric field.

Question 4.
Derive an expression for the capacitance of a parallel plate capacitor.
Answer:
Capacitance of a parallel plate capacitor :
Consider a parallel plate capacitor consisting of two parallel plates of area’ A, separated by a small distance ‘d’.
Let V be the potential difference between two plates.
The charges of the plates are Q and -Q.
The plate 1 has uniform surface charge density and the plate 2 has uniform surface charge density -σ.

The electric field due to plate 1 is \(\frac{\sigma}{2 \varepsilon_0}\). The electric field due to plate 2 is \(\frac{\sigma}{2 \varepsilon_0}\)
Electric field between two charged plates is E = \(\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}=2\left(\frac{\sigma}{2 \varepsilon_0}\right)=\frac{\sigma}{\varepsilon_0}\)
We know a = Q/A
∴ The field inside the capacitor is E = \(\frac{\mathrm{Q}}{\mathrm{~A} \varepsilon_0}\) ……….. (1)
But, E = \(\frac{V}{d}\) and Q = CV From (1), \(\frac{V}{d}=\frac{C V}{A \varepsilon_0}\) ⇒ C = \(\frac{\varepsilon_0 A}{d}\)
∴ Capacitance of the parallel plate capacitor is given by C = \(\frac{\varepsilon_0 A}{d}\)

Question 5.
Explain the behaviour of dielectrics in an external field.
Answer:
Dielectrics are non-conducting substances.
The molecules of a dielectric may be polar or non-polar.
Polarization: When a dielectric with non-polar molecules is placed in an external electric field, the positive and negative charges of non-polar molecules are displaced in opposite directions. Thus induced dipole moments are developed in the dielectric by the external field.

Hence the dielectric is said to be polarised by the external field. All the induced dipole moments of the non-polar molecules add up to give a net dipole moment to the dielectric.
In dielectric with polar molecules, in the absence of external electric field, the different permanent dipoles are oriented randomly due to thermal agitation. Hence, the total dipole moment becomes zero. When an external electrical field is applied, the individual dipole moments tend to align with the field. As a result, a net dipole moment develops in the dielectric in the direction of external field. Hence the dielectric gets polarised.

Thus in either case, whether polar or non-polar, a dielectric develops a net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation (P).

For linear isotropic dielectrics, P = 𝜒_e E

Here 𝜒_e is electric susceptibility and E is intensity of external field.

Question 6.
Derive an expression for the equivalent capacitance of a combination when three capacitors are connected in series.
Answer:
Series Combination: In series combination, the second plate of first capacitor is connected to first plate of second capacitor and so on. Also, first plate of first capacitor and second plate of last capacitor are connected to opposite terminals of a battery.

Capacitors in Series:
Let us suppose that three capacitors of capacitances C₁, C₂, C₃ are connected in series to a potential difference V.
Let Q be charge in each capacitor which remains constant in each capacitor.
Let V₁, V₂, V₃ be the potential differences between the plates of the capacitors C₁, C₂,C₃ and C be the equivalent capacitance Total potential.differences is V = V₁ + V₂ + V₃ ………… (1)
⇒ V = \(\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\) [∵ V₁ = \(\frac{Q}{C_1}\), V₂ = \(\frac{Q}{C_2}\), V₃ = \(\frac{Q}{C_3}\)]
\(\frac{V}{Q}=\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]\) ⇒ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\) ………… (2)
For n number of capacitors in series \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\ldots \ldots \ldots+\frac{1}{C_n}\)

Question 7.
Derive an expression for the equivalent capacitance of a combination when three capacitors are connected in parallel.
Answer:
Capacitors in Parallel:
In parallel combination, the first plates of all the capacitors are connected to a common terminal and second plates of all the capacitors are connected to common terminal and these terminals are connected to opposite terminals of the battery.
Consider three capacitors of capacitances C₁, C₂, C₃ connected in parallel to a potential difference V.
In parallel combination of capacitors, potential difference V remains the same on all the capacitors.

Let Q₁, Q₂, Q₃ be the charges stored on the plates of capacitors C₁,C₂,C₃.
In parallel combination, the total charge(Q) is equal to sum of charges stored in the capacitors.
∴ Q = Q₁ + Q₂ + Q₃ …………… (1)
Q = C₁V + C₂V + C₃V [∵ Q₁ = C₁V, Q₂ = C₂V, Q₃ = C₃V]
\(\frac{\text { Q }}{\text { V }}\) = C₁ + C₂ + C₃ ⇒ C = C₁ + C₂ + C₃ ………. (2)
This is the equivalent capacitance of a combination.
For n number of capacitors in parallel, C = C₁ + C₂ + C₃ + ……… + C_n

Question 8.
Derive an expression for the energy stored in a capacitor.
Answer:
Energy Stored in a Capacitor :
When a capacitor of capacitance C is connected to a battery of potential difference V, the capacitor gets charged to Q. The work done to move the charge is stored in the capacitor in the form of potential energy called energy stored (U).
If dW is the small work done to store a charge of dQ at the potential drop V, we have
dW = V dQ ……………. (1)
The total work done to store the charge can be obtained by integrating it in the limits from 0 to Q.
W = \(\int_0^Q\) VdQ But we know C = \(\frac{\mathrm{Q}}{\mathrm{V}}\) ⇒ V = \(\frac{\mathrm{Q}}{\mathrm{C}}\)
∴ W = \(\int_0^Q \frac{Q}{C} d Q=\frac{1}{C} \int_0^Q Q d Q=\frac{1}{C}\left(\frac{Q^2}{2}\right)=\frac{Q^2}{2 C} .\) ………. (2) (∵ ∫ x dx = \(\frac{x^2}{2}\))
This work done is stored in the form of energy U.
∴ U = \(\frac{Q^2}{2 C}=\frac{(C V)^2}{2 C}=\frac{C V^2}{2}\) (∵ \(\frac{\mathrm{Q}}{\mathrm{C}}\) = V ⇒ Q = CV)
∴ Energy Stored in a Capacitor is U = \(\frac{1}{2}\)CV²

Question 9.
Define electric field intensity and electric potential and then derive the relation between them.
Answer:
Electric field intensity (E) at a point in an electric field is defined as the force experienced by unit positive test charge (q₀) kept at that point. Formula: E = \(\frac{F}{q_0}\)
Electric potential (V) is defined as the work done per unit positive charge in bringing it from infinity to that point against the electric field.

Relation between E and V:
Consider two points separated by a distance ‘d’ in a uniform electric field of intensity E.
Let V be the potential difference between these points.
The amount of work done in moving a charge q is W = Vq ……. (1)
The force acting on the charge q is F = Eq.
Work done W = Fd = (Eq)d …… (2)
From (1) & (2), Vq = Eqd
∴ V = Ed ⇒ E = \(\frac{V}{d}\). This is the required relation between E and V.

VI. Long Answer Questions

Question 1.
Define electric potential. Derive an expression for the electric potential due to an electric dipole and hence the electric potential at a point (a) on the axial line of electric dipole (b) on the equatorial line of the electric dipole. P
Answer:
Electric potential (V) is the work done per unit positive charge in bringing it from infinity to that point against the electric field.
Formula: V = \(\frac{W}{q}\)

Potential due to an Electric Dipole:
Consider a dipole of charges -q and +q separated by a distance 2a. Its dipole moment p = (2a)q .
Let P be a point at a distance r from the centre O of the dipole. Let OP make an angle θ with the dipole.
The distances of point P from -q and q are r₁ and r₂.
From figure, as r >> a, it can be taken as r₁ ≈ r + a cos θ and r₂ ≈ r – a cos θ
The potential at P due to the charge -q is V₁ = –\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_1}\)
⇒ V₁ = –\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{r}+\mathrm{a} \cos \theta)}\) ………. (1) [∵ r₁ ≈ r + a cos θ]
The potential at P due to the charge q is V₂ = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}_2}\)
∴ V₂ = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(\mathrm{r}-\mathrm{a} \cos \theta)}\) ……….. (2) [∵ r₂ ≈ r + a cos θ]
The net potential at P due to the dipole is V = V₁ + V₂ From (1) & (2),
From (1) & (2), V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{r-a \cos \theta}-\frac{q}{r+a \cos \theta}\right]=\frac{q}{4 \pi \varepsilon_0}\left[\frac{(r+a \cos \theta)-(r-a \cos \theta)}{r^2-a^2 \cos ^2 \theta}\right]\)
V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{2 \mathrm{aq} \cos \theta}{\mathrm{r}^2-\mathrm{a}^2 \cos ^2 \theta}\right]\) [∵ a² cos²θ is very small and it can be neglected.]
∴ V = \(\frac{1}{4 \pi \varepsilon_0}\left[\frac{2 \mathrm{aq} \cos \theta}{\mathrm{r}^2}\right]\) But q(2a) = p = dipole moment
∴ V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p} \cos \theta}{\mathrm{r}^2}\)
a) Potential on the Axial line:
When the point lies on axial line of the dipole, θ = 0 or 180°.
∴ V_ax = ±\(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\mathrm{r}^2}\)

b) Potential on the Equatorial line:
When the point lies on equatorial line of the dipole, θ = 90°.
∴ V_eq = 0
Thus the potential on equatorial line of a dipole is zero.

Question 2.
Explain series and parallel combination of capacitors. Derive the formula for equivalent capacitance in each combination.
Answer:
Series Combination: In series combination, the second plate of first capacitor is connected to first plate of second capacitor and so on. Also, first plate of first capacitor and second plate of last capacitor are connected to opposite terminals of a battery.

Capacitors in Series:
Let us suppose that three capacitors of capacitances C₁, C₂, C₃ are connected in series to a potential difference V.
Let Q be charge in each capacitor which remains constant in each capacitor.
Let V₁, V₂, V₃ be the potential differences between the plates of the capacitors C₁, C₂,C₃ and C be the equivalent capacitance Total potential.differences is V = V₁ + V₂ + V₃ ………… (1)
⇒ V = \(\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}\) [∵ V₁ = \(\frac{Q}{C_1}\), V₂ = \(\frac{Q}{C_2}\), V₃ = \(\frac{Q}{C_3}\)]
\(\frac{V}{Q}=\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]\) ⇒ \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\) ………… (2)
For n number of capacitors in series \(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\ldots \ldots \ldots+\frac{1}{C_n}\)

Capacitors in Parallel:
In parallel combination, the first plates of all the capacitors are connected to a common terminal and second plates of all the capacitors are connected to common terminal and these terminals are connected to opposite terminals of the battery.
Consider three capacitors of capacitances C₁, C₂, C₃ connected in parallel to a potential difference V.
In parallel combination of capacitors, potential difference V remains the same on all the capacitors.

Let Q₁, Q₂, Q₃ be the charges stored on the plates of capacitors C₁,C₂,C₃.
In parallel combination, the total charge(Q) is equal to sum of charges stored in the capacitors.
∴ Q = Q₁ + Q₂ + Q₃ …………… (1)
Q = C₁V + C₂V + C₃V [∵ Q₁ = C₁V, Q₂ = C₂V, Q₃ = C₃V]
\(\frac{\text { Q }}{\text { V }}\) = C₁ + C₂ + C₃ ⇒ C = C₁ + C₂ + C₃ ………. (2)
This is the equivalent capacitance of a combination.
For n number of capacitors in parallel, C = C₁ + C₂ + C₃ + ……… + C_n

Question 3.
Derive an expression for the energy stored in a capacitor. What is the energy stored when the space between the plates is filled with a dielectric (a) With charging battery disconnected (b) With charging battery connected to the circuit.
Answer:
Energy Stored in a Capacitor: The work done to move the charge is stored in the capacitor in the form of potential energy called energy stored (U).
Consider a capacitor of capacitance C is connected to a battery of potential difference V, so the capacitor gets charged to Q.
If dW is the work done to store a charge of dQ at the potential drop V then dW = V dQ …………… (1)
The total work done to store the charge can be obtained by integrating it in the limits from 0 to Q.
W = \(\int_0^Q\) V dQ But V = \(\frac{\text { Q }}{\mathrm{C}}\)
∴ W = \(\int_0^Q \frac{Q}{C} d Q=\frac{1}{C} \int_0^Q Q d Q=\frac{1}{2}\left(\frac{Q^2}{2}\right)\) ((∵ ∫ x dx = \(\frac{x^2}{2}\))
This work is stored in the form of energy.
So it can be written as U = \(\frac{Q^2}{2 C}=\frac{(C V)^2}{2 C}\) = \(\frac{1}{2}\)CV² (Q = CV)
U = \(\frac{1}{2}\)CV²

Effect of Dielectric on Energy Stored in a Capacitor :
i) With Battery disconnected:
If the dielectric is filled in the capacitor after disconnecting the battery its C increases,

Q remains constant and V decreases.
Energy stored without dielectric U₀ = \(\frac{Q^2}{2 C}\) ….(1)
Energy stored with dielectric U = \(\frac{Q^2}{2 C^{\prime}}\)
But C’ = KC ∴ U = \(\frac{Q^2}{2 K C}\) ………… (2)
Dividingeq(2)by eq(1), \(\frac{U}{U_0}=\frac{1}{K}\) ⇒ U = \(\frac{\mathrm{U}_0}{\mathrm{~K}}\)
Thus if the dielectric is filled in a capacitor energy stored in it decreases to 1/K times.

ii) With Battery connected:
If the dielectric is filled in the capacitor keeping the battery connected, its C increases,

V remains constant and charge Q increases.
Energy stored without dielectric U₀ = \(\frac{1}{2}\)CV² ….(3)
Energy stored in the capacitor with dielectric U = \(\frac{1}{2}\)C’V²
But C ‘ =KC ∴ U = \(\frac{1}{2}\)(KQ)V² ………. (4)
Dividing eq (4) by eq (3), we get \(\frac{U}{U_0}\) = K ⇒ U = KU₀
Thus if dielectric is filled in a capacitor, energy stored in it increases by K times.

Textual Solved Problems

Question 1.
Calculate the potential at a point P due to a charge of 4 × 10^-7 C located 9cm away.
Solution:
Given charge Q =4 × 10^-7C, distance r = 9 cm = 0.09 m,
we know \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹ Nm²C^-2, V = ?
V = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}\) = 9 × 10⁹ × \(\frac{4 \times 10^{-7}}{0.09}\) = 4 × 10⁴ V

Question 2.
What is the work done in bringing a charge of 2 × 10^-9 C from infinity to the point of potential 4 × 10⁴ V ? Does this work depend on the path along which the charge is brought?
Solution:
Given q = 2 × 10^-9C, V= 4 × 10⁴
∴ Work W = qV = (2 × 10^-9)( 4 × 10⁴) = 8 × 10^-5 J
No. Work done is path independent.

Question 3.
Two charges 3 × 10^-8 C and -2 × 10^-8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Solution:
If the point of zero potential P lies at distance x from q₁ in between the charges q₁ and -q₂, then V₁ + V₂ = 0

⇒ \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{x}+\frac{q_2}{r-x}\right)\) = 0 ⇒ \(\left(\frac{q_1}{x}+\frac{q_2}{r-x}\right)\) = 0
Here, q₁ = 3 × 10^-8 C, q₂ = -2 × 10^-8 C, r = 15 cm = 0.15 m, x = ?
Now \(\frac{3 \times 10^{-8}}{x}-\frac{2 \times 10^{-8}}{0.15-x}\) = 0 ⇒ 3(0.15 – x) = 2x ⇒ x = 0.09 m = 9 cm
The point of zero potential may lie outside the charges also. Then V₁ – V₂ = 0
If so, \(\left(\frac{q_1}{x}+\frac{q_2}{r-x}\right)\) = 0
∴ \(\frac{3 \times 10^{-8}}{x}-\frac{2 \times 10^{-8}}{x-0.15}\) = 0 ⇒ 3(x – 015) = 2x ⇒ x = 0.45 m = 45 cm

Question 4.
A slab of material of dielectric constant K has the same area as the plates of a parallel- plate capacitor but has a thickness(3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates ?
Solution:
When there is no dielectric in the capacitor, let the potential difference be V₀, capacitance be C₀.
The electric field in it is E₀ = V₀/d.
The electric field with dielectric is E = E₀/K.
We know potential difference field × distance
∴ V = E₀ \(\left(\frac{d}{4}\right)\) + \(\frac{\mathrm{E}_0}{\mathrm{~K}}\left(\frac{3 \mathrm{~d}}{4}\right)\) = E₀d\(\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)\)
V = V₀\(\left(\frac{1}{4}+\frac{3}{4 \mathrm{~K}}\right)\) …………… (1) (∵ E₀d = V₀)
But the charge Q₀ in the capacitor remains unchanged.
∴ Q₀ = CV = C₀V₀ ⇒ V = \(\frac{\mathrm{C}_0 \mathrm{~V}_0}{\mathrm{C}}\)
Putting this in eqn (1), we get \(\frac{\mathrm{C}_0 \mathrm{~V}_0}{\mathrm{C}}\) = V₀\(\left(\frac{1}{4}+\frac{3}{4 K}\right)\) ⇒ C = C₀\(\left(\frac{4 \mathrm{~K}}{\mathrm{~K}+3}\right)\)

Question 5.
A 900 pF capacitor is charged by 100 V battery . How much electrostatic energy is stored in the capacitor?
Solution:
Given C = 900 × 10^-12 F, V = 100V, U = ?
U = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)(900 x 10^-12)(100)² = 4.5 × 10^-6 J

Question 6.
(a) A comb run through one’s dry hair attracts small bits of paper. Why ?What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.)
(b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary?
(c) Schieles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why?
(d) A bird perches on a bare high power line, and nothing happens to the bird.
A man standing on the ground touches the same line and gets a fatal shock. Why ?
Solution:
(a) When a comb runs through dry hair, it gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. Then the comb does not get charged and thus it will not attract small bits of paper.

(b) To enable the tyres to conduct charge to the ground; because much of static electricity accumulated may result in spark and result in fire.

(c) Reason similar to (b).

(d) Current passes only when there is difference in potential.

Exercise Problems

Question 1.
The work in moving a charge of AC’ from a point A, which is at a potential of -2V’ to another point B is 18.1. Find potential at point B?
Solution:
Given charge q = 3C, Work done W = 18J Initial potential V_A = -2V, Final potential V_B = ?
Work done W = q ∆V = q(V_B – V_A)
⇒ 18 = 3[V_B -(-2)] ⇒ 6 = V_B + 2
⇒ V_B = 4V

Question 2.
The electric potential at 0.9m from a point charge is +50V. What is the magnitude and sign of the charge?
Solution:
Given potential V = 50V, distance r = 0.9m
Potential V = K\(\frac{\mathrm{q}}{\mathrm{r}}=\) ⇒ q = \(\frac{V r}{K}=\frac{50 \times 0.9}{9 \times 10^9}\) = 5 × 10^-9
Since the given potential is positive, the charge must also be positive.

Question 3.
The electric field at a point due to a point charge is 20 N/C and the electric potential at that point is 10 J/C. Calculate the distance of the point from the charge and the magnitude of the charge.
Solution:
Electric field intensity E=20 N/C, Electric potential V = 10 J/C, Couloumb’s constant K = 9 × 10⁹
(i) Distance of the point from the charge is r = \(\frac{V}{E}=\frac{10}{20}\) = 0 5 m
(ii) Magnitude of the charge, q = \(\frac{\mathrm{Vr}}{\mathrm{~K}}=\frac{10 \times 0.5}{9 \times 10^9}\) = – 0-55 × 10^-9 C

Question 4.
A uniform electric field of 30 N/C’ exists along the X-axis. Calculate the potential difference V_B – V_A between the points A(4m, 2m) and B(10m, 5m).
Solution:
Electric field intensity \(\vec{\mathrm{E}}=30 \hat{\mathrm{i}}\) N / C
Displacement vector from A to B is \(\vec{r}=\overrightarrow{r B}-\overrightarrow{r A}=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}\)
\(\vec{\mathrm{r}}=(10-4) \hat{\mathrm{i}}+(5-2) \hat{\mathrm{j}}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\)
∴ Potential difference ∆V = V_B – V_A = \(\vec{\mathrm{E}} \cdot \vec{\mathrm{r}}=-(30 \hat{\mathrm{i}}) \cdot(6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})\) = -180 V
∴ ∆V = -180 V

Question 5.
Two point charges +10 μC and -10 μC are separated by a distance of 2.0 cm in air. Calculate the potential energy of the system, assuming the zero of the potential energy to be at infinity.
Solution:
Given charge q₁ = +10 μC = 10 × 10^-6 C , Charge q₂ = -10 μC = -10 × 10^-6 C
distance r = 2cm = 2 × 10^-2 m
Potential energy U = k\(\frac{q_1 q_2}{r}=\frac{9 \times 10^9\left(10 \times 10^{-6}\right)\left(-10 \times 10^{-6}\right)}{2 \times 10^{-2}}=\frac{-0.9}{0.02}\) = -45 J

Question 6.
Two charges of magnitude +q each are kept ‘2a’ distance apart. A Third charge ‘-2q’ is placed midway between them. What is the potential energy of the system?
Solution:
Given that q₁ = +q, q₂ = +q, q₃ = -2q

Potential energy of the system U = U_AB + U_BC + U_AC
= \(\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q})(-2 \mathrm{q})}{\mathrm{a}}+\frac{1}{4 \pi \varepsilon_0} \frac{(-2 \mathrm{q}) \mathrm{q}}{\mathrm{a}}+\frac{1}{4 \pi \varepsilon_0} \frac{(\mathrm{q})(\mathrm{q})}{2 \mathrm{a}}\)
= \(\frac{1}{8 \pi \varepsilon_0}\left[\frac{-8 q^2+q^2}{a}\right]\) = \(\frac{-7 q^2}{8 \pi \varepsilon_0 a}\)
∴ Potential energy of the system U = \(\frac{-7 q^2}{8 \pi \varepsilon_0 a}\)

Question 7.
A parallel plate capacitor has plates of area 200 cm2 and separation between the plates 1.0 mm. What potential difference will be developed if a charge of 1.0 NC is given to the capacitor.
Solution:
Given Area A = 200 Cm² = 200 × 10^-4 m² = 0.02 m² ; Separation d = 1 mm = 10^-3m
Charge q = 1 nC = 10^-9 C
Capacity C = \(\frac{\varepsilon_0 A}{d}=\frac{8.854 \times 10^{-12} \times 0.02}{10^{-3}}\) = 0.177 × 10^-9 F
Potential V = \(\frac{q}{C}=\frac{10^{-9}}{0.177 \times 10^{-9}}\) = 5.65 V

Question 8.
If C₁ = 3 μF and C₂ = 2 μF (fig.).

Calculate the equivalent capacitance of the given network between points A and B.
Solution:
Capacity C₁ = 3 μF, Capacity C₂ = 2μF
Initially C₁, C₂, C₁ are in series
⇒ \(\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1}=\frac{1}{3}+\frac{1}{2}+\frac{1}{3}=\frac{2+3+2}{6}\)
∴ C_S = \(\frac{6}{7}\) pF
Now C_S and C₂ are in parallel.
C_P = C_S + C₂ = \(\frac{6}{7}\) + 2 = \(\frac{6+14}{7}\) = \(\frac{20}{7}\) pF
Then C₁, C_P, C₁ are in series.
\(\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_P}+\frac{1}{C_1}=\frac{1}{3}+\frac{7}{20}+\frac{1}{3}=\frac{20+21+20}{60}=\frac{61}{60}\)
∴ Resultant capacity C = \(\frac{60}{61}\)pF

Question 9.
In the circuit shown in fig. the charge on the capacitor of 4 μF is 16 μC. Calculate the energy stored in the capacitor of 12 μF capacitance.

Solution:
Given that, charge on the capacitor of 4 μF is of 16 μC.
Voltage across 4 μF capacitor. V₂ = \(\frac{q}{C}=\frac{16}{4}\) = 4 V
∴ Voltage across 12 μF capacitor V₁ = V₁ – V₂ = 12 – 4 = 8V
∴ Energy stored in the capacitor 12 μF is U = \(\frac{1}{2}\)CV₁² = \(\frac{1}{2}\)(12)(8)² = 384 μJ

Question 10.
In parallel plate capcitor, the capacitance increases from 4 μF to 80 μF on introducing a dielectric medium between plates. What is the dielectric constant of the medium.
Solution:
Initial capacitance in air C₀ = 4 μF, Final capacitance with dielectric medium C = 80 μF
when a dielectric is introduced new capacity C = KC₀
∴ K = \(\frac{\mathrm{C}}{\mathrm{C}_0}=\frac{80}{4}\) = 20
∴ Dielectric constant K = 20

Question 11.
In a hydrogen atom the electron and proton are at a distance of 0.5 Å. The dipole moment of the system is
Solution:
Given 2a = 0.5Å = 0.5 × 10^-10 m, Charge q = 1.6 × 10^-19 C, p = ?
Formula; Dipole moment p = q(2a) = 1.6 × 10^-19 0.5 × 10^-10 = 0.8 × 10^-29 = 8 × 10^-30 C-m

Question 12.
There is a uniform electric field in the XOY plane represented by \((40 \hat{\mathrm{i}}+40 \hat{\mathrm{j}})\) Vm^-1
If the electric potential at the origin is 200V, the electric potential at the point with coordinates (2 m, 1 m) is
Solution:
Given \(\vec{\mathrm{E}}=(40 \hat{\mathrm{i}}+40 \hat{\mathrm{j}})\) Vm^-1, \(\)m,V₁ = 200V, V₂ = ? dV = V₂ – V₁
Formula: dV = –\(\vec{\mathrm{E}} \cdot \mathrm{~d} \vec{\mathrm{r}}\) = V₂ – V₁ = –\(\vec{\mathrm{E}} \cdot \mathrm{~d} \vec{\mathrm{r}}\)
⇒ V₂ – 200 = – \((40 \hat{\mathrm{i}}+40 \hat{\mathrm{j}}) \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}})\)
⇒ V₂ – 200 = – (80 + 40) = -120
⇒ V₂ = – 120 + 200 = 80 V

Question 13.
An electric dipole of moment P is placed in a uniform electric field E, with P parallel to E. It is then rotated by an angle θ. What is the work done?
Solution:
Torque on a dipole 𝜏 = pE sin θ
The work done to rotate the dipole from 0 to θ is given by W = \(\int_0^\theta\) pE sin θdθ
W = pE (cos 0 – cos θ) = pE (1 – cos θ)

Objective Questions

Question 1.
In a certain region of space with volume 0.2 m³ the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is
1) zero
2) 0.5 N/C
3) 1 N/C
4) 5 N/C
Answer:
1) zero

Question 2.
A bullet of mass 2 g is having a charge of 2μC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of 10 m/s ?
1) 5 kV
2) 50 kV
3) 5 V
4) 50 V
Answer:
2) 50 kV

Question 3.
Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.
1) 1980 V
2) 660 V
3) 1320 V
4) 1520 V
Answer:
1) 1980 V

Question 4.
Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁ , σ₂) is
1) \(\frac{\mathrm{R}_1^2}{\mathrm{R}_2^2}\)
2) \(\frac{R_1}{R_2}\)
3) \(\frac{R_2}{R_1}\)
4) \(\sqrt{\left(\frac{R_1}{R_2}\right)}\)
Answer:
3) \(\frac{R_2}{R_1}\)

Question 5.
A hollow metallic sphere of radius 10 cm is charged such that potential of its surface is 80 V. The potential at the centre of the sphere would be
1) 80 V
2) 800V
3) zero
4) 8 V
Answer:
1) 80 V

Question 6.
The diagrams below show regions of equipotentials.

A positive charge is moved from A to B in each diagram.
1) In all the four cases the work done is the same.
2) Minimum work is required to move q in figure (I).
3) Maximum work is required to move q in figure (II).
4) Maximum work is required to move q iN figure (III).
Answer:
1) In all the four cases the work done is the same.

Question 7.
Charge q₂ is at the centre of a circular path with radius r. Work done in carrying charge q₁, once around this equipotential path, would be
1) \(\frac{1}{4 \pi \varepsilon_0} \times \frac{q_1 q_2}{r^2}\)
2) \(\frac{1}{4 \pi \varepsilon_0} \times \frac{q_1 q_2}{r}\)
3) zero
4) infinite
Answer:
3) zero

Question 8.
In bringing an electron towards another electron, the electrostatic potential energy of the system
1) becomes zero
2) increases
3) decreases
4) remains same
Answer:
2) increases

Question 9.
A dipole is placed in an electric field as shown. In which direction will it move?

1) towards the right as its potential energy will increase.
2) towards the left as its potential energy will increase.
3) towards the right as its potential energy will decrease.
4) towards the left as its potential energy will decrease.
Answer:
3) towards the right as its potential energy will decrease.

Question 10.
An electric dipole of moment \(\vec{\mathrm{p}}\) is lying along a uniform electric field \(\vec{\mathrm{E}}\). The work done in rotating the dipole by 90° is
1) pE
2) \(\sqrt{2}\)pE
3) pE/2
4) 2pE
Answer:
1) pE

Question 11.
There is an electric field F. in x-direction. If the work done on moving a charge of 0.2 C through a distance of 2 in along a line making an angle 60° with x-axis is 4 .1, then what is the value of K?
1) 5 N/C
2) 20 N/C
3) \(\sqrt{3}\) N/C
4) 4 N/C
Answer:
2) 20 N/C

Question 12.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2C and 5 × 10^-2 respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
1) 2 × 10^-2C
2) 3 × 10^-2C
3) 4 × 10^-2C
4) 1 × 10^-2C
Answer:
2) 3 × 10^-2C

Question 13.
A parallel plate air capacitor has capacity C. distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is
1) \(\frac{\mathrm{CV}^2}{\mathrm{~d}}\)
2) \(\frac{C^2 V^2}{2 d^2}\)
3) \(\frac{C^2 V^2}{2 d}\)
4) \(\frac{\mathrm{CV}^2}{2\mathrm{~d}}\)
Answer:
4) \(\frac{\mathrm{CV}^2}{2\mathrm{~d}}\)

Question 14.
The capacitance of a parallel plate capacitor with air as medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permitivity of the medium is
(ε₀ = 8.85 × 10^-12C²N^-1 m²)
1) 0.44 × 10^-13C²N^-1 m²
2) 1.77 × 10^-12C²N^-1 m²
3) 0.44 × 10^-12C²N^-1 m²
4) 5.00 C²N^-1 m²
Answer:
3) 0.44 × 10^-12C²N^-1 m²

Question 15.
Two parallel metal plates having charges +Q and -Q face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will
1) become zero
2) increase
3) decrease
4) remain same
Answer:
3) decrease

Question 16.
The equivalent capacitance of the combination shown in the figure is

1) 3C/2
2) 3C
3) 2C
4) C/2
Answer:
3) 2C

Question 17.
Two metallic spheres of radii 1 cm and 2 cm are given charges 10^-2;C and 5 × 10^-2 C respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is
1) 3 × 10^-2C
2) 4 × 10^-2C
3) 1 × 10^-2 C
4) 2 × 10^-2C
Answer:
4) 2 × 10^-2C

Question 18.
Three capacitors each of capacitance C and I of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be
1) 3C, V/3
2) C/3, 3V
3) 3C, 3V
4) C/3,V/3
Answer:
2) C/3, 3V

Question 19.
A network of four capacitors of capacity equal to C₁ = C, C₂ = 2C, C₃ = 3C and lo a battery as shown in the figure. The ratio of the charges on C2 and C4 is

1) 4/7
2) 3/22
3) 7/4
4) 22/3
Answer:
2) 3/22

Question 20.
Three capacitors each of capacity 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by
1) connecting all of them in series
2) connecting them in parallel
3) connecting two in series and one in parallel
4) connecting two in parallel and one in series.
Answer:
3) connecting two in series and one in parallel

Question 21.
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system
1) decreases by a factor of 2
2) remains the same
3) increases by a factor of 2
4) increases by a factor of 4.
Answer:
1) decreases by a factor of 2

Question 22.
A parallel plate capacitor has a uniform electric field F. in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is
1) \(\frac{1}{2}\) ε₀E²
2) \(\frac{\mathrm{E}^2 \mathrm{Ad}}{\varepsilon_0}\)
3) \(\frac{1}{2}\)ε₀E²Ad
4) ε₀EAd
Answer:
3) \(\frac{1}{2}\)ε₀E²Ad

Question 23.
Energy per unit volume for a capacitor ! having area A and separation d kept at potential difference V is given by
1) \(\frac{1}{2} \varepsilon_0 \frac{\mathrm{~V}^2}{\mathrm{~d}^2}\)
2) \(\frac{1}{2 \varepsilon_0} \frac{\mathrm{~V}^2}{\mathrm{~d}^2}\)
3) \(\frac{1}{2} \mathrm{CV}^2\)
4) \(\frac{Q^2}{2 C}\)
Answer:
1) \(\frac{1}{2} \varepsilon_0 \frac{\mathrm{~V}^2}{\mathrm{~d}^2}\)

Question 24.
A capacitor is charged with a battery and energy stored is U. After disconnecting battery another capacitor of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is
1) U/2
2) U/4
3) 4U
4) 2U
Answer:
2) U/4

Question 25.
The energy stored in a capacitor of capacity C and potential V is given by
1) \(\frac{\text { CV }}{2}\)
2) \(\frac{\mathrm{C}^2 \mathrm{~V}^2}{2}\)
3) \(\frac{\mathrm{C}^2 \mathrm{~V}}{2}\)
4) \(\frac{\mathrm{CV}^2}{2}\)
Answer:
4) \(\frac{\mathrm{CV}^2}{2}\)

Question 26.
An infinite number of charges of equal magnitude q, but of opposite sign are placed along the x-axis at x = 1, x = 2, x = 4, x = 8… and so on. The electric potential at the point x = 0 due to these charges will be (in multiples of \(\frac{1}{4 \pi \varepsilon_0}\))
1) q/2
2) q/3
3) 2q/3
4) 3q/2
Answer:
3) 2q/3

Question 27.
A 20 F capacitor is charged to 5V and isolated. It is then connected in parallel with an uncharged 30F capacitor. The decrease in the energy of the system will be
1) 25 J
2) 100 J
3) 125 J
4) 150 J
Answer:
4) 150 J

Question 28.
A 100 μF capacitor is charged to 200 volt. It is discharged through a 2 ohm resistance. The amount of heat generated will be
1) 2 J
2) 4 J
3) 0.2 J
4) 0.4 J
Answer:
1) 2 J

Question 29.
Three condensers whose capacities are 3 μF, 9 μF and 18 μF respectively are first connected in series and then in parallel. The ratio of resultant capacities in two cases will be
1) 1 : 15
2) 15 : 1
3) 1 : 1
4) 1 : 3
Answer:
1) 1 : 15

Question 30.
A condenser of capacity 500pF is charged at the rate of 50 μC/s. The time taken for charging the condenser to 10V will be
1) 100 s
2) 50 s
3) 25 s
4) 10 s
Answer:
1) 100 s

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 3 Chemical Kinetics Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 3rd Lesson Chemical Kinetics Questions and Answers

I. Multiple Choice Questions

Question 1.
The unit of rate of a chemical reaction is generally expressed as: [ ]
1. mol L^-1 s^-1
2. mol L^-1
3. mol
4. s^-1
Answer:
1. mol L^-1 s^-1
Rate change in concentration / time
Unit = mol L^-1 / s = mol L^-1 / s^-1 (or) mol L^-1 / s^-1

Question 2.
The instantaneous rate of reaction can be determined by using [ ]
1. average time
2. total time
3. slope of the concentration-time curve
4. product concentration and time
Answer:
3. slope of the concentration-time curve
Instantaneous rate = rate at a particular moment
It is given by the slope of tangent to concentration-time graph
Slope of the concentration-time curve

Question 3.
Which of the following does not affect the rate of reaction? [ ]
1. Temperature
2. Catalyst
3. Pressure
4. Color of reactants and products
Answer:
4. Color of reactants and products
Rate depends on physical conditions like temperature, pressure, catalyst
Color has no effect on rate of reaction. Color of reactants and products

Question 4.
The rate law for a given reaction is determined from [ ]
1. balanced chemical equation
2. thermodynamic data
3. experimental data
4. catalyst used
Answer:
3. experimental data
Rate law cannot be predicted from equation. It must be determined experimentally Experimental data

Question 5.
Molecularitv of a reaction can be [ ]
1. zero
2. always integer
3. negative
4. fractional
Answer:
2. always integer
Molecularity = No. of atoms, ions or molecules in rate determine step.
It is always a whole number (never fraction/zero). Always integer

Question 6.
What is the overall order of a reaction which has the rate expression rate = k|A|²|B|? [ ]
1. One
2. Two
3. Three
4. Zero
Answer:
3. Three
Overall order = sum of powers in rate law. k[A]²[B]¹ = 2 + 1 = 3

Question 7.
Radioactive decay is an example of which order kinetics? [ ]
1. Zero
2. First
3. Second
4. Third
Answer:
2. First
Radioactive decay rate depends on only one nucleus at a time. So it follows first-order kinetics

Question 8.
What is the slope of the |A| vs. t graph for a first order reaction? [ ]
1. -k
2. k
3. -1/k
4. ln k
Answer:
1. -k
For first order:ln[A] = ln[A₀]-kt
Slope of graph = -k
ln[A] = ln[A₀] – kt

Question 9.
The unit of rate constant of a first order reaction is [ ]
1. s^-1
2. Lmol^-1s^-1
3. mol L^-1s^-1
4. mol s^-1
Answer:
1. s^-1
For first order: rate = k[A]. So k must have unit s^-1

Question 10.
The half-life of a first-order reaction is given by [ ]
1. t_1/2 = 1/k
2. t_1/2 = [A]₀ 2k
3. t_1/2 = 0.693/k
4. t_1/2 = k/0.693
Answer:
3. t_1/2 = 0.693/k
Standard half-life formula for first order reaction: t_1/2 = \(\frac{0.693}{k}\)

Question 11.
The hydrolysis of methyl acetate in aqueous acid is considered as pseudo first-order reaction because [ ]
1. acid acts as catalyst
2. water is in large excess
3. methyl acetate is in excess
4. reaction occurs slowly
Answer:
2. water is in large excess
In pseudo first-order reactions, one reactant is in large excess.
Its concentration remains almost constant. Here water is in excess

Question 12.
With the rise in temperature by 10 °C, the rate constant of a reaction
1. decreases by 2 or 3 times
2. remains same
3. increases by 2 or 3 times
4. increases by 10 times
Answer:
3. increases by 2 or 3 times
Rule of thumb in kinetics: For every 10°C rise, rate constant roughly doubles or triples. This is due to more molecules crossing activation energy. Increases by 2 or 3 times

Question 13.
Which of the following represents Arrhenius equation?
1. k = A/T
2. k = A + E_a/RT
3. k = Ae^-Ea/RT
4. k = ln A – E_a
Answer:
1. k = A/T
Arrhenius equation relates rate constant with temperature and activation energy:
k = Ae^-Ea/RT Only option matching exponential form is correct.

Question 14.
Catalyst increases the rate of a reaction by [ ]
1. decreasing activation energy
2. increasing temperature
3. increasing activation energy
4. increasing concentration
Answer:
1. decreasing activation energy
Catalyst provides an alternative pathway with lower energy barrier. It does NOT change temperature or concentration. Decreasing activation energy

Question 15.
Identify the correct statements among the following. [ ]
a. A catalyst does not alter Gibbs energy
b. A catalyst does not change equilibrium constant
c. A catalyst can catalyze a nonspontaneous reaction
1. a only
2. a & b only
3. b & c only
4. a & c only
Answer:
2. a & b only
(a) Catalyst does NOT change ΔG → True
(b) Catalyst does NOT change equilibrium constant → True
(c) Cannot make non-spontaneous reaction spontaneous → x False
Catalyst only speeds up equilibrium attainment. a & b only

II. Fill in the Blanks

Question 1.
The sum of powers of the concentration terms in the rate law expression is called ____ of that chemical reaction.
Answer:
order

Question 2.
The overall rate of the reaction is controlled by the slowest step of the reaction and is called the ____
Answer:
rate determining step

Question 3.
The number of collisions per second per unit volume of the reaction mixture is known as ____
Answer:
collision frequency

Question 4.
k = A e^-Ea/RT where A is called ____
Answer:
Arrhenius factor or frequency factor

Question 5.
For a first order reaction the following graph is obtained. From this graph slope is equal to ____
Answer:
k/2.303

III. One Word Answer Questions

Question 1.
What is the molecularity of the elementary reaction: A + B → Products?
Answer:
Molecularity of the given elementary reaction is 2 (two)

Question 2.
Which reactions takes place in one step?
Answer:
Elementary reactions takes place in one step.

Question 3.
In a zero order reaction, if initial concentration of reactant is 0.4 mol/L and rate constant is 0.1 mol L^-1s^-1, what is its t_1/2 ?
Answer:
Given that [A₀] = 0.4 m/L, K = 0.1 m.L^-1 s^-1.
For Zero order reaction t_1/2 = \(\frac{\left[\mathrm{A}_0\right]}{2 \mathrm{~K}}\) = \(\frac{0.4}{2 \times 0.1}\) = 2 s

Question 4.
What is the unit of rate constant for a zero-order reaction?
Answer:
Unit of rate constant for a zero-order reaction is mol L^-1 s^-1

Question 5.
Collision theory is based on which theory of gases?
Answer:
Collision theory is based on Kinetic molecular theory of gases.

IV. Very Short Answer Questions

Question 1.
Write the equation for the rate of the reaction 5Br^–(aq) + BrO₃^– (aq) + 6H⁺ (aq) → 3 Br₂(aq) + 3 H₂O (l)
Answer:

Question 2.
What is rate law? Illustrate with an example.
Answer:
Rate Law: It is an experimentally determined expression which relates the rate of reaction with concentration of reactants. For a hypothetical reaction, aA + bB → Products
Rate ∝ [A]^m [B]ⁿ, where ‘m’ and ‘n’ may or may not be same as the stoichiometric coefficients of the reaction in a balanced chemical equation.

Question 3.
What are elementary reactions?
Answer:
The reactions taking place in one step are called elementary reactions.
Ex: 2HI → H₂ + I₂

Question 4.
What are complex reactions? Name one complex reaction.
Answer:
Complex reactions: These are the reactions which involve sequence of elementary reactions for the conversion of reactants into products.
Ex 1: Oxidation of ethane to, CO₂ and H₂O passes through a series of intermediate steps. Here alcohol, aldehyde and acid are formed.
Ex 2: 2H₂O₂ → 2H₂O + O₂

Question 5.
Define molecularity of a reaction with an example.
Answer:
Molecularity of a reaction: It is the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.
Examples:

• NH₄NO₂ → N₂ + 2H₂O (Unimolecular reaction)
• 2HI_(g) → H_2(g) + I_2(g) (Bimolecular reaction)
• 2NO_(g) + O_2(g) → 2NO_2(g) (Trimolecular reaction)

Question 6.
Give two examples for zero Order reactions.
Answer:

Question 7.
Give two examples for gaseous first order reactions.
Answer:

1. N₂O_5(g) → 2NO_2(g) + \(\frac{1}{2}\) O_2(g), order = 1 [ Decomposition of N₂O₅]
2. C₂H_4(g) + H_2(g) → C₂H_6(g), order =1 [Hydrogenation of ethene]

Question 8.
What is half-life of a reaction? Illustrate your answer with an example.
Answer:
Half-life of a reaction (t_1/2):
It is the time required to reduce half of the initial concentration of reactant.

1) For a zero order reaction, rate constant is given by k = \(\frac{\left[\mathrm{R}_0\right]-[\mathrm{R}]}{\mathrm{t}}\)

2) For a first order reaction, t_1/2 = \(\frac{0.693}{k}\)

Question 9.
What are pseudo first order reactions? Give one example.
Answer:
Pseudo first order reactions: It is a reaction in which one of the reactants, is present in large amount and its concentration does not get altered during the course of the reaction and it behaves as first order reaction.
Ex: Hydrolysis of Ethyl acetate:

Rate of reaction =k[CH₃COOC₂H₅]
Here the H₂O is taken in excess. So, the rate doesn’t depend on [H₂O]

Question 10.
Explain the term ‘activation energy’ of a reaction with a suitable diagram.
Answer:
Activation energy: It is the minimum energy required to initiate a chemical reaction to form an intermediate activated complex (C).

E_a = Activation energy of forward reaction [E_r-E_R]
E’_a = Activation energy of backward reaction [E_r-E_p]
E_p = Energy of products,
E_R = Energy of reactants
E_T = Threshold energy
Also E_a = E_T – E_R
Ex 1: Lighting a match stick using match box.
Ex 2: The spark produces the energy to start the combustion of fuel and air.

V. Short Answer Questions

Question 1.
What is ‘molecularity’ of a reaction? How is it different from the ‘order’ of a reaction? Name one bimolecular and one trimolecular gaseous reactions.
Answer:
1) Molecularity of a reaction: It is the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction.

2) Differences between order and molecularity of a reaction:

1. Order of reaction is an experimental quantity. It can be zero and even a fraction. but molecularity can not be zero or a non integer.
2. Order is applicable to elementary as well as complex reactions where as molecularity is applicable only for elementary reactions.
3. For a complex reaction, order given by the slowest step and molecularity of the slowest step is taken as the order of the overall reaction.

3) Bimolecular reaction:
Ex: 2HI → H₂ + I₂ [Dissociation of HI into H₂ and I₂.]

4) Trimolecular reaction:
Ex: 2NO + O₂ → 2NO₂

Question 2.
Derive the integrated rate equation for a zero order reaction.
Answer:
Consider a zero order reaction: R→ product
∴ Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R]⁰ = k ⇒ d[R] = – kdt
Integrating both sides, we get [R] = −kt + I….(1), I is the constant of integration At t = 0, we have [R] = [R]_o, where [R]₀ is initial concentration of the reactant.
Equation (1) can be written as [R]_o = -k × 0 + I ⇒ [R]₀ = I
Substituting the value of I in the equation (1), we have [R] = -kt + [R]_o
[This is in the form y = mx + C of a line, with slope m = -k]
⇒ kt = [R]_o – [R] ⇒ k = \(\frac{[\mathrm{R}]_0-[\mathrm{R}]}{\mathrm{t}}\). This is the integrated equation for zero order reaction.
Here, k= rate constant ;
[R] = Initial concentration of the reactant
t = time,
[R] = Final concentration of the time t.

Question 3.
Derive an integrated rate equation for a first order reaction.
Answer:
Consider a first order reaction: R→ Product
Rate = \(\frac{-\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R] ⇒ \(\frac{d[R]}{[R]}\) = -kdt
Integrating this equation, we get ln[R] = -kt + I……(1), where I is the constant of integration When t=0 we have [R] = [R]_o, where [R]_o is the initial concentration of the reactant.
∴ Equation (1) can be written as ln[R]_o = k × 0 + I ⇒ ln[R]_o = I
Substituting the value of I in equation (1), we have
ln[R] = -kt + ln[R]_o ⇒ kt = ln[R]_o – ln[R] = ln\(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}\) ⇒ k = \(\frac{1}{t} \ln \frac{[R]_0}{[R]}\) ⇒ k = \(\frac{2.303}{t}\)log\(\frac{[\mathrm{R}]_0}{[\mathrm{R}]}\)
This is the integrated rate equation for first order reaction.
Here, k = rate constant ; [R]_o = Initial concentration of the reactant
t = time, [R] = Final concentration of the time t.

Question 4.
What is Arrhenius equation? Derive an equation which describes the effect of rise of temperature (T) on the rate constant (k) of a reaction.
Answer:
Arrhenius equation gives the temperature dependence of the rate of a chemical reaction. Arrhenius equation: Rate constant k = A e^-Ea/RT ….. (1)
Where A is frequency factor, E_a activation energy.
Taking natural logarithm on both sides of equation (1), we get ln k = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) + ln A …… (2)
At temperature T₁, equation (2) is ln k₁ = \(\frac{-E_a}{R T_1}\) + ln A …… (3)
At temperature T₂, equation (2) is ln k₂ = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}\) + ln A …..(4)
Here K₁ and K₂ are the values of rate constants at temperature T₁ and T₂ respectively.
Subtracting equation (3) from (4), we get lnk₂ – Ink₁ = \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT} 1}\) – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}\)
ln \(\frac{\mathbf{k}_2}{\mathbf{k}_1}\) = \(\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\) ⇒ 2.303l0g \(\frac{\mathbf{k}_2}{\mathbf{k}_1}\) = \(\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\) ⇒ 2.303log[] \(\frac{k_2}{k_1}\) = \(\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\)

Question 5.
Discuss the effect of catalyst on the kinetics of a chemical reaction with a suitable diagram.
Answer:
A catalyst is a substance which increases the rate of a reaction without itself undergoing any permanent chemical change.
Ex: MnO₂ acts as catalyst in the following reaction to increase its rate.


The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants. It results in the formation of an intermediate complex. This has a transitory existence and decomposes to yield products and the catalyst.
Catalyst takes the reaction in a new path having low activation energy. When the activation energy decreases, the rate of reaction increases.

Question 6.
Describe the salient features of the collision theory of reaction rates of bimolecular reactions.
Answer:
Molecular collision theory was proposed by Arrhenius.
It explains the rate of gaseous bimolecular reactions.

Main postulates :

1. A reaction takes place only when reactant molecules collide with proper orientation.
2. All collisions do not lead to the formation of products.
3. Threshold energy (E_T): It is the minimum energy that the reactant molecules should possess to give products.
4. Under STP conditions, very few molecules possess E_T. But most of the molecules have less energy than threshold energy.
5. Activation energy (E_a): It is the minimum excess energy that reactant molecules should acquire to participate in chemical reaction or to cross threshold energy barrier. E_a = E_T – E_R
6. The molecules possessing the threshold energy are called activated molecules and collisions between these activatedmolecules alone lead to the formation of products. Hence these collisions are called activated collisions or fruitful collisions or effective collisions.
7. Number of binary collisions per unit time is given by Z = \(\Pi \sigma_{A B}^2 \sqrt{\frac{8 k T}{\pi \mu}} n_A \cdot n_B\)
   where \(\sigma_{\mathrm{AB}}\) = collision diameter, μ= reduced mass.
8. Specific rate, k = PZ.e^-Ea/RT (or) k = Ae^-Ea/RT

Graphical representation of activation energy :

E_a= Activation energy of forward reaction [E_T – E_R]
E’_a = Activation energy of backward reaction [E_T – E_P]
E_P = Energy of products
E_R = Energy of reactants
E_T = Threshold energy

Question 7.
Define average rate of a reaction. How is the rate of reaction expressed in terms of change in the concentration of reactants and products for the following reactions.
1) 2Hl_(g) → H_2(g) + I_2(g)
2) Hg_(l) + Cl_2(g) → HgCl_2(s)
3) \(\mathrm{Br}_{(\mathrm{aq})}^{-}\) + \(\mathrm{BrO}_{3(\mathrm{aq})}^{-}\) + \(6 \mathrm{H}_{(\mathrm{aq})}^{+}\) → 3Br_2(aq) + 3H₂O_(l)
Answer:
Average rate of a reaction is defined as the rate of change of concentration per unit time.

Question 8.
What is rate equation? How is it obtained? Write the rate equations for
1) 2NO_(g) + O_2(g) → 2NO_2(g)
2) CHCl₃ + Cl₂ → CCl₄ + HCl
3) CH₃COOC₂H₅ + H₂O → CH₃COOH(aq) + C₂H₅OH_(aq)
Answer:
Rate equation (rate law) is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species, in a balanced chemical equation.
Rate equation is always obtained from the experimental data.
1) 2NO_(g) + O_2(g) → 2NO_2(g); Rate = k[NO]²[O₂]
2) CHCl₃ + Cl₂ → CCl₄ + HCl; Rate = [CHCl₃][Cl₂]^1⁄2.
3) CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH; Rate = k[CH₃COOC₂H₅]

Question 9.
What is half-life (t_1/2) of a reaction? Derive the equations for the ‘half-life’ value of zero and first order reactions.
Answer:
Half-life of a reaction (t_1/2) is the time in which the concentration of a reactant is reduced to one half of its initial concentration.

Question 10.
Explain the terms
a) Activation energy (Ea)
b)Collision frequency (Z)
c) Probability factor (P) with respect to Arrhenius equation.
Answer:
a) Activation Energy: It is the minimum excess energy that reactant molecules should acquire to participate in chemical reaction or to cross threshold energy barrier. Ea_a= E_T – E_R.
In other words, activation energy is the difference between threshold energy and the average kinetic energy of the reactant molecules.

b) Collision Frequency: The number of collisions that take place per second per unit volume of the reaction mixture is called collision frequency.

c) Probability Factor: For a collision to be effective, the colliding molecules must have energy greater than threshold energy and also must have proper orientations.
For the collision to be effective, another factor, P is called probability factor has to be introduced. Thus Arrhenius equation is modified to the form k = PZ_AB e^-Ea/RT
where P = Probability factor

VI. Long Answer Questions

Question 1.
Explain the following terms with suitable examples.
a) Average rate of a reaction
b) Slow and fast reactions
c) Order of a reaction
d) Molecularity of a reaction
e) Activation energy of a reaction
Answer:
a) Average rate of a reaction : The average rate of a reaction is defined as the rate of change of concentration per unit time.
Average time \(=\frac{\text { Change in concentration in give time }}{\text { Time taken }}\)
= \(\frac{\Delta x}{\Delta t}\) or –\(\frac{\Delta[\mathrm{R}]}{\Delta \mathrm{t}}\) = +\(\frac{\Delta[\mathrm{P}]}{\Delta \mathrm{t}}\)

b) Slow reactions: These reactions involve high activation energy making them proceed slowly.
Ex:Rusting of Iron
Fast reactions:These reactions involve low activation energy making them proceed quickly. Ex Reactions between ionic compounds NaCl + AgNO₃ → NaNO₃ + AgCl ↓

c) Order of a reaction “The sum of powers of the concentration terms of reactants in the rate law expression” is called order of a reaction.
Consider a hypothetical reaction:
aA + bB → products.
The rate law expression for this reaction is, Rate = k[A]^m[B]ⁿ.
The order of the above reaction is (m+n).

d) Molecularity of a reaction: The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction.
The reaction can be unimolecular when one reacting species is involved.
Ex:decomposition of ammonium nitrite into N₂ and H₂O.
NH₄NO₂ → N₂ + 2H₂O

e) Activation Energy (E): It is the minimum excess energy that reactant molecules should acquire to participate in chemical reaction or to cross threshold energy barrier. E_a = E_T – E_R
In other words, activation energy is the difference between threshold energy and the average kinetic energy of the reactant molecules.

Question 2.
Give two examples for each of zero order and first order reactions. Write the equations for the rate of a reaction in terms of concentration changes of reactants and products for the following reactions:
i) A_(g) + B_(g) → C_(g) + D_(g)
ii) A_(g) → B_(g) + C_(g)
iii) A_(g) + B_(g) → C_(g)
Answer:

Question 3.
Discuss the effect of temperature on the rate of a reaction. Derive necessary equations in this context.
Answer:
Effect of temperature on rate of reaction: In general for every 10°C rise of temperature the rate of a reaction is nearly doubled and in some cases it is tripled.
The ratio of specific rates that differ by 10°C is known as temperature coefficient.
\(\frac{K_{t+10^{\circ} \mathrm{C}}}{K_{t^{\circ} \mathrm{C}}}\) = 2(or)3
Effect of temperature on rate of reaction is explained by Arrhenius in the form of an equation known as Arrhenius Equation.

1) Arrhenius equation: The arrhenius equation gives accurately the temperature dependence of the rate of a chemical reaction.
2) Rate constant k = A e^-Ea/RT ….. (i)
Where A is frequency factor, E_a is activation energy
3) Taking natural logarithm of both sides of equation (i), we get ln k = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\) + ln A ……. (ii)
4) The plot of ln k vs 1/T is a straight line.
The slope of the line = \(-\frac{E_a}{R}\)
y-intercept = lnA.
Using these values, E, and A can be calculated.

5) At temperature T₁, equation (ii) is lnk₁ = \(\frac{-E_a}{R T_1}\) + ln A …..(iii)
6) At temperature T₂, equation (ii) is ln k₂ = \(\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}_2}\) + ln A …..(iv)
K₁ and K₂ are the values of rate constants at temperature T₁ and T₂ respectively.
7) Subtracting equation (iii) from (iv) we get, ln k₂ – ln k₁ = \(\frac{E_a}{R T_1}\) – \(\frac{E_a}{R T_2}\) [∵ T₂ > T₁]
8) ln \(\frac{k_2}{k_1}\) = \(\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\) ⇒ 2.303log\(\frac{k_2}{k_1}\) ⇒ \(\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\) ⇒ log\(\frac{k_2}{k_1}\) = \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]\)

Question 4.
Give a detailed account of the collision theory of reaction rates of bimolecular gaseous reactions.
Answer:
Molecular collision theory was proposed by Arrhenius.
It explains the rate of gaseous bimolecular reactions.

Main postulates :

1. A reaction takes place only when reactant molecules collide with proper orientation.
2. All collisions do not lead to the formation of products.
3. Threshold energy (E_T): It is the minimum energy that the reactant molecules should possess to give products
4. Under STP conditions, very few molecules possess E. But most of the molecules have less energy than threshold energy.
5. Activation Energy (E_a): It is the minimum excess energy that reactant molecules should acquire to participate in chemical reaction or to cross threshold energy barrier. E_a = E_T – E_R
   
6. The molecules possessing the threshold energy are called activated molecules and collisions between these activated molecules alone lead to the formation of products. Hence these collisions are called activated collisions or fruitftil collisions or effective collisions.
7. Number of binaiy collisions per unit time is given by Z = \(\Pi \sigma_{A B}^2 \sqrt{\frac{8 k T}{\pi \mu}} n_A \cdot n_B\)
   where σ_AB = collision diameter, µ = reduced mass.
8. Specific rate, k = PZe^-Ea/RT (or) k = AeT

Formation of methanol from bromoethane depends upon the orientation of reactant molecules as shown below.

Diagram showing molecules having proper and improper orientation
The Proper orientation of reactant molecules lead to bond formation whereas improper orientation makes them simply bounce back and no products are formed.
To account for effective collisions, another factor P, called the probability factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented
Rate = PZ e^-Ea/RT.
Thus in collision theory, activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction.

Question 5.
a) Write the differences between order & molecularity.
b) Describe the salient features of collision theory.
Answer:
Differences between order & molecularity.

Salient features of collision theory:

Molecular collision theory was proposed by Arrhenius. It explains the rate of gaseous bimolecular reactions.

Main postulates :

1. A reaction takes place only when reactant molecules collide with proper orientation.
2. All collisions do not lead to the formation of products.
3. Threshold energy (E_T): It is the minimum energy that the reactant molecules should possess to give products
4. Under STP conditions, very few molecules possess E_T. But most of the molecules have less energy than threshold energy.
5. ctivation Energy (E_a): It is the minimum excess energy that reactant molecules should acquire to participate in chemical reaction or to cross threshold energy barrier. E_a = E_T – E_R
6. The molecules possessing the threshold energy are called activated molecules and collisions between these activatedmolecules alone lead to the formation of products. Hence these collisions are called activated collisions or fruitful collisions or effective collisions.
7. Number of binary collisions per unit time is given by Z = \(\Pi \sigma_{A B}^2 \sqrt{\frac{8 k T}{\pi \mu}} n_A \cdot n_B\)
   where σ_AB collision diameter, μ = reduced mass.
8. Specific rate, k = PZ.e^-Ea/RT (or) k = Ae^-Ea/RT

Graphical representation of activation energy :

E_a = Activation energy of forward reaction [E_r – E_R]
E’_a= = Activation energy of backward reaction [E_T – E_P]
E_p = Energy of products E_R = Energy of reactants
E_T = Threshold energy. Also E_a = E_T – E_R

Textual Solved Problems

Question 1.
The decomposition of N₂O₅ in CCl₄ at 318K has been studied by monitoring the concentration of N₂O₅ in the solution. Initially the concentration of is 2.33 mol L^-1 and after 184 minutes, it is reduced to 2.08mol L^-1. The reaction takes place according to the equation.
2N₂O₅ → 4NO_2(g) + O_2(g)
Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO₂ during this period?
Answer:
Average Rate = \(\frac{1}{2}\left\{-\frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}\right\}\) = \(-\frac{1}{2}\left[\frac{(2.08-2.33) \mathrm{mol} \mathrm{~L}^{-1}}{184 \mathrm{~min}}\right]\)
= 6.79 × 10^-4 mol L^-4/min = (6.79 × 10^-4 mol L^-1/min^-1)(60 min/1h)
= 4.07 × 10^-2 mol L^-1/h = 6.79 × 10^-4 mol L^-1/1min/60s = 1.13 × 10^-5 mol L^-1/s^-1
It may be remembered that Rate = \(\frac{1}{4}\left\{\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\right\}\)
⇒ \(\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}\) = 4 × Rate = 6.79 × 10^-44 mol L^-1/m = 2.72 × 10^-3mol L^-1min^-1

Question 2.
Calculate the overall order of a reaction which has the rate expression
(a) Rate = k[A]^1/2[B]^3/2
(b) Rate = k[A]^3/2|B|^-1
Answer:
(a) Rate = k[A]^x[B]^y; Order = x + y
∴ Order = \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2 i.e., second order
(b) Order = \(\frac{3}{2}\) + (-1) = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) i.e., half order

Question 3.
The initial concentration of N₂O₅ in the following first order reaction N₂O₅ → 2NO₂(g) + 1/2O₂(g) was 1.24 ×10^-2 mol L^-1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10^-2 mol L^-1. Calculate the rate constant for the reaction at 318K.
Answer:
For a first order reaction, rate constant k = \(\frac{2.303}{\left(t_2-t_1\right)}\)log\(\frac{[\mathrm{R}]_1}{[\mathrm{R}]_2}\) = \(\frac{2.303}{(60-0)}\)log\(\frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}}\)
= \(\frac{2.303}{60}\) log 6.2 min^-1 = \(\frac{2.303}{60}\) × 0.793 = 0.0304 min^-1
∴ k = 0.0304 min^-1.

Question 4.
A first order reaction is found to have a rate constant k = 5.5 × 10^-14 s^-1. Find the half-life of the reaction.
Answer:
Half-life for a first order reaction is t_1/2 = \(\frac{0.693}{k}\) = \(\frac{0.693}{5.5 \times 10^{-14} \mathrm{~s}^{-1}}\) = 1.26 × 10¹³s

Question 5.
The first order rate constant for the decomposition of ethyl iodide by the reaction.
C₂H₅I(g) → C₂H₄(g) + HI_(g) at 600K is 1.60 × 10^-5 s^-1. Its energy of activation is 209 kJ/mol Calculate the rate constant of the reaction at 700K.
Answer:
Given data E_a = 209 kJ/mol = 209000 J/mol, k₁ = 1.60 × 10^-5 s^-1, T₁ = 600K, T₂ = 700K We have to find k₂
We know that logk2 – logk1

Objective Questions

Question 1.
Which of the following expressions is correct for the rate of reaction given below?

Answer:

Question 2.
The role of a catalyst is to change
1) gibbs energy of reaction.
2) enthalpy of reaction.
3) activation energy of reaction.
4) equilibrium constant.
Answer:
3) activation energy of reaction.

Question 3.
Rate law for the reaction A + 2B → C is found to be Rate = k [A][B]. Concentration of reactant ‘B’ is doubled, keeping the concentration of ‘A’ constant, the value of rate constant will be
1) the same
2) doubled
3) quadrupled
4) halved
Answer:
2) doubled

Question 4.
In a reaction A + B → Product, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written as
1) Rate = k [A] [B]
2) Rate = k[A]² [B]
3) Rate = k [A] [B]²
4) Rate = k [A]² [B]²
Answer:
2) Rate = k[A]² [B]

Question 5.
For the reaction
N₂O₂(g) → 2NO₂(g) + \(\frac{1}{2}\)O₂(g)
The value of rate of disappearance of N₂O₅ is given as 6.25 × 10^-3 mol L^-1s^-1. The rate of formation of NO₂ and O₂ is given respectively as
1) 6.25 × 10^-3molL^-1s^-1&6.25×10^-3molL^-3s^-1
2) 1.25 × 10^-2molL^-1s^-1&3.125×10^-3molL^-1s^-1
3) 6.25 × 10^-3molL^-1s^-1&3.125×10^-3molL^-1s^-1
4) 1 .25 × 10^-2molL^-1 s^-1 & 6.25 × 10^-3molL^-1s^-1
Answer:
2) 1.25 × 10^-2molL^-1s^-1&3.125×10^-3molL^-1s^-1

Question 6.
Consider a first order gas phase decomposition reaction given below :
A(g) → B(g) + C(g)
The initial pressure of the system before decomposition of A was p. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘p_t’ The rate constant k for the reaction is given as ________

Answer:
2) k = \(\frac{2.303}{t}\)log\(\frac{p_i}{2 p_i-t}\)

Question 7.
In the presence of a catalyst, the heat evolved or absorbed during the reaction ___
1) increases
2) decreases
3) remains unchanged.
4) may increase or decrease.
Answer:
3) remains unchanged.

Question 8.
Activation energy of a chemical reaction can be determined by
1) determining the rate constant at standard temperature.
2) determining the rate constants at two temperatures.
3) determining probability of collision.
4) using catalyst.
Answer:
2) determining the rate constants at two temperatures.

Question 9.
Consider the Arrhenius equation given below and mark the correct option. k = \(\mathrm{Ae}^{-k_a / \mathrm{RT}}\)
1) Rate constant increases exponentially with increasing activation energy and decreasing temperature.
2) Rate constant decreases exponentially with increasing activation energy and decreasing temperature.
3) Rate constant increases exponentially with decreasing activation energy and decreasing temperature.
4) Rate constant increases exponentially with decreasing activation energy and increasing temperature.
Answer:
4) Rate constant increases exponentially with decreasing activation energy and increasing temperature.

Question 10.
A first order reaction is 50% completed in 1.26 × 10¹⁴ s. How much time would it take for 100% completion?
1) 1.26 × 10¹⁵ s
2) 2.52 × 10¹⁴ s
3) 2.52 × 10²⁸ s
4) infinite
Answer:
4) infinite

Question 11.
The value of rate constant of a pseudo first order reaction is
1) depends on the concentration of reactants present in small amount.
2) depends on the concentration of reactants present in excess.
3) is independent of the concentration of reactants.
4) depends only on temperature.
Answer:
2) depends on the concentration of reactants present in excess.

Question 12.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314 J mol^-1K^-1)
1) 15.1 kJ mol^-1
2) 342kJ mol^-1
3) 269 kJ mol^-1
4) 34.7 kJ mol^-1
Answer:
4) 34.7 kJ mol^-1

Question 13.
A first order reaction has a rate constant of 2.303 × 10^-3s^-1. The time required for 40g of this reactant to reduce 10g will be
[Given that log₁₀2=0.3010]
1) 230.3 s
2) 301s
3) 2000s
4) 602s
Answer:
4) 602s

Question 14.
If the rate constant for a first order reaction is k, the time(t) required for the completion of 99% of the reaction is given by
1) t = 2.303/k
2) t = 0.693/k
3) t = 6.909/k
4) t = 4.606/k
Answer:
4) t = 4.606/k

Question 15.
An increase in the concentration of the reactants of a reaction leads to change in
1) activation energy
2) heat of reaction
3) threshold energy
4) collision frequency
Answer:
4) collision frequency

Question 16.
The slope of Arrhenius plot \(\left(\ln k \text { vs } \frac{1}{1}\right)\) of first order reaction -5×103 K. The value of E_a, of the reaction is
| given R=8.314 JK^-1mol^-1]
1) -83 kJmol^-1
2) 41.5 kJmol^-1
3) 83.0 kJmol^-1
4) 166kJmol^-1
Answer:
2) 41.5 kJmol^-1

Question 17.
In a zero-order reaction for every 10o rise of temperature, the rate is doubled. If the temperature is increased from 10oC to 100°C, the rate of the reaction will become
1) 64 times
2) 128 times
3) 256 times
4) 512 times
Answer:
4) 512 times

Question 18.
Which of the following statements is not correct about order of a reaction.
1) The order of a reaction can be a fractional number.
2) Order of a reaction is experimentally determined quantity.
3) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
4) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.
Answer:
3) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.

Question 19.
Which of the following statements is incorrect about the collision theory of chemical reaction?
1) It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
2) Number of effective collisions determines the rate of reaction.
3) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
4) Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer:
3) Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 1 Electric Charges and Fields Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 1st Lesson Electric Charges and Fields Questions and Answers

I. Multiple Choice Questions

Question 1.
By quantisation, charge q of a body is given by (wheren is an integer):
1) q = \(\frac{\mathrm{n}}{2 \mathrm{e}}\)
2) q = \(n\left(\frac{e}{3}\right)\)
3) q = \(\mathrm{n}\left(\frac{2 \mathrm{e}}{3}\right)\)
4) q = ne
Answer::
4) q = ne
The total charge of a body is always an integral multiple of the unit charge.

Question 2.
The value of permittivity of free space is : [ 3 ]
1) 9 × 10⁹ N m²/C²
2) 8.85 × 10^-12 \(\frac{\mathrm{Nm}^2}{\mathrm{C}^2}\)
3) 8.85 × 10^-12 \(\frac{\mathrm{C}^2}{\mathrm{Nm}^2}\)
4) 9 × 10⁹ \(\frac{\mathrm{C}^2}{\mathrm{Nm}^2}\)
Answer:
3) 8.85 × 10^-12 \(\frac{\mathrm{C}^2}{\mathrm{Nm}^2}\)
ε₀ is derived from Coulomb’s constant K = 9 × 10⁹ Nm²/ C²
We know K = \(\frac{1}{4 \pi \varepsilon_0}\) ⇒ ε₀ = \(\frac{1}{4 \pi \mathrm{~K}}=\frac{1}{4 \pi \times 9 \times 10^9}\) = 8.85 × 10^-12 \(\frac{\mathrm{C}^2}{\mathrm{Nm}^2}\)

Question 3.
If two bodies are rubbed and one of them acquires q₁ charge and another acquires q₂ charge, then the ratio q₁ : q₂ is:
1) 1: 2
2) 2 : 1
3) -1 : 1
4) 1 : 4
Answer:
3) -1 : 1
When two bodies are rubbed together, the bodies get equal and opposite charges
so that q₁ := – q₂
Ratio q₁ : q₂ = -1 : 1

Question 4.
Two charges of equal magnitudes and at a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is
1) F/8
2) F/4
3) 4F
4) F/ 16
Answer:
4) F/ 16
From Coulomb’s law F ∝ \(\) ⇒ F = \(\)
Here charges are halved and distance is doubled
F¹ = \(\frac{K\left(\frac{q_1}{2}\right)\left(\frac{q_2}{2}\right)}{(2 r)^2}=\frac{K}{16}\left[\frac{q_1 q_2}{r^2}\right]=\frac{F}{16}\)
∴ F¹ = \(\frac{F}{16}\)

Question 5.
Two charges +2C and +6C are repelling each other with a force of 12N. If each charge is added with -2 C, then the value of the force will be
1) 4 (Attractive)
2) 4 (Repulsive)
3) 8 (Repulsive)
4) Zero
Answer:
4) Zero
From Coulomb’s law F ∝ q₁q₂.
Given q₁ = 2e, q₂ = 6e. When each charge is added with -2c.
New charge \(\mathrm{q}_1^1\) = +2c – 2 = 0, \(\mathrm{q}_2^1\) = 6e – 2c = 4c
∴ F’ = 0 (4c) = 0

Question 6.
Figure shows the electric lines of force emerging from a charged body. If the electric field at A and B are E_A and E_B respectively and the displacement between A and B is r, Then

1) E_A > E_B
2) E_A < E_B
3) E_A = \(\frac{E_B}{r}\)
4) E_A = \(\frac{E_B}{r^2}\)
Answer:
1) E_A > E_B
The electric field lines are more concentrated at A than at B.
So that the magnitude of electric field at A ¡s greater than at B.
∴ E_A > E_B.

Question 7.
The figure shows some of the electric field lìes corresponding to an electric field. The figure suggests

1) E_A > E_B > E_C
2) E_A = E_B = E_C
3) E_A = E_B = E_C
4) E_A = E_B < E_C
Answer:
3) E_A = E_B = E_C
When the electric lines are closer together, the field is stronger, when they are spread apart, the field is weaker.
At points A & C, the field lines are equally crowded. At point B, the field lines are spread out.
S0, E_A = E_C > E_B

Question 8.
The intensity of an electric field E due to a short dipole depends on distance r as :
1) E ∝ \(\frac{1}{r^4}\)
2) E ∝ \(\frac{1}{r^3}\)
3) E ∝ \(\frac{1}{r^2}\)
4) E ∝ \(\frac{1}{r}\)
Answer:
2) E ∝ \(\frac{1}{r^3}\)
E_axial = \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{P}}{\mathrm{r}^3}\), E_eq = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{P}}{\mathrm{r}^3}\)
In both the cases, E ∝ \(\frac{1}{\mathrm{r}^3}\)

Question 9.
The angle between the electric dipole moment \(\overrightarrow{\mathrm{p}}\) and the direction of electric field strength due to it on the equatorial line is :
1) 0°
2) 90°
3) 180°
4) 60°
Answer:
3) 180°
On the equatorial plane, the resultant electric field E always directed antiparallel to the direction of the dipole moment.When they are in opposite, θ = 180°.

Question 10.
For the surface S shown in the figure (a closed surface) if the charge enclosed is q, then the net outward flux is:

1) \(\frac{q}{\varepsilon_0}\)
2) \(\frac{2 q}{\varepsilon_0}\)
3) 0
4) \(\frac{q}{2 \varepsilon_0}\)
Answer:
1) \(\frac{q}{\varepsilon_0}\)
From Gauss’s Law, the total electric flux Φ through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times q.
∴ Φ = \(\frac{q}{\varepsilon_0}\)

II. Fill in the Blanks

Question 1.
When a glass rod is rubbed with silk, then the charge on glass rod becomes ________.
Answer:
positive.
When we rub glass with silk, electrons are transferred from glass to silk.
Since glass loses electrons, it becomes positive.

Question 2.
A simple apparatus to detect charge on a body is the __________.
Answer:
Gold – Leaf electroscope.
A gold leaf electroscope consists of gold leaves. When a charged object touches the at the top, the charge spreads to the leaves. Since the like charges repel, the gold leaves get diverge.

Question 3.
The fact that the electric charge is always an integral multiple of ‘e’ is termed as _________
Answer:
Quantization of charge.
The total charge is given by q = ne . This is the quanitzation of charge. Here n is an integer.

Question 4.
A soap bubble is given a negative charge then its radius _________
Answer:
increases.
When we add positive or negative charges, the charges spread over the surface and repel each other. Due to this, the repulsion acts outwards. Then the bubble expand hence radius increases.

Question 5.
The value of basic unit of charge is __________
Answer:
1.6 × 10^-19C
e = 1.6 × 10^-19 C. This is the magnitude of the charge on a single electron.

Question 6.
The SI unit of electric field is _________
Answer:
N/C torV Volt/m.
The SI unit of electric field N/C or V/m
(∵ E = \(\frac{F}{q}\) ⇒ \(\frac{N}{C}\)) or (E = \(\frac{V}{d}\) ⇒ \(\frac{V}{m}\))

Question 7.
The direction of electric field lines of a isolated negative charge are _________.
Answer:
radially inward.
Isolated negative charge directs inward because negative charge attracts the positive.

Question 8.
The net force exerted on electric dipole placed in a uniform external electric field is ________
Answer:
zero.
In a uniform electric field, the force on the positive charge (+E_q) and negative charge (-E_q) are equal and opposite. So, they cancel each other.

Question 9.
The electrostatic field inside a charged conductor is __________
Answer:
zero.
In electrostatic equilibrium, charges resides only on the outer surface of the conductor, hence inside conductor is zero.

Question 10.
The magnitude of torque on electric dipole of dipole moment P in a uniform external field E is _________.
Answer:
pE sinθ.
In vector form 𝜏 = P × E = PE sinθ.

Question 11.
SI unit of volume charge density is __________.
Answer:
C/m³.
Volume charge density ρ = \(\frac{q}{v}\) ⇒ \(\frac{\mathrm{C}}{\mathrm{~m}^3}\)

Question 12.
SI unit of linear charge density is ________.
Answer:
C/m.
Linear charge density λ = \(\frac{\mathrm{q}}{l}=\frac{\mathrm{C}}{\mathrm{~m}}\)

III. One Word Answer Questions

Question 1.
What is the SI unit of electric charge?
Answer:
SI unit of electricicharge is coulomb (C).

Question 2.
What is the nature of force between two like charges?
Answer:
The nature of the force between two like charges is repulsive.

Question 3.
How many number of electrons are in 1 coulomb charge?
Answer:
Given that q = 1C, e = 1.6 ×10^-19 C, n = ?
Formula: q = ne ⇒ n = \(\frac{\mathrm{q}}{e}\) ⇒ n = \(\frac{1}{1.6 \times 10^{-19}}\)
= 0.625 × 10¹⁹ electrons

Question 4.
What happens to the weight of a body when it is charged positively?
Answer:
When a body is charged positively, its weight decreases very slightly. Because, some electrons are removed from the body.

Question 5.
Give the expression for net force on charge q₁ due to system of charges q₂, q₃, q₄,… are located at r₁₂, r₁₃,… distance from charge q₁.
Answer:
The net force on q₁, due to charges q₂, q₃, q₄… is F_net = F₁₂ + F₁₃ + F₁₄ + ……
∴ F_net = \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1 q_2}{r_{12}^2}+\frac{q_1 q_3}{r_{13}^2}+\frac{q_1 q_4}{r_{14}^2}+\ldots\right)\)

Question 6.
What is the direction of electric Held due to a single positive charge?
Answer:
The direction of electric field due to a single positive charge is radially outward in all directions.

Question 7.
Define electric field intensity at a point.
Answer:
Electric field intensity at a point is the force experienced by unit positive test charge placed at that point. Formula: E = \(\frac{F}{q}\)

Question 8.
What is the SI unit of electric flux?
Answer:
SI unit of electric flux is the Volt-meter (V-m). It is also equal to Nm² /C.

Question 9.
What is the SI unit of electric dipole moment?
Answer:
SI unit of dipole moment is coulomb -meter (C-m)

Question 10.
What is the total charge of electric dipole?
Answer:
Zero. Total charge of an electric dipole is (+q) + (-q) = 0

IV. Very Short Answer Questions

Question 1.
What is meant by the statement charge is quantised ?
Answer:
Quantisation of charge: The electric charge(q) is always equal to an integral multiple of the electron charge (e).
Thus q = ne. Here n = 0, ± 1, ±2, ±3, … and e = 1.6 × 10^-19C.

Question 2.
How many electrons constitute 1C of charge ?
Answer:
Given that q= 1C, e = 1.6 × 10^-19 C, n = ?
Formula: q = ne ⇒ n = \(\frac{\mathrm{q}}{e}\) ⇒ n = \(\frac{1}{1.6 \times 10^{-19}}\)
= 0.625 × 10¹⁹ electrons

Question 3.
What happens to the force between two charges if the distance between them is a) halved b) doubled
Answer:
From coulomb’s law, we have F ∝ \(\frac{1}{r^2}\) ⇒ \(\frac{F_2}{F_1}=\frac{r_1^2}{r_2^2}\)
a) When distance is halved r₂ = \(=\frac{\mathrm{r}_1}{2}\) ; New force F₂ = ?
\(\frac{F_2}{F_1}=\frac{r_1^2}{r_2^2}=\frac{r_1^2}{\left(r_1 / 2\right)^2}=\frac{r^4}{r^4 / 4}\) = 4 ⇒ F₂ = 4F₁.
Thus, force increases to 4 times
b) When distance is doubled r₂ = 2r₁, F₂ = ?
\(\frac{F_2}{F_1}=\frac{r_1^2}{r_2^2}=\frac{r_1^2}{\left(2 r_1\right)^2}=\frac{r^2}{4 r^2}=\frac{1}{4}\) ⇒ F₂ = F₁/4.
Thus, force decreases to 1/4

Question 4.
The electric lines of force do not intersect. Why ?
Answer:
If electric lines of force intersect, the electric field at the point of intersection will have two directions simultaneously, which is physically impossible. So, for electric lines of force do not intersect

Question 5.
Consider two charges +q and -q placed at B and C of an equilateral triangle ABC. For this system, the total charge is zero. But the electric field (intensity) at A which is equidistant from B and C is not zero. Why ?
Answer:
At point A, the electric field vectors due to +q and -q
are not in opposite direction, (actually 120° ).
So they do not cancel each other.
Hence the resultant vector is non-zero.
So, intensity of electric field is not zero at A.

V. Short Answer Questions

Question 1.
State and explain Coulomb’s inverse square law in electricity.
Answer:
Coulomb’s inverse square law:
The force of attraction or repulsion between two electric charges is directly proportional to product of their charges and is inversely proportional to the square of distance between them and acts along the line joining the charges.
Explanation: Consider two point charges q₁, q₂ separated by a distance r .

From Coulomb’s law: i) F ∝ q₁q₂ (ii) F ∝ \(\frac{1}{\mathrm{r}^2}\)
∴ F ∝ \(\frac{q_1 q_2}{r^2}\) ⇒ F = k \(\frac{q_1 q_2}{r^2}\) ⇒ F = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\)
Here ε₀ is the permittivity of free space and \(\frac{1}{4 \pi \varepsilon_0}\) = 9 x 109 Nm2C~2
If the charges are in a medium with permittivity ε then F_med = \(\frac{1}{4 \pi \varepsilon} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\)

Question 2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a point charge.
Answer:
Intensity of Electric Field (E):
The intensity of electric field at’ a point in space is defined as the force (F) experienced by a unit positive test charge (q) placed at that point in the field.
Formula: E = \(\frac{F}{q}\) SI Unit: N/C
Intensity due to a Point Charge :
Consider a point charge q kept at point O.
Let a test charge q₀ be kept at P, which is at a distance r from O.

The force between the two charges is F = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q} q_0}{\mathrm{r}^2}\) ⇒ \(\frac{\mathrm{F}}{\mathrm{q}_0}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\)
E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\) (∵ E = \(\frac{\mathrm{F}}{\mathrm{q}_0}\))

Question 3.
Derive an equation tor the couple acting on an electric dipole in a uniform electric field.
Answer:
Consider an electric dipole consisting of two equal and opposite charges +q & -q separated by a distance 2a.
Let the dipole be placed in a uniform electric field of intensity E at an angle θ to the direction of electric field. The electrostatic forces acting on the charges q and-q are Eq and-Eq respectively.

These two forces form a couple.
couple 𝜏 = Force × Perpendicular distance between the forces
𝜏 = Eq × BC ………….. (1)
From ∆ ABC, sin θ = \(\frac{B C}{A B}=\frac{B C}{2 a}\) ⇒ BC = 2a sin θ ………….. (2)
From (1) and (2), x = Eq (2a sin 0) = (2aEq) sin θ = q(2a)E sin θ
But we know dipole moment p= q(2a)
∴ 𝜏 = pE sin θ …………… (3)
In vector form \(\vec{\tau}=\vec{\mathrm{p}} \times \vec{\mathrm{E}}\)
Direction of the couple \(\vec{\tau}\) is perpendicular to the plane containing \(\vec{\mathrm{p}}\) and \(\vec{\mathrm{E}}\).

Question 4.
Derive an expression for the intensity of the electric field at a point on the axial line of a dipole.
Answer:
Consider an electric dipole AB consisting of two equal and opposite charges -q & q separated by a distance 2a.
Let P be a point on the axial line of the dipole at a distance r from its centre O.

Then distance PB = (r – a) and PA = (r+a).
Intensity of electric field at P due to +q is E₁ = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{(r-a)^2}\) …………… (1)
Intensity of electric field at P due to -q is E₂ = \(\frac{1}{4 \pi \varepsilon_0} \frac{q}{(r+a)^2}\) ………….. (2)
The resultant intensity of electric field due to the dipole at point P is E = E₁ – E₂

For shorter dipole we have (a<<<r) so that a² can be neglected.
∴ E = \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{pr}}{\mathrm{r}^4}=\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{p}}{\mathrm{r}^3}\)
Thus, the intensity of electric field on the axial line is E = \(\frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{p}}{\mathrm{r}^3}\)

Question 5.
Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole.
Answer:
Consider an electric dipole AB consisting of two equal and
opposite charges -q & q separated by a distance 2a.
Let P be a point on the equatorial plane of the dipole at a distance r from its centre.
Electric field at P due to +q is E₁ = \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{q}}{\mathrm{r}^2+\mathrm{a}^2}\right) .\) ……….. (1) (∵ AP² = r² + a²)
Electric field at P due to -q is E₂ = \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{q}}{\mathrm{r}^2+\mathrm{a}^2}\right) .\) ………… (2) (∵ BP² = r² + a²)
From figure, it is clear that the

y-components of E₁ and E₂ are equal and opposite.
Hence they cancel each other.
But the sum of x-components gives the resultant field E at P.
Magnitude of resultant electric field is E= E₁ cosθ + E₂ cosθ
= \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{q}}{\mathrm{r}^2+\mathrm{a}^2}\right) \cos \theta+\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{q}}{\mathrm{r}^2+\mathrm{a}^2}\right) \cos \theta\) = 2 × \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{q}}{\mathrm{r}^2+\mathrm{a}^2}\right) \cos \theta\)
From ∆ PAO, cos θ = \(\frac{A O}{A P}=\frac{a}{\sqrt{r^2+a^2}}\)
∴ E = \(2 \frac{1}{4 \pi \varepsilon_0}\left(\frac{q}{r^2+a^2}\right) \frac{a}{\sqrt{r^2+a^2}}\) = \(\frac{2 \mathrm{qa}}{4 \pi \varepsilon_0\left(\mathrm{r}^2+\mathrm{a}^2\right)^{3 / 2}}\)
But we know dipole moment p = q(2a)
∴ E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\left(\mathrm{r}^2+\mathrm{a}^2\right)^{3 / 2}}\) …………. (3)
For shorter dipole we have (a<<r) so that a² can be neglected.
From (3), for shorter dipole, E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\mathrm{r}^3}\)

Question 6.
State Gauss’s law in electrostatics and prove it.
Answer:
Gauss’s Law: The total electric flux (Φ) through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge (q) enclosed by the closed surface.
Thus Φ = \(\frac{1}{\varepsilon_0}\) (q)
Proof of Gauss’s Law
Suppose an isolated positive point charge q is situated at the centre O of a sphere of radius r.
Consider a point P on the sphere enclosed in a small closed surface area dS.
The electric field intensity at P is E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2} .\) ………… (1)
Total electric flux over the entire surface of sphere is Φ_E = \(\oint_S \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\oint_{\mathrm{S}} \mathrm{EdS} \cos \theta\)
Here, \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) are parallel. Hence θ = 0° ⇒ cos θ = cos 0° = 1
∴ Φ_E = \(\oint\) EdS = E\(\oint\)dS = E[S] = E × 4πr² …………. (2) [∵ Surface area of sphere S = 4πr²]
From (1) & (2), Φ_E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{r^2}\) × 4 πr² = \(\frac{q}{\varepsilon_0}=\frac{1}{\varepsilon_0}\)(q)
∴ Φ = \(\frac{1}{\varepsilon_0}\) (q)
Hence Gauss law is proved.

VI. Long Answer Questions

Question 1.
Define electric flux. Applying Gauss’s law, derive the expression for electric intensity due to an infinite long straight charged wire. (Assume that the electric field is everywhere radial and depends only on the radial distance r of the point from the wire.)
Answer:
Electric flux is the total number of electric field lines passing normally through a given area.
The electric flux through the surface area S due to electric field E is Φ = \(\vec{E} \vec{S}\) = ES cosθ
Here θ is the angle between \(\vec{E}\) and \(\vec{S}\).
Intensity (E) due to an infinite long wire :
Consider an infinite long straight wire having uniform linear charge density ‘λ’.
Let P be a point at a radial distance r from the wire.
Imagine a cylindrical Gaussian surface of length l and radius r coaxial with the wire.

The electric field is radial everywhere,.
Flux through the two end flat surfaces is zero.
(∵ Φ = \(\vec{E} \vec{S}\) = EScosθ and θ = 90°)
At the curved surface of the cylinder at every point, its magnitude of E is constant which depends only one r.
Curve surface area of the cylindrical surface is S= 2πrl
Total Electric flux Φ = EScos0 = ES (1)= E (2πrl) ………… (1)
The charge enclosed by the Gaussian surface is q = λ l
From Gauss’ law, Φ = \(\frac{\mathrm{q}}{\varepsilon_0}=\frac{\lambda l}{\varepsilon_0}\) ……….. (2)
From (1) & (2), E(2πrl) = \(\frac{\lambda l}{\varepsilon_0}\) ⇒ E = \(\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}}\)
∴ Intensity due to infinite long wire is E = \(\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}}\)
In the vector form, E = \(\vec{\mathrm{E}}=\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}} \hat{\mathrm{n}}\)
Here \(\hat{\mathrm{n}}\) is the radial unit vector in the plane normal to the wire.

Question 2.
State Gauss’s law in electrostatics. Applying Gauss’ law derive the expression for electric intensity due to an infinite plane sheet of charge.
Answer:
Gauss’s Law: The total electric flux (Φ) through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the net charge (q) enclosed by the closed surface.
Φ = \(\frac{1}{\varepsilon_0}\) (q)
Electric Intensity at a point due to an infinite plane sheet of charge:

Consider an infinite plane sheet has a uniform surface charge density ‘σ’.
Let us imagine a cylindrical Gaussian surface piercing through the vertical plane sheet.
Let P be a point on the cylindrical surface.
As the field is perpendicular to the plane sheet, the flux through the curved surface of the cylinder is zero. θ = 90° ⇒ Φ = EScosθ = EScos90° = ES(0) = 0
Total flux through the two flat surfaces of the Gaussian cylinder Φ = ES cos0° + EScos0°
⇒ Φ = ES + ES = 2ES ……..(1)
From Gauss’ law, Φ = \(\frac{q}{\varepsilon_0}=\frac{\sigma S}{\varepsilon_0}\) …….. (2) (∵ Surface charge density σ = \(\frac{q}{S}\) ⇒ q = σS)
From (1) & (2), E(2 S) = \(\frac{\sigma {S}}{\varepsilon_0}\) ⇒ E = \(\frac{\sigma}{2 \varepsilon_0}\)
∴ The electric intensity due an infinite plane sheet of charge is E = \(\frac{\sigma}{2 \varepsilon_0}\)
Its vector form is \(\vec{\mathrm{E}}=\frac{\sigma}{2 \varepsilon_0} \hat{\mathrm{n}}\)
where \(\hat{\mathrm{n}}\) is the unit vector normal to the plane and going away from it.

Question 3.
Applying Gauss’s law derive the expression for electric intensity due to a charged conducting spherical shell at (i) a point outside the shell (ii) a point on the surface of the shell and (iii)a point inside the shell.
Answer:
Intensity at a point due to a charged spherical shell:
Case(i): Point P, outside the shell:
Consider a charged conducting spherical shell of surface charge density σ and radius R.
Let P be a point outside the shell at a distance r from the centre of the shell. Imagine a Gaussian spherical surface of radius r such that its centre coincides with the centre of the shell.

Total Electric flux = Intensity × surface area of the sphere
Φ = E (4πr²) ………….. (1)
From Gauss’s law, Φ = \(\frac{q}{\varepsilon_0}\) ……….. (2)
The charge enclosed by Gaussian surface is q = σ (4πR²)
From (1) & (2), E(4πr²) = \(\frac{\mathrm{q}}{\varepsilon_0}=\frac{\sigma 4 \pi \mathrm{R}^2}{\varepsilon_0}\)⇒ E = \(\frac{\sigma 4 \pi \mathrm{R}^2}{4 \pi \mathrm{r}^2 \varepsilon_0}\)
⇒ E = \(\frac{\sigma \mathrm{R}^2}{\varepsilon_0 \mathrm{r}^2}=\frac{\mathrm{q} \mathrm{R}^2}{4 \pi \mathrm{R}^2 \varepsilon_0 \mathrm{r}^2}\) ⇒ E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\) ………… (3) [∵ σ = \(\frac{\mathrm{q}}{4 \pi \mathrm{R}^2}\)]
Case (ii): Point P, on the surface of the shell
In this case r = R,
From (3) E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\)
E = \(\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi R^2}{R^2}=\frac{\sigma}{\varepsilon_0}\) [∵ q = σ (4πR²)

Case (iii): Point P, inside the Shell
Electric flux Φ = Intensity × surface area = E (4πr²) ………. (1)
The charge enclosed by Gaussian surface inside the sphere is q = 0
From Gauss’ law, Φ = \(\frac{q}{\varepsilon_0}=\frac{0}{\varepsilon_0}\) = 0 ……….. (2)
From (1) & (2), E(4πr²) = 0 ⇒ E = 0
Thus the field due to a charged conducting shell is zero at all points inside the shell.

Textual Solved Problem

Question 1.
If 10⁹ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?
Solution:
Given n = 10⁹, we know e = 1.6 × 10^-19
∴ The charge that moves out in 1 second is q = ne = 10⁹ × 1.6 × 10^-19 = 1.6 × 10^-10C
∴ The time required for 1 C is t = \(\frac{1}{1.6 \times 10^{-10}}\) = 6.25 × 10⁹s = \(\frac{6.25 \times 10^9}{365 \times 24 \times 3600}\) ≅ 198 years

Question 2.
How much positive and negative charge is there in a cup of water?
Solution:
Suppose the mass of one cup of water is 250 g. The molecular mass of water is 18g.
Thus, mass of one mole (= 6.02 × 10²³ molecules) of water 18 g.
∴ the number of molecules in one cup of water is (250/18) × 6.02 × 10²³.
Each molecule of water (H₂O) contains 10 electrons and 10 protons.
Also the total positive and total negative charge has same magnitude.
∴ Total charge = (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10^-19 C = 1.34 × 10⁷ C.

Question 3.
Consider three charges q₁, q₂, q₃ each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in figure.
Solution:
Let F₁, F₂, F₃ be the forces on Q at centroid due to charges q at each vertices. The distance of the centroid from any vertex is l/\(\sqrt{3}\)

Thus, |F₁| = |F₂| = |F₃| = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Qq}}{(l / \sqrt{3})^2}\)
As the three forces of equal magnitude with angle between each pair 120° are acting simultaneously, symmetrically, the net force of the charge Q is zero.

Question 4.
Consider the charges q, q, and -q placed at the vertices of an equilateral triangle, as shown in figure. What is the force on each char.

Solution:
Force on +Q at A due to -q at C is, F_AC = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{l^2}\)
Force on +Q at A due to +q at B is, F_AB = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{l^2}\)
These two forces act at 120°. So, the resultant force, F_res = 2Fcos \(\frac{\theta}{2}\)
⇒ F_res = 2 × \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{l^2}\) × cos 60°
= 2 × \(\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{l^2} \times \frac{1}{2}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}^2}{l^2}\)

Exercise Problem

Question 1.
Calculate the charge on an alpha particle (Given charge on a proton = +1.6 × 10^-19C)
Answer:
An α particle consists of two protons and two neutrons.
The charge on each proton is = 1.602 × 10^-19 C.
Total charge Q = 2 × charge of proton= 2 × 1.602 × 10^-19 = 3.204 × 10^-19C

Question 2.
Calculate the distance between two protons such that the electrical repulsive force between them is equal to the weight of either. (Given mass of the proton = +1.67 × 10^-27 kg, Charge of the proton = +1.6 × 10^-19 C )
Answer:
From Coulomb’s law,F = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}^2}\) ⇒ r² = \(\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{~F}}\) …………… (1)
We know \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹, q₁ = q₂ = 1.602 × 10^-19 C m = 1.67 × 10^-27Kg
Repulsive force between two protons = Weight of the proton
⇒ F = mg = 1.67 × 10^-27 × 9.8 = 1.64 × 10^-26 N [∵ g = 9.8]
From (1), r² = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{~F}}=\frac{9 \times 10^9 \times\left(1.602 \times 10^{-19}\right)^2}{1.64 \times 10^{-26}}\) = 0.014078
∴ Distance r = \(\sqrt{0.014078}\) = 0.118 m = 11.8 cm

Question 3.
Charges q₁ = 6 mc, q₂ = 20 μC and q₃ = -8 mc are placed at the points A, R and C respectively as shown in figure. If r₁ = 6 m and r₂ = 6m. Calculate the magnitude of resultant force on q₂.

Answer:
Given q₁ = 6mc = 6 × 10^-3 C, q₂ = 20 mc = 20 × 10^-6 C, r₁ = 6m, r₂ = 6m,
Force on q₂ due to q₁ is F₂₁ = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}_1^2}=\frac{9 \times 10^9 \times 6 \times 10^{-3} \times 20 \times 10^{-6}}{6^2}=\frac{1080}{36}\) = 30N
Also q₃ = -8 _-13C, r₂ = 6m
Force on q₂ due to q₃ is F₂₃ = \(\frac{1}{4 \pi \varepsilon_0} \frac{q_2 q_3}{r_2^2}=\frac{9 \times 10^9 \times 20 \times 10^{-6} \times 8 \times 10^{-3}}{6^2}=\frac{1440}{36}\) = 40 N
As the two charges are at right angle, net force F_R = \(\sqrt{\mathrm{F}_{21}^2+\mathrm{F}_{23}^2}=\sqrt{30^2+40^2}\) = 50N

Question 4.
Calculate the electric field strength required to just support a water drop of mass 10^-3 kg and having a charge of 1.6 × 10^-19 C .
Answer:
Given Mass = 10^-3 kg, charge q = 1.6 × 10^-19 C , g = 9.8 m/s²
When the drop is at equilibrium then we have Electric force = Gravitational force
Eq = mg ⇒ E = \(\frac{\mathrm{mg}}{\mathrm{q}}=\frac{10^{-3} \times 9.8}{1.6 \times 10^{-19}}\) = 6.125 × 10¹⁶ N/C

Question 5.
Determine the electric field produced by a helium nucleus at a distance of lA from it.
Answer:
Charge of the nucleus q = 2e = 2 × 1.6 × 10^-19 = 3.2 × 10^-19 C,
Distance r = 1 Å = 10^-10 m
Electric field intensity E = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{q}{r^2}=\frac{9 \times 10^9 \times 3.2 \times 10^{-19}}{\left[10^{-10}\right]^2}\) = 2.88 × 10¹¹ N/C [∵ \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10⁹]
∴ E = 2.88 × 10¹¹ N/C

Question 6.
Two charges, one +5 μC and another -5 μC are placed 1mm apart. Calculate the dipole moment.
Answer:
Given charge q = +5 μC = 5 × 10^-6 C, distance 2a = 1 mm = 10^-3 m
∴ dipole moment p = (2a) q = (10^-3)(5 × 10^-6) = 5 × 10^-9 C – m

Question 7.
An electric dipole of dipole moment 4 × 10^-5 Cm is located in a uniform electric field of 10^-3 N/C making an angle of 30° with the direction of the field. Calculate the Torque exerted by electric field on the dipole.
Answer:
Given dipole moment p = 4 × 10^-5 C, Electric field intensity E = 10^-3 N/C, Angle θ = 30°
Torque 𝜏 = pE sin θ = 4 × 10^-5× 10^-3 × sin 30°= 4 × 10^-8 × \(\frac{1}{2}\) = 2 × 10^-8 Nm

Question 7.
An electric dipole of dipole moment 4 × 10^-5 Cm ¡s located in a uniform electric field of N/C making an angle of 30U with the direction of the field. Calculate the Torque exerted by electric field on the dipole.
Answer:
Given dipole moment p = 4 × 10^-5 C, Electric field intensity E = 10^-3 N/C, Angle θ = 30°
Torque 𝜏 = pE sinθ = 4 × 10^-5 × 10^-3 × sin30° = 4 × 10^-8 × \(\frac{1}{2}\) = 2 × 10^-8 Nm

Question 8.
If the electric field is given by E = \(3 \hat{i}+4 \hat{j}+3 \hat{k}\) N / C . Calculate the electric flux through a surface area 100m² lying in the X-V plane.
Answer:
Given electric field intensity \(\vec{\mathrm{E}}=8 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) N / C Area \(\vec{\mathrm{E}}\) = 100 \(\hat{k}\) m²;
Flux Φ = \(\vec{E} \cdot \vec{A}=(8 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(100 \hat{k})\) = 300
∴ Φ = 300 Nm²C^-1.

Question 9.
A spherical Gaussian surface enclosed by a charge of 8.85 × 10^-8C.
(i) Calculate the electric flux passing through the surface.
(ii) If the radius of the Gaussian surface is doubled, how would the flux change?
Answer:
Charge q = 8.85 × 10^-8C, Permittivity ε₀ = 8.85 × 10^-12 C²/Nm/²
i) Total electric flux Φ = \(\frac{1}{\varepsilon_0}\)q = \(\frac{8.85 \times 10^{-8}}{8.85 \times 10^{-12}}\) = 10⁴ Nm²C^-1
ii) The electric flux depends only on the charge enclosed and is independent of the radius of the surface. “So, flux doesnot change.”

Question 10.
A point charge of 17.7 μC is located at the centre of a cube of side 0.03m. Find the electric flux through each face of the cube.
Answer:
Given charge q = 17.7 μC = 17.7 × 10^-6C, ε₀ = 8.85 × 10^-12 C²/Nm²
Total electnc flux Φ = \(\frac{1}{\varepsilon_0}\)q = \(\frac{17.7 \times 10^{-6}}{8.85 \times 10^{-12}}\) = 2 × 10⁶ Nm² /C
Flux through each face of cube Φ_face = \(\frac{\phi}{6}=\frac{2 \times 10^{\circ}}{6}\) = 3.3 × 10⁵ Nm²C^-1

Question 11.
An infinite number of charges each of magnitude q are placed on x – axis at distances of 1. 2, 4, 8 meter from the origin respectively. Find intensity of the electric field at origin.
Answer:
E = \(\frac{1}{4 \pi \varepsilon_0}\left(\frac{q}{r_1^2}+\frac{q}{r_2^2}+\frac{q}{r_3^2}+\frac{q}{r_4^2}+\ldots .+\infty\right)\)
= \(\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{2^0}+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\ldots \infty\right)\)
Here, a = 1, r = 1/4 and S_∞ = \(\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
∴ E = \(\frac{q}{4 \pi \varepsilon_0}\left(\frac{4}{3}\right)=\frac{q}{3 \pi \varepsilon_0}\) towards origin.

Question 12.
The electric field in a region is given by \(\vec{\mathrm{E}}=\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}\). Here a and b are constants. Find the net flux passing through a square area of side L parallel to y – z plane.
Answer:
Given \(\vec{\mathrm{E}}=\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}\)
The area parallel to YZ plane means the normal to the area is parallel to x-axis.
∴ \(\vec{\mathrm{A}}=\mathrm{L}^2 \hat{\mathrm{i}}\)
Net flux Φ_E = \(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}}=(\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}) \cdot \mathrm{L}^2 \hat{\mathrm{i}}\) = aL² Weber

Question 13.
Two infinitely long thin straight wires having uniform linear charge densities λ and 2λ are arranged parallel to each other at a distance r apart. The intensity of the electric field at a point midway between them is
Answer:
Field due to an infinite long wire of linear charge density λ at a distance r is given by \(\frac{\lambda}{2 \pi \varepsilon_0 \mathrm{r}}\)
Here, E = E₂ – E₁
E = \(\frac{2 \lambda}{2 \pi \varepsilon_0(\mathrm{r} / 2)}-\frac{\lambda}{2 \pi \varepsilon_0(\mathrm{r} / 2)}=\frac{2 \lambda}{\pi \varepsilon_0 \mathrm{r}}-\frac{\lambda}{\pi \varepsilon_0 \mathrm{r}}=\frac{\lambda}{\pi \varepsilon_0 \mathrm{r}}\)

Objective Questions

Question 1.
Two electrons separated by distance V experience a force ‘F’ between them. The force between a proton and a singly ionised helium atom separated by distance 2r is
1) 4F
2) 2F
3) F/2
4) F/4
Answer:
4) F/4

Question 2.
The magnitude of electric intensity at a distance V from a charge ‘q’ is E. An identical charge is placed at a distance ‘2x’ from it. Then the magnitude of the force its experiences is
1) Eq
2) 2Eq
3) \(\frac{\mathrm{Eq}}{2}\)
4) \(\frac{\mathrm{Eq}}{4}\)
Answer:
4) \(\frac{\mathrm{Eq}}{4}\)

Question 3.
Two electric charges of 9μC and -3μC are placed 0.16m apart in air. There will be a point P at which electric potential is zero on the line joining the two charges and in between them. The distance of P from 9μC charge is
1) 0.14m
2) 0.12 m
3) 0.08 m
4) 0.06 m
Answer:
2) 0.12 m

Question 4.
The electric intensity in air at a point 20cm from a point charge Q coulombs is 4.5 × 10⁵ N/C. The magnitude of Q is
1) 0.1 μC
2) 0.2 μC
3) 1 μC
4) 2 μC
Answer:
4) 2 μC

Question 5.
Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes
1) 4F/3
2) F
3) 9F/16
4) 16F/9
Answer:
3) 9F/16

Question 6.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
1) increases K times
2) remains unchanged
3) decreases K times
4) increases K^-1 times
Answer:
3) decreases K times

Question 7.
A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to
1) -Q/4
2) Q/4
3) -Q/2
4) Q/2
Answer:
1) -Q/4

Question 8.
A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is
1) qEy
2) qE²y
3) qEy²
4) q²Ey
Answer:
1) qEy

Question 9.
A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (mdch larger than the size of the dipole), then the electric field at Q is proportional to
1) p² and r³
2) p and r²
3) p^-1 and r²
4) p and r³
Answer:
4) p and r³

Question 10.
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 10⁵ NC^-1. It experiences a torque equal to 4 Nm. The charge on the dipole, if the dipole length is 2 cm, is
1) 8 mC
2) 2 mC
3) 5 mC
4) 7 μC
Answer:
2) 2 mC

Question 11.
A dipole of dipole moment \(\vec{\mathrm{p}}\) is placed in uniform electric field \(\vec{\mathrm{E}}\), then torque acting on it is given by
1) \(\vec{\tau}=\vec{\mathrm{p}} \cdot \vec{\mathrm{E}}\)
2) \(\vec{\tau}=\vec{\mathrm{p}} \times \vec{\mathrm{E}}\)
3) \(\vec{\tau}=\vec{\mathrm{p}}+\vec{\mathrm{E}}\)
4) \(\vec{\tau}=\vec{\mathrm{p}}-\vec{\mathrm{E}}\)
Answer:
2) \(\vec{\tau}=\vec{\mathrm{p}} \times \vec{\mathrm{E}}\)

Question 12.
A spherical conductor of radius 10 cm has a charge of 3.2 × 10^-7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the
centre of the sphere? [\(\frac{1}{4 \pi {\varepsilon_0}}\) = 9 × 10⁹ Nm²/C²)
1) 1.28 × 10⁴ N/C
2) 1.28 × 10⁵ N/C
3) 1.28 × 10⁶ N/C
4) 1.28 × 10⁷ N/C
Answer:
2) 1.28 × 10⁵ N/C

Question 13.
What is the flux through a cube of side a if a point charge of q is at one of its corner?
1) \(\frac{2 q}{\varepsilon_0}\)
2) \(\frac{q}{8 \varepsilon_0}\)
3) \(\frac{q}{\varepsilon_0}\)
4) \(\frac{q}{2 \varepsilon_0} 6 a^2\)
Answer:
2) \(\frac{q}{8 \varepsilon_0}\)

Question 14.
A charge Q μC is placed at the centre of a cube, the flux coming out from each face will be
1) \(\frac{\mathrm{Q}}{6 \varepsilon_0}\) × 10^-6
2) \(\frac{\mathrm{Q}}{6 \varepsilon_0}\) × 10^-3
3) \(\frac{\mathrm{Q}}{24 \varepsilon_0}\)
4) \(\frac{\mathrm{Q}}{8 \varepsilon_0}\)
Answer:
1) \(\frac{\mathrm{Q}}{6 \varepsilon_0}\) × 10^-6

Question 15.
A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is
1) Q/6ε₀
2) Q/8ε₀
3) Q/ε₀
4) Q/2ε₀
Answer:
2) Q/8ε₀

Question 16.
A point charge+q is placed at the centre of a cube of side l. The electric flux emerging from the cube is
1) \(\frac{6 q \ell^2}{\varepsilon_0}\)
2) \(\frac{q}{6 \ell^2 \varepsilon_0}\)
3) zero
4) \(\frac{q}{\varepsilon_0}\)
Answer:
4) \(\frac{q}{\varepsilon_0}\)

Question 17.
A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 metres?
1) 20 μCm^-2
2) 5 μCm^-2
3) zero
4) 8 mCm^-2
Answer:
3) zero

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 4 The d- and f-Block Elements Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 4th Lesson The d- and f-Block Elements Questions and Answers

I. Multiple Choice Questions

Question 1.
The elements in which the differentiating electron enter into the penultimate energy level are
1. s-block elements
2. p-block elements
3. d-block elements
4. f-block elements
Answer:
3. d-block elements
Differentiating electron enters (n-1)d orbital, not outer shell.
That’s the key identity of d-block elements.

Question 2.
Which of the following represents the general electronic configuration of d-block elements?
1. (n-1)d^1-10 ns^1-2
2. (n-1)d^1-9ns^0-1
3. (n-1)d^1-8 ns^1-2
4. (n-2)d^1-10 ns^1-2
Answer:
1. (n-1)d^1-10 ns^1-2
d-block involves filling of (n-1)d and ns orbitals.General form: d¹– d¹⁰ and s¹– s² (n-1)d^1-10ns^1-2

Question 3.
The correct electronic configuration of Cr is
1. [Ar]3d⁴ 4s²
2. [Ar]3d⁵ 4s¹
3. [Ar]3d⁶ 4s⁰
4. [Ar]3d³ 4s²
Answer:
2. [Ar]3d⁵ 4s¹
Chromium shows exception due to half-filled stability. ,
One electron shifts → stable 3d5 Cr 24 : [Ar] 4s¹ 3d⁵

Question 4.
Which of the following is not regarded as a transition element?
1. Cr
2. Mn
3. Fe
4. Zn
Answer:
4. Zn
Transition elements must have partially filled d-orbitals.
Zn = d¹⁰ (fully filled) → not transition. Zn 30 : [Ar] 4s² 3d¹⁰

Question 5.
Which block elements in the periodic table have high ductility and malleability
1. s-block
2. p-block
3. d-block
4. f-block
Answer:
3. d-block
Strong metallic bonding + delocalized electrons → ductility & malleability.
Seen strongly in transition metals.

Question 6.
The common oxidation state of d-block elements is
1. +3
2. +2
3. +1
4. +4
Answer:
2. +2
ns electrons are lost first → gives +2 oxidation state commonly.

Question 7.
Which of the following is the only element of 3d series with positive standard electrode potential(M²⁺/M)?
1. Mn
2. Cr
3. Co
4. Cu
Answer:
4. Cu
Positive E° → metal is less reactive (noble).Only Cu shows this in 3d series.
S.E.P of Cu is +0.34 V

Question 8.
The transition metal which shows the highest oxidation state in its compounds is
1. Mn
2. Cr
3. V
4. Co
Answer:
1. Mn
Highest oxidation state depends on total valence electrons.
Mn 25 : [Ar] 4s² 3d⁵ → +7 maximum.

Question 9.
Identify the option in which the given species is correctly arranged in the increasing order of oxidizing power.
1. VO₂⁺ < MnO₄^– < Cr₂O₇^2-
2. Cr₂O₇^2- < VO₂⁺ < MnO₄^–
3. VO₂⁺ < Cr₂O₇^2- < MnO₄^–
4. Cr₂O₇^2- < MnO₄^– < VO₂⁺
Answer:
3. VO₂⁺ < Cr₂O₇^2- < MnO₄^–
Oxidizing power order (strong → weak):

Reverse for increasing order. VO₂⁺ < Cr₂O₇^2- < MnO₄^–

Question 10.
The spin only formula used for calculation of magnetic moment of a substance is
1. μ = \(\sqrt{n(n+1)}\)
2. μ = \(\sqrt{(n+2)}\)
3. μ = \(\sqrt{n(n+0)}\)
4. μ = \(\sqrt{n(n+2)}\)
Answer:
4. μ = \(\sqrt{n(n+2)}\)
Magnetic moment depends on unpaired electrons (n):
μ = \(\sqrt{n(n+2)}\)

Question 11.
Which of the following set of elements represent coinage metals?
1. Cu, Ag, Au
2. Cu, Fe, Zn
3. Ag, Fe, Cr
4. Au, Fe, Zn
Answer:
1. Cu, Ag, Au
Coinage metals = Group 11 metals used in coins.Cu, Ag, Au

Question 12.
Transition metals form large number of complex compounds. This is due to
1. small size
2. high nuclear charge
3. presence of vacant d-orbitals
4. All the above
Answer:
4. All the above
Complex formation depends on:
Small size → strong attraction; High charge → strong bonding
Vacant d-orbitals → accept lone pair of electrons from ligands . All contribute.

Question 13.
The catalyst used in Haber’s process is
1. Ni
2. Fe
3. Pt
4. Mo
Answer:
2. Fe
Haber process uses iron catalyst. With promoters like K₂O, Al₂O₃.

Question 14.
Which of the following element is not involved in the formation of Interstitial compounds with transition elements?
1. H
2. C
3. N
4. Cl
Answer:
4. Cl
Interstitial compounds form with small atoms like H, C, N.
Cl is too large → cannot fit.

Question 15.
In long form of periodic table which elements have high melting and boiling points?
1. Representative elements
2. Transition elements
3. Lanthanides
4. Actinides
Answer:
2. Transition elements
High melting/boiling points due to strong metallic bonding.
Maximum in transition elements.

II. Fill in the Blanks

Question 1.
The atomic radii of 5d series of elements is same as that of _______ series of elements.
Answer:
4d series

Question 2.
In aqueous solution the colour of the Cu²⁺ ion is ________
Answer:
blue

Question 3.
Ce⁴⁺ is a strong ________ agent.
Answer:
oxidizing

Question 4.
Most of the radioactive elements are present in __________ series/ block of periodic j table.
Answer:
5f series / f-block

Question 5.
The common oxidation state of lanthanides is __________
Answer:
+3

III. One Word Answer Questions

Question 1.
Name the transition element which does not exhibit variable oxidation states?
Answer:
Sc (Scandium)does not exhibit variable oxidation states.

Question 2.
Which substances are attracted very strongly by applied magnetic Held?
Answer:
Ferro Magnetic substances are attracted very strongly.

Question 3.
What is the other name of f-block elements?
Answer:
Inner transition elements is the other name of f-block elements.

Question 4.
What is the catalyst used in the manufacture of high density polythene?
Answer:
Zeigler – Natta Catalyst is used in the manufacture of high-density polythene.

Question 5.
Which metals are involved in the formation of an alloy called Brass?
Answer:
Cu and Zn are used in the formation of Brass.

IV. Very Short Answer Questions

Question 1.
What are transition elements? Give examples.
Answer:
Transition elements: These are the elements with partially filled d-orbital either in their atoms or ions.
Ex: Chromium (Cr 24) : [Ar] 4s¹3d⁵ ; for Cr⁺³ : [Ar]4s⁰3d³.
Manganese (Mn 25) : [Ar]4s²3d⁵. for Mn⁺²: [Ar]4s⁰3d⁵.

Question 2.
Why is it difficult to obtain M³⁺ oxidation state in Ni,Cu and Zn
Answer:
The element Ni, Cu and Zn are stable in +2 oxidation state and hence difficult to obtain +3 oxidation state.
Reason for +2 oxidation stability:
Ni and Cu in +2 oxidation state (Ni⁺²) are stable due to highest negative enthalpy of hydration.
Zn in +2 oxidation state (Zn⁺²) is stable due to completely filled d-subshell.

Question 3.
Why Zn²⁺ is diamagnetic whereas Mnup>2+ is paramagnetic?
Answer:
Electronic configuration of Zn⁺² is [Ar] 4s⁰3d¹⁰
It has no unpaired electrons. So it is diamagnetic.
Atomic numbers
Zn = 30
Mn = 25
Ar = 18
E.C of Mn⁺² is [Ar]4s⁰3d⁵ It has five impaired electrons. So it is paramagnetic
Note: E.C of Cr⁺³ is [Ar]4s⁰3d³
It has three unpaired electrons. So it is paramagnetic.

Question 4.
Calculate the ‘spin only’ magnetic moment of Fe²⁺_(aq) ion.
Answer:
Electronic configuration of Fe⁺² ion = [Ar] 3d⁶4s⁰

The number of unpaired electrons is n = 4.
∴ Spin only’ magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{4(4+2)}\) = \(\sqrt{24}\) = 4.9 BM

Question 5.
Aqueous Cu²⁺ ions are blue in colour, where as Aqueous Zn²⁺ ions are colourless. Why?
Answer:
E.C of Cu⁺² is [Ar] 4s⁰3d⁹. It contains one unpaired electron.
Hence it exhibits blue colour in aqueous solution.
E.C of Zn⁺² is [Ar] 4s⁰3d¹⁰
It contains no unpaired electron. Hence it is colourless in aqueous solution.
Atomic numbers
Cu = 29
Zn = 30

Question 6.
Why do the transition metals form a large number of complex compounds?
Answer:
Transition metals form a large number of complex compounds due to their small size, higher effective nuclear charge and presence of partially filled d-orbitals.

Question 7.
How do transition metals exhibit catalytic activity?
Answer:
Transition elements exhibit catalytic activity due to
(i) Variable oxidation states
(ii) Free valnecies on their surfaces
(iii) presence of partially filled d-orbitals.
Ex: Fe is the catalyst used in Haber’s process in the manufacture of ammonia.
V₂O₅ is the catalyst used in Contact process in the manufacture of H₂SO₄.
Pt is the catalyst used in Ostwald’s process in the manufacture of nitric acid.

Question 8.
What is an alloy? Give example.
Answer:
Alloy: It is a homogeneous mixture of a metal with other metals or metalloids or non-metals, having similar physical properties to that of the metal.
Ex: Brass: Cu (60% to 70%) and Zn (30% to 40%).
Bronze: Cu (75% to 90%) and Sn (10% to 25%).

Question 9.
What is mischmetal? Give its composition and uses.
Answer:
Mischmetal is a well known alloy of Lanthanoid metal.
Composition: 95% Lanthanoid metals (Ce, La, Nd, Pt), 5% iron and traces of S, C, Ca, Al.
Uses: Ignitium sources – Cigarette lighter, fire starter, bullets.

Question 10.
What is actinoid contraction
Answer:
Actinoid contraction: It is the gradual decreases in the size of atoms of actinoid series.
It is due to poor screening effect of 5f orbitais.

V. Short Answer Questions

Question 1.
Explain why Ni and Fe exhibit zero oxidation state in [Ni(CO)₄] and [Fe(CO)₅] respectively.
Answer:
In [Ni(CO)₄] and [Fe(CO)₅] the oxidation state of Ni and Fe is zero.
CO is a neutral ligand, hence the oxidation states of Ni and Fe are zero.
Low oxidation states are found when the complex compound has ligands capable of π-acceptor character in addition to the σ bonding.

Question 2.
What are interstitial compounds? How are they formed? Give two examples.
Answer:
Interstitial compounds are compounds formed when small atoms (H,C,B,N) occupy the empty spaces (interstitial sites-voids) in the metals of crystal lattice structure.
Formation: In metals such as Fe, Ti atoms are arranged in a closely packed structures with tiny gaps (voids). Small atoms can fit in this voids without disturbing the overall structure much.
Ex: Fe-C (Iron- carbon steel), Ti-H (Titanium – Hydride), Fe₃H

Question 3.
Write the characteristics properties of transition elements.
Answer:
Characteristics properties of transition elements:

1. Transition elements show variable oxidation states.
2. They exhibit paramagnetic character.
3. They form complex compounds.
4. They form coloured hydrated ions and salts.
5. They show catalytic activity.
6. They form alloys like Brass, Bronze.

Question 4.
Write down the electronic configuration of
(i) Cr³⁺
(ii) Cu⁺
(iii) Co²⁺
(iv) Mn²⁺
Answer:
i) E.C of Cr³⁺ : [Ar]3d³
Cr(24) : Cr³⁺ (21 = 18 + 3)

ii) E.C of Cu⁺ :[Ar]3d¹⁰
Cu(29) : Cu⁺ (28 = 18 + 10)

iii) E.C of Co²⁺ : [Ar]3d⁷
Co(27) = Co²⁺ (25 = 18 + 7)

iv) E.C of Mn²⁺ : [Ar]3d⁵
Mn(25) : Mn²⁺ (23 = 18 + 5)

Question 5.
What is lanthanoid contraction?What are the consequences of lanthanoid contraction?
Answer:
Lanthanoid contraction: It is the steady decrease in the atomic radii even with increase in atomic number from Lanthanum to Lutetium due to poor screening effect of 4f-electrons. Consequences of Lanthanoid contraction:
i) Basic character of oxides and hydroxides: Due to the lanthanoid contraction, the covalent nature of La-OH bond increases and thus, the basic character of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₃.

ii) Due to lanthanide contraction, size of 4d series and 5d series elements remains the same. Thus, pairs of elements such as Zr/Hf, Nb/Ta and Mo/W are almost identical in size. Due to almost similar size, such pairs have very similar properties which makes their separation difficult.

iii) Separation of lanthanoids: Due to lanthanoid contraction, there is a difference in some properties of lanthanoid like solubility, degree of hydration and complex formation.
These differences enable the separation of lanthanoids by ‘ion exchange method’.

Question 6.
Describe the preparation of potassium dichromate from iron chromite ore.
Answer:
Preparation of potassium dichromate (K₂Cr₂O₇) from chromile ore (FeCr₂O₄):
1) Potassium dichromate (K₂Cr₂O₇) is obtained by the fusion of chromite ore (FeCr₂O₄) with sodium (or potassium) carbonate (Na₂CO₃) in the presence of excess of air.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂. (yellow solution)
2) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a
solution from which orange sodium dichromate (Na₂Cr₂O₇.2H₂O) can be crystallised.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
3) Na₂Cr₂O₇ is more soluble than potassium dichromate. So the solution of sodium dichromate is treated with potassium chloride to prepare K₂Cr₂O₇.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Orange crystals of potassium dichromate crystallise out.

Question 7.
Why do the transition metal ions exhibit characteristic colours in aqueous solution. Explain giving examples.
Answer:
Transition metal ions exhibit colour property in aqueous solution due to the presence of unpaired d-electrons.

These unpaired d-electrons from a lower energy are excited to a higher energy d-orbitals of the same ‘n’ value. The energy of excitation corresponds to the frequency of light absorbed and this frequency lies in visible region. The colour observed corresponds to the complementary colour of light absorbed. The frequency of light absorbed is determined by the nature of the ligand. In aqueous solutions water molecules are the ligands.
Ex 1: Ti²⁺ [Ar] 3d¹ Purple
V²⁺ [Ar] 3d³ Violet
These have unpaired electrons in their d orbitals and hence they are coloured.
Ex 2: Sc³⁺ [Ar] 3d⁰ colourless
Cu⁺ [Ar] 3d¹⁰ colourless
These have paired electrons in their d orbitals and hence they are colourless.

Question 8.
Compare the stability of +2 oxidation state of the elements of the first transition series.
Answer:
The sum I.E₁ + I.E₂ of first transition series increases due to increased nuclear charge.
Thus, the standard potentials become less and less negative.
So the tendency to form M⁺² ion decreases.
The greater stability of +2 state of Mn is due to half-filled d-subshell (d⁵).
For zinc, it is due to completely filled d-subshell (d¹⁰).
For nickel, it is due to highest negative enthalpy of hydration.

Question 9.
Compared to the changes in atomic and ionic sizes of elements of 3d and 4d series, the change in radii of elements of 4d and 5d series is virtually the same. Comment.
Answer:
Compared to the change in atomic and ionic sizes of elements 3d and 4d series, the change in radii of elements 4d and 5d series is virtually the same.
This is due to the intervention of the 4f orbitals which must be filled before the 5d series of elements. This filling of 4f before 5d orbitals results in a regular decrease in atomic radii called Lanthanoid contraction. Due to Lanthanoid contraction the 4d and 5d series elements exhibit same size.

Question 10.
Write the characteristics of interstitial compounds.
Answer:
Characteristics of Interstitial compounds:

1. They are very hard and rigid. Ex: Some borides approach diamond in hardness.
2. They have high melting points, higher than those of their pure metals.
3. They retain metallic conductivity
4. They are chemically inert.

VI. Long Answer Questions

Question 1.
Explain giving reasons:
i) Transition metals and many of their compounds show paramagnetic behaviour.
ii) The enthalpies of atomisation of the transition metals are high.
iii) The transition metals generally form coloured compounds.
iv) Transition metals and their many compounds act as good catalysts.
Answer:

1. Transition metals and their compounds show paramagnetic behaviour. This is due to the presence of unpaired d-electrons. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum.
2. The enthalpies of atomisation of the transition metals are high. This is due to the presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them.
3. The transition metals generally form coloured compounds.This is due to absorption of visible light. The electron absorbs the radiation of a particular frequency (of visible region) and jumps into next orbital and the colour of compounds is complementary colour of observed colour.
4. Transition metals and their many compounds act as good catalysts. Due to Variable oxidation states (ii) Free valencies on their surfaces (iii) presence of partially filled d-orbitals.

Ex: Consider the reaction

Mechanism of catalysing action of Fe⁺³ in the above reaction is as follows:
a) 2Fe⁺³ + 2I^– → 2Fe⁺² + I₂
b) 2Fe⁺² + S2O₈^-2 2Fe⁺³ + 2SO₄^-2

Question 2.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with
(i) iron (II) ions
(ii) SO₂ and
(iii) oxalic acid? Write the ionic equations.
Answer:
1) Preparation of KMnO₄: Potassium permanganate is prepared by the fusion of MnO₂ with an alkalimetal hydroxide and an oxidising agent like KNO₃. It forms dark green, K₂MnO₄ which disproportionate in a neutral or acidic solution to give permanganate.

3K₂MnO₄ + 2H₂SO₄ → 2KMnO₄ + 2K₂SO₄ + MnO₂ + 2H₂O

2) Reactions of KMnO₄ in acidic medium MnO₄^– + 8H⁺ + 5e^– -» Mn⁺² + 4H₂O

1. Iron (II) ions: Ferrous ion is oxidised to Ferric ion.
   5Fe⁺² + MnO₄^– + 8H⁺ → Mn⁺² + 4H₂O + 5Fe⁺³
2. SO₂: It is oxidised to SO₄^-2 by acidified KMnO₄.
   5SO₂ + 2MnO₄^– + 2H₂O → 2Mn⁺² + 4H⁺ + 5SO₄^-2
3. Oxalic acid:
   5C₂O₄^-2 + 2MnO₄^– + 16H⁺ → 2Mn⁺² + 8H₂O + 10CO₂

Question 3.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
i) Electronic configuration
ii) Oxidation state
iii) Atomic and ionic sizes
iv) Chemical reactivity
Answer:
i) Electronic configuration (E.C):
E.C of lanthanoids: [Xe]4f^1-145d^0-16s².
E.C of actinoids: [Rn] 5f^1-146d^0-17s²,
Thus, lanthanoids belong to 4f series whereas actinoids belong to Sf series.

ii) Oxidation state: Lanthanoids show limited oxidation states (+2,+3,+4) out of which +3 is most common. This is because of large energy gap between 4f and Sd subshells.
But actinoids show a large number of oxidation states because of small energy gap between 5f, 6d, 7s subshells.

iii) Atomic and ionic size: Both show decrease in size of their atoms or ions in +3 oxidation state. In Lanthanoids, the decrease is called lanthanoid contraction whereas in actinoids, it is called actinoid contraction. However, the contraction is greater from element to element in actinoids due to poorer shielding by Sf electrons than that by 4f electrons in lanthanoids.

iv) Chemical reactivity: Lanthanoids show less tendency towards complex formation than actinoids. Lanthoids except promethium are non-radioactive. All the actinoids are radioactive. Lanthanoids do not form oxocations. Actinoids form oxo cations like UO₂⁺², PuO₂⁺²,UO⁺. Oxides and hydroxides of Lanthanoids are less basic whereas actinoids are more basic.

Question 4.
How would you account for the following:
(i) among d⁴ species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) is stable in aqueous solution but in the presence of compiexing reagents it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.
Answer:
i) Electronic configuration ofCr⁺² is [AT] 4s⁰3d⁴
Electronic configuration of Mn⁺³ is [Ar]4s⁰3d⁴
Though Cr⁺² and Mn⁺³ have the same electronic configuration, Cr⁺² acts as reducing agent but Mn⁺³ acts as oxidizing agent.
Reason: E<sup0 value for Cr⁺³/Cr⁺² is negative (-0.41V), where as E⁰ value for Mn⁺³/Mn⁺² is positive (+ 1.57 V)
Thus Cr⁺² ions can easily undergo oxidation to give Cr⁺³ ions and hence act as strong reducing agent. But Mn⁺³ can easily undergo reduction to give Mn⁺² and hence act as oxidizing agent.

ii) Cobalt (III) has greater tendency to form coordination complexes than Co(II). Thus, in the presence of ligands, Co(II) changes to Co(III) and easily gets oxidised.
Hence Co(II) is more stable in aqueous solution.

iii) The ions with d1 configuration have the tendency to lose the only electron present in d-subshell to acquire stable d⁰ configuration. Hence, they are unstable and undergo oxidation or disproportionation. Hence the d1 configuration is unstable.

Objective Questions

Question 1.
Transition elements belong to:
1. s-block
2. p-block
3. d-block
4. f-block
Answer:
3. d-block

Question 2.
The general electronic configuration of d-block elements is:
1. (n-1)d^1-10 ns^1-2
2. ns² np⁶
3. ns² np^1-6
4. ns¹
Answer:
1. (n-1)d^1-10 ns^1-2

Question 3.
Gadolinium belongs to 4f series. It’s atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?
1. [Xe] 4f⁷5d¹6s²
2. [Xe] 4f⁶5d²6s²
3. [Xe]4f⁸6d²
4. [Xe] 4f⁹5s¹
Answer:
1. [Xe] 4f⁷5d¹6s²

Question 4.
Transition elements show variable oxidation states due to:
1. Small size
2. Participation of ns and (n-1)d electrons
3. High ionisation energy
4. Presence of p-electrons
Answer:
2. Participation of ns and (n-1)d electrons

Question 5.
Which element shows the highest oxidation state?
1. Sc
2. Ti
3. Mn
4. Zn
Answer:
3. Mn

Question 6.
The colour of transition metal compounds is due to:
1. Presence of neutrons
2. d-d transitions
3. Proton transfer
4. Radioactivity
Answer:
2. d-d transitions

Question 7.
Which of the following is colourless?
1. Cu²⁺
2. Fe³⁺
3. Zn²⁺
4. Mn²⁺
Answer:
3. Zn²⁺

Question 8.
KMnO₄ acts as:
1. Reducing agent
2. Oxidising agent
3. Catalyst only
4. Acid
Answer:
2. Oxidising agent

Question 9.
Lanthanoids belong to:
1. 3d series
2. 4f series
3. 5d series
4. 6d series
Answer:
2. 4f series

Question 10.
The electronic configuration of Cu(II) is 3d⁹ whereas that of Cu(I) is 3d¹⁰. Which of the following is correct?
1. Cu(II) is more stable
2. Cu(II) is less stable
3. Cu(I) and Cu(II) are equally stable
4. Stability of Cu(I) and Cu(II) depends on nature of copper salts
Answer:
1. Cu(II) is more stable

Question 11.
The common oxidation state of lanthanoids is:
1. +1
2. +2
3. +3
4. +4
Answer:
3. +3

Question 12.
Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d⁵. What is its atomic number?
1. 25
2. 26
3. 27
4. 24
Answer:
2. 26

Question 13.
Which element among the following is radioactive?
1. Ce
2. Nd
3. U
4. La
Answer:
3. U

Question 14.
The first transition series starts from:
1. Sc
2. Ti
3. Cr
4. Fe
Answer:
1. Sc

Question 15.
Which of the following has highest number of unpaired electrons?
1. Sc
2. Mn
3. Cu
4. Zn
Answer:
2. Mn

Question 16.
Cu²⁺ compounds are generally:
1. Coloured
2. Colourless
3. Radioactive
4. Acidic
Answer:
2. Colourless

Question 17.
When acidified K₂Cr₂O₇ solution is added to Sn²⁺ salts then Sn²⁺ changes to
1. Sn
2. Sn³⁺
3. Sn⁴⁺
4) Sn⁺
Answer:
3. Sn⁴⁺

Question 18.
KMnO₄ acts as an oxidising agent in acidic medium. The number of moles of KMnO₄ that will be needed to react with one mole of sulphide ions in acidic solution is
1. \(\frac{2}{5}\)
2. \(\frac{3}{5}\)
3. \(\frac{4}{5}\)
4. \(\frac{1}{5}\)
Answer:
3. \(\frac{4}{5}\)

Question 19.
Which transition metal shows maximum oxidation state?
1. Fe
2. Mn
3. Cu
4. Zn
Answer:
2. Mn

Question 20.
Which of the following statements about the interstitial compounds is incorrect?
1) They retain metallic conductivity.
2) They are chemically reactive
3) They are much harder than the pure metal
4) The have higher melting points than the pure metals
Answer:
2) They are chemically reactive

Question 21.
Which ion is diamagnetic?
1. Fe³⁺
2. Mn²⁺
3. Zn²⁺
4. Cu²⁺
Answer:
3. Zn²⁺

Question 22.
Zr(Z = 40) and HF(Z=72) have similar atomic and ionic radii because of
1) having similar chemical properties
2) belonging to same group
3) diagonal relationship
4) lanthanoid contraction
Answer:
4) lanthanoid contraction

Question 23.
Which of the following pairs has the same size?
1) Fe²⁺, Ni²⁺
2) Zr⁴⁺, Ti⁴⁺
3) Zr⁴⁺, Hf⁴⁺
4) Zn²⁺, Hf⁴⁺
Answer:
3) Zr⁴⁺, Hf⁴⁺

Question 24.
Reason of lanthanide contraction is
1) negligible screening effect of f-orbitals
2) increasing nuclear charge
3) decreasing nuclear charge
4) decreasing screening effect
Answer:
1) negligible screening effect of f-orbitals

Question 25.
Which ion is strongly oxidising?
1. MnO₄^–
2. Fe²⁺
3. Cr²⁺
4. Cu⁺
Answer:
1. MnO₄^–

Question 26.
Lanthanoid contraction is due to:
1. Increase in nuclear charge
2. Poor shielding by 4f electrons
3. Presence of d-electrons
4. High density
Answer:
2. Poor shielding by 4f electrons

Question 27.
Which element shows only +3 oxidation state?
1. Sc
2. Mn
3. Fe
4. Cr
Answer:
1. Sc

Question 28.
Which metal is used as catalyst in hydrogenation of oils?
1. Cu
2. Fe
3. Ni
4. Zn
Answer:
3. Ni

Question 29.
The highest oxidation state shown by Mn is:
1. +4
2. +5
3.+6
4.+7
Answer:
4.+7

Question 30.
Which of the following forms coloured ions?
1. Sc³⁺
2. Zn²⁺
3. Cu²⁺
4. Ca²⁺
Answer:
3. Cu²⁺

Question 31.
Transition elements form interstitial compounds because:
1. They are non-metals
2. They have large atomic size
3. Small atoms occupy empty spaces in lattice
4. They are radioactive
Answer:
3. Small atoms occupy empty spaces in lattice

Question 32.
Magnetic moment. 2.83 BM is given by which of the following ions?
(At.No Ti = 22, Cr = 24, Mn = 25, Ni = 28)
1) Ti³⁺
2) Ni²⁺
3) Cr³⁺
4) Mn²⁺
Answer:
2) Ni²⁺

Question 33.
Mischmetal contains mainly:
1. Actinoids
2. Lanthanoids
3. Transition metals
4. Alkali metals
Answer:
2. Lanthanoids

Question 34.
Which compound is used in volumetric estimation of Fe²⁺ ?
1. KMnO₄
2. NaCl
3. NaOH
4. NH₄Cl
Answer:
1. KMnO₄

Question 35.
Which of the following has completely filled d-orbitals?
1. Cu²⁺
2. Zn²⁺
3. Fe²⁺
4. Mn²⁺
Answer:
2. Zn²⁺

Question 36.
Which transition metal compound is used as bleaching agent?
1. KMnO₄
2. K₄Cr₂O₇
3. TiO₂
4. FeSO₄
Answer:
1. KMnO₄

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 5 Coordination Compounds Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 5th Lesson Coordination Compounds Questions and Answers

I. Multiple Choice Questions

Question 1.
Which of the following coordination compound contains cobalt metal?
1. Chlorophyll
2. Hemoglobin
3. Vitamin B₁₂
4. Myoglobin
Answer:
3. Vitamin B₁₂
Check biological molecules containing metals:
Chlorophyll → Mg, Hemoglobin/Myoglobin → Fe.
Vitamin B₁₂ contains Co (cobalt)

Question 2.
According to Werner theory of coordination compounds, the secondary valence of metal ion in CoCl₃. 4NH₃ is
1. 6
2. 4
3. 3
4. 7
Answer:
1. 6
Secondary valence = coordination number (ligands attached). [Co(NH₃)₄ Cl₂] Cl
In CoCl₃. 4NH₃ is Co is bonded to 4 NH₃ & 2Cl^– .
Here coordination number is 4 + 2 = 6

Question 3.
Which of the following complex produces 2 moles of AgCl as precipitate, when 1 mole of complex reacts with excess of AgNO₃?
1. C0Cl₃. 4NH₃
2. NiCl₂. 6H₂O
3. PtCl₂. 2NH₃
4. CoCl₃. 6NH₃
Answer:
2. NiCl₂. 6H₂O
[Ni(H₂O)₆]Cl₂ ⇌ [Ni(H₂O)₆] + 2Cl^–. It gives 2 chloride ions.

Question 4.
Which of the following is not a double salt?
1. FeSO₄. (NH₄)2SO₄. 6H₂O
2. KCl.MgCl₂6H₂O
3. PtCl₄. 2HCl
4. KA1(SO₄)₂.12H₂O
Answer:
3. PtCl₄. 2HCl
Double salt dissociates completely into simple ions.
PtCl₄ . 2HCl does NOT behave like a double salt → complex nature. PtCl₄ . 2HCl

Question 5.
Lewis acid in the complex ion [CoCI(NH₃)₅|²⁺ is
1. Cl^–
2. NH₃
3. Co²⁺
4. Co³⁺
Answer:
4. Co³⁺
Lewis acid → Electron air acceptor. Metal ion accepts electron pairs from ligands.
So Co³⁺ acts as Lewis acid.

Question 6.
Which among the following is a combination of amhidentate ligand and neutral ligand?
1. CN^– & CO
2. EDTA & CO
3. H₂O & OH^–
4. Ethylene diamine (en) & oxalato
Answer:
1. CN^– & CO
Ambidentate ligand = can bind through different atoms (Ex: CN). It can bind through C &N.
Neutral ligand = no charge (Ex: CO).CN^– (ambidentate) + CO (neutral).

Question 7.
Which among the following is a homoleptic complex with primary valence of 3 and secondary valence of 6?
1. COCl₃.6NH₃
2. NiCl₂.6H₂O
3. PdCl₂.4NH₃
4. COCl₃. 4NH₃
Answer:
1. COCl₃.6NH₃
Homoleptic = only one type of ligand.
Primary valence = oxidation state, secondary = coordination number.
In COCl₃.6NH₃ Primary valence is 3 & secondary valency is 6.

Question 8.
Which ligand gives a stable complex?
1. SCN^–
2. Cl^–
3. H₂O
4. Ethylene diamine(en)
Answer:
4. Ethylene diamine(en)
Chelating ligands (multidentate) form most stable complexes.
Ethylene diamine (en) is bidentate → forms chelate rings.Ethylene diamine (en)

Question 9.
The correct formula for the given IUPAC name: tris (ethane-1,2-diamine) cobalt (III) sulphate is
1. [CO(NH₃)₆]₂(SO₄)₃
2. [CO(NH₃)₄(H₂O)Cl]Cl₂
3. [CO(H₂NCH₂CH₂NH₂)₃](SO₄)₂
4. [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃
Answer:
4. [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃
Charge of complex ion is +3. Charge of SO₄^-2 is -2. Balance by Criss-cross method.

Question 10.
Which complex shows both linkage and optical isomerism?
1. [Co(en)₃]Cl₃
2. [CoCl₂(en)₂]Br
3. [Co(en)₂(NO₂)Cl]Cl
4. [PdCl₂(NH₃)₂]
Answer:
3. [Co(en)₂(NO₂)Cl]Cl
Optical isomeristn requires non-superimposable mirror images. Typically seen in octahedral complexes with bidentate ligands. [Co(en)₂]³⁺ shows optical isomerism.

Question 11.
Which of the following shows colour?
1. K₃[Cu(CN)₄]
2. [Ti(H₂O)₆]⁴⁺
3. CuF₂
4. [Cu(CH₃CN)₄]BF₄
Answer:
3. CuF₂
Due to the presence of unpaired electrons.

Question 12.
Which complex shows geometrical, optical and also structural isomerism?
1. [Co(en)₃]Cl₃
2. [CoCl₂(en)₂]Br
3. [Co(en)(NH₃)₂Cl₂]Cl
4. [PdCl₂(NH₃)₂
Answer:
2. [CoCl₂(en)₂]Br
Geometrical → cis/trans possible .Optical → chiral (no plane of symmetry)
Structural → ionisation [CoCl₂(en)₂] Br . satisfies all cis/trans (geometrical), optical (due to en ligands), ionisation isomerism possible

Question 13.
Which among them has highest magnetic moment?
1. [CO(NH₃)₆]Cl₃
2. [NiCl₄]^2-
3. [COF₆]^3-
4. [Ni(CO)₄]
Answer:
3. [COF₆]^3-
Co⁺³ – [Ar]4s⁰3d⁶
More no.of unpaired electrons.

Question 14.
Among them, which complex shows ∆₀ > P, (∆₀ – relative magnitude of the crystal field splitting, P = pairing energy)
1. [CO(H₂O)₆]³⁺
2. [CO(NH₃)₆]³⁺
3. [MnF₆]^3-
4. [CoF₆]^3-
Answer:
2. [CO(NH₃)₆]³⁺
Strong field ligands → large ∆₀ → pairing occurs NH₃ is stronger ligand than H₂O, F^– , NH₃ – strong field ligand.

Question 15.
In which of the following metal ion is with 5 unpaired electrons ?
1. [CO(CN)₆]^3-
2. [CO(NH₃)₆]³⁺
3. [MnF₆]^4-
4. [Ni(en)Cl₄]^2-
Answer:
3. [MnF₆]^4-
In [MnF₆]^4-, Mn shows +2 oxidation state. E.C of Mn^{+2 is [Ar] 4s03d5
More no. of unpaired electrons. (n = 5)}

II. Fill in the Blanks

Question 1.
According to Werner’s theory of coordination compounds primary valences of metal ion are satisfied by _______.
Answer:
negative ions

Question 2.
Mohr’s salt formula is ________
Answer:
FeSO₄(NH₄)₂SO₄.6H₂O

Question 3.
The coordination sphere in potassium ferrocyanide complex is _________
Answer:
[Fe(CN)₆]^4-

Question 4.
The IUPAC name of K₂[PdCl₄] is ________
Answer:
Potassium tetrachloridopalladate (II)

Question 5.
M-CO bond in metal carbonyls is very strong due to _______ effect, (or) π-back bonding.
Answer:
Synergic

III. One Word Answer Questions

Question 1.
Which platinum complex can effectively inhibit the growth of tumors?
Answer:
Cis-platin can effectively inhibit the growth of tumors. Ex: Cis [Pt(NH₃)₂Cl₂]

Question 2.
Facial and Meridional isomers come under which category of isomerism?
Answer:
Facial and Meridional isomers show Geometrical isomerism.

Question 3.
What is the geometry of Fe(CO)₅ complex?
Answer:
Geometry of Fe(CO)₅ is Trigonal bipyramidal.

Question 4.
Which complex ion is formed when undecomposed AgBr of the film dissolves in hypo solution.
Answer:
[Ag(S₂O₃)₂]^3- is formed when AgBr film dissolves in hypo solution.

Question 5.
How many unpaired electrons are there in |NiCl₄]^2- ion?
Answer:
Two unpaired electrons are present in [NiCl₄]^2- ion.
Ni²⁺: [Ar]4s⁰3d⁸

IV. Very Short Answer Questions

Question 1.
What are coordination compounds? Give two examples.
Answer:
Coordination compounds are the metal atoms that are bond to few anions or neutral molecules by sharing electrons through coordinate covalent bonds.
Ex: [Co(NH₃)₆]Cl₃, K₄[Fe(CN)₆]

Question 2.
What is a double salt? Give example.
Answer:
Double salts contain two or more compounds which are stable in solid state. But they dissociate into its constituent ions when dissolved in water . In this process, the individual properties of constituent ions are not lost.
Ex: Mohr salt – FeSO₄, (NH₄)₂SO₄. 6H₂O
Potash alum – K₂SO₄.Al₂(SO₄)₃.24H₂O
Camallite – KCl.MgCl₂.6H₂O

Question 3.
What is the difference between a double salt and a complex compound?
Answer:
A double salt differs from complex compound in its dissociation.
A double salt dissociates completely into its constituent ions when dissolved in water, whereas a complex compound do not dissociate into its constituent ions when dissolved in water. Ex: Double salt: Camalite KCl.MgCl₂.6H₂O → K⁺+Mg⁺²+2Cl^–
Ex: Complex salt: Potassium ferrocyanide K₄[Fe(CN)₆] → 4K⁺ +[Fe(CN)₆]^4-

Question 4.
What is a chebte ligand ? Give example.
Answer:
Chelate ligand: It is a ring-like structure obtained when a bidentate or a polydentate ligand uses its two or more donor atoms to bind a single metal ion.
Ex: Oxalate C₂O₂, en-ethylene diammine

Question 5.
What is an ambidentate ligand? Give example.
Answer:
Ambidentate ligand: It is a ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal atom / ion.
M ← SCN
thiocyanato
(S donor atom)
M ← NCS
isothiocyanato
(N donor atom)

Question 6.
CuSO₄.5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why?
Answer:
In CuSO₄.5H₂O, water acts as ligand. As a result it causes crystal field splitting.
Hence, d-d transition is possible in CuSO₄.5H₂O and it shows colour.
In the anhydrous CuSO₄ due to the absence of water (ligand), crystal field splitting is not possible. Hence it is colourless.

Question 7.
How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C₂O₄)₃]^3-
(ii) [CO(NH₃)₃Cl₃]
Answer:
i) No geometrical isomers are possible for [Cr(C₂O₄)₃]^3-
ii) Two geometrical isomers are possible for [CO(NH₃)₃Cl₃]

Question 8.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Why?
Answer:
[Cr(NH₃)₆]³⁺ is paramagnetic due to the presence of three unpaired electrons.
[Ni(CN)₄]^2- is diamagnetic due to the absence of unpaired electrons.

Question 9.
[Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:
In both the complexes, Fe has +2 oxidation state . It has four unpaired electrons.
In the presence of weak H₂O ligand, the unpaired electrons do not pair up.
In the presence of strong CN^– ligand, the unpaired electrons pair up .
They have different colours due to the difference in the number of unpaired electrons.

Question 10.
What is the oxidation state of cobalt in
(i) K[Co(CO)₄] and (ii) [Co(NH₃)₆]³⁺
Answer:
i) K[Co(CO)₄] : Oxidation state of cobalt is-1
ii) [Co(NH₃)₆]³⁺ Oxidation state of cobalt is +3

V. Short Answer Questions

Question 1.
Explain Werner’s theory of coordination compounds with suitable examples.
Answer:
Werner’s theory explains the structure, bonding, and isomerism of complex compounds,
1) Every complex has a central metal atom (or) ion bonded with ligands and some anions.

2) The central metal atom/ion shows two types of valences: Primary & Secondary valences.
a) Primary valence (1°): It is equal to the oxidation state of the central atom. It is satisfied by the negative anions. These are ionisable and non-directional. The anion bonds connected to central metal are shown with dotted lines ( – – – – -) in the diagram.

b) Secondary valence (2°): It is equal to co-ordination number of the central atom.
It is the number of coordinate covalent bonds formed by central metal atom/ ion with ligands. The species satisfying secondary valence are not ionisable in solutions. The ligand bonds connected to the central atom are shown with continuous lines (————-) in the diagram.
Every metal has a fixed coordination number in a given oxidation state. This coordination number defines the shape of complex.

3) Some ligands satisfy both primary and secondary valence of the metal.
Such ligands are non-ionisable.
The following examples explain Werner’s theory.
Ex 1: CoCl₃.6NH₃ = [Co(NH₃)₆]Cl₃
The secondary valency of the central ion Co³⁺ is 6.
That means Co is bonded with 6 NH₃.
The primary valency of Co³⁺ is 3.
That means Co is bonded with 3Cl^– ions.

Ex 2: CoCl₃.5NH₃ = [Co(NH₃)₅Cl]Cl₂
The secondary valency of the central ion Co³⁺ is 6.
That means Co is bonded with 5 NH3 and one Cl^–.
Here Cl satisfies both 1° & 2° valencies.
The primary valency of Co³⁺ is 3.
Two ionisable and one Cl^– satisfies both valencies which is non-ionidable.

Question 2.
Give the geometrical shapes of the following complex entities
(i) [CO(NH₃)₆]³⁺
(ii) [Ni(CO)₄]
(iii) [PtCl₄)]^2-
(iv) [Fe(CN)₆]^4-
Answer:
i) [CO(NH₃)₆]³⁺ → octahedral
ii) [Ni(CO)₄] → tetrahedral
iii) [PtCl₄)]^2- → square planar
iv) [Fe(CN)₆]^4- → octahedral

Question 3.
Using IUPAC norms write the formulas for the following :
(i) Tetrahydroxozincate (II) ion
(ii) Hexaaminecobalt (III) sulphate
(iii) Potassium tetrachioropalladate(II)
(iv) Potassium tri(oxoalato)chromate(III)
Answer:
i) Tetrahydroxozincate(II) ion: [Zn(OH)₄]^-2
ii) Hexaaminecobalt (III) sulphate : [Co(NH₃)₆]₂(SO₄)₃
iii) Potassium tetrachloropalladate(II): K₂[PdCl₄]:
iv) Potassium trioxalatochromate(III): K₄[Cr(C₂O₄)₃]

Question 4.
Using IUPAC norms write the systematic names of the following:
(i) [Co(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂ CH₃)]Cl
(iii) [Ti(H₂O)₆]³⁺
(iv) [NiCl₄]^2-
Answer:
i) [Co(NH₃)₆]Cl₃ : Hexaammine Cobalt (III)chloride
ii) [Pt(NH₃)₂Cl(NH₂ CH₃)]Cl : Diammine chloro Methyl amtye platinum(II)chloride
iii) [Ti(H₂O)₆]³⁺ : Hexaaquotitanium (III) ion
iv) [NiCl₄]^2- : Tetrachloronickelate (II) ion

Question 5.
What are homoleptic and heteroleptic complexes? Give one example for each.
Answer:
Homoleptic complexes: These are the complexes in which a metal is bound by only one kind of ligands. Ex: [Co(NH₃)₆]⁺³
Heteroleptic complexes: These are the complexes in which a metal is bound by more than one kind of ligands. Ex: [Co(NH₃)₄Cl₂]⁺

Question 6.
Explain geometrical isomerism in coordination compounds giving suitable examples.
Answer:
Geometrical isomerism: This type of isomerism is due to different possible geometric arrangements of the ligands. Here, the two types of isomers are cis isomer and trans isomer.
1) In the square planar complex of the form [MX₂L₂], the two X ligands arranged adjacent to each other in the same side are cis isomers and which are arranged opposite to each other are trans Isomers.
Ex: [Pt(NH₃)₂Cl₂]

2) In the Octahedral complex of the form [MX₂L₄] , the two X ligands are oriented cis or trans to each other. Ex: [CoCl₂(NH₃)₄]⁺

Question 7.
What is meant by chelate effect ? Give example.
Answer:
Chelate effect: It is the stabilization of complex compounds due to ring formation (chelation) by bidentate or polydentate ligands. Complexes containing chelating ligands are more stable than complexes containining unidentate ligands.
Ex: [Co(en)₃]³⁺ is more stable than [Co(NH₃)₆]³⁺
Note: en= Ethylenediamine ⇒ bidentate ligand (chelating ligand)

Question 8.
Explain the following terms by giving an example
i) Unidentate ligand
ii)Bidentate ligand
iii) Polydentate ligand
iv)Ambidentate ligand
Answer:
i) Unidentate: It is a ligand with One donor atom. Ex: : NH₃
ii) Bidentate: It is a ligand with Two donor atoms Ex: NH₂-NH₂
iii) Polydentate: It is a ligand with more than two donor atoms Ex: EDTA
iv) Amhidentate ligand: It is a ligand with two donor atoms, but only one of them forms a coordinate bond at a time with central metal atom / ion.

Question 9.
Give the oxidation numbers of the central metal atoms in the following complex entities
(i) [Ni(CO)₄]
(ii) [Co(NH₃)₆]⁺³
(iii) [Fe(CN)₆]^4-
(iv) [Fe(C₂O₄)₃]^3-
Answer:
i) [Ni(CO)₄]
x + 4(0) = 0
⇒ x = 0

(ii) [Co(NH₃)₆]⁺³
x + 6(0) = +3
⇒ x = +3
In metal carbonyls metal oxidation state is equal to zero, (synergic effect)
iii) [Fe(CN)₆]^4-
x + 6(-1) = -4
⇒ x = 6 – 4 = +2

(iv) [Fe(C₂O₄)₃]^3-
x + 3(-2) = -3
⇒ x = +3

Question 10.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central melai atom/’ ion
Answer:
i) Ligand: The ions or molecules bound to the central atom / ion in the complex by donating lone pair of electrons are called ligands. These may be
a) simple ions such as Cl^–
b) small molecules such as H₂O
c) large molecules such as H₂N-CH₂-CH₂-NH₂
d) macro molecules such as proteins.

ii) Coordination Number (CN) : It is the number of ligands to which the metal is directly bonded. Ex: In [Pt Cl₆]^-2, C.N of Pt = 6 .
iii) Coordination entity: It is the central metal atom/ ion together with the ligands directly bonded to it. It is enclosed between square brackets.
Ex: [COCl₃(NH₃)₃], [Ni(CO)₄]
iv) Central metal atom /ion:The atom / ion to which a fixed number of ligands/ ions/ molecules are bound in a complex is called central metal atom or ion.
Ex: In K₄[Fe(CN)₆] the central metal is Fe

VI. Long Answer Questions

Question 1.
Explain different types of isomerism exhibited by coordination compounds, giving suitable examples.
Answer:
Coordination compounds exhibit two types of isomerism
I) Stereoisomerism
II) Structural Isomerism

I) Stereoisomerism: Stereoisomerism is a form of isomerism in which two substances have the same composition and structure but differ in the relative spatial positions of the ligands.

1) Geometrical isomerism: It is due to different geometric arrangements of the ligands. Here, the two types of isomers are cis isomer and trans isomer.

i) In the square planar complex of the form [MX₂L₂], the two X ligands arranged adjacent to each other in the same side are cis isomers and which are arranged opposite to each other are trans isomers. Ex: [Pt(NH₃)₂Cl₂]

ii) In the Octahedral complex of the form [MX₂L₄], the two X ligands are oriented cis or trans to each other. Ex: [CoCl₂(NH₃)₄]⁺
In octahedral coordination entities of the type [Ma₃b₃] we have two types of isomers a) facial (fac) isomer b) meridional (mer) isomer Ex: [Co(NH₃)₃(N0₃)₃].

2) Optical Isomerism: Optical isomers are mirror images that cannot be superimposed on one another. One isomer that rotates the plane polarised light in clock wise direction is d-isomer and other isomer that rotates plane polarised light in anticlock wise direction is/-isomer.
Ex: [CoCl₂(en)₂]⁺ [Co(en)₃],sup>3+

II) Structural Isomerism: Two (or) more compoundshaving same empirical formula but different structural arrangements are called structural isomers. They are 4 types.
1) Linkage lsomerism:Isomers which differ in point of attachment when ambidentate ligands are present in coordination sphere are known as linkage isomers and the phenomenon is known as linkage isomerism. ,
Ex: [ Co(NH₃)₅(NO₂)] Cl₂ & [Co(NH₃)₅(ONO)] Cl₂

2) Coordination Isomerism:Isomers formed due to interchange of ligands between cationic and anionic entities of a complex are known as coordination isomers and the phenomenon is known as coordination isomerism.
Ex: [Co(NH₃)₆][Cr(CN)₆], [Co(CN)₆][Cr(NH₃)₆]

3) Ionisation Isomerism: It is due to exchange of ligands between coordination and ionisation sphere. Hence compounds give different ions in solution. These compounds are known as isomerisation isomers and the phenomenon is known as Ionisation Isomerism
Ex: [ Co(NH₃)₅SO₄] Br & [Co(NH₃)₅Br] SO₄

4) Hydrate Isomerism: Isomers which differ in number of water molecules in coordination sphere are known hydrate isomer and the phenomenon is known as hydrate isomerism.
Ex: [Cr(H₂O)g]Cl₃ (violet) and its solvate isomer [Cr(H₂O)5Cl]Cl₂.H₂O (grey-green).

Question 2.
Discuss the nature of bonding and magnetic behaviour in the following coordination entities on the basis of valence bond theory
(i) [Fe(CN)₆]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3-
(iv) [CoF₆]^3-
Answer:
(i) [Fe(CN)₆]^4- : The oxidation state of Fe in this complex is +2.
Fe (Z = 26) = [Ar] 3d⁶4s²

Fe⁺²(Z = 24) = [Ar]3d⁶4s⁰

CN^– is a strong ligand. So forced pairing happens and hence sp³d² hybridization takes place.

Now six empty d²sp³ hybrid orbitals overlap with 6 ligand orbitals (CN^–) to form the complex ion.

This complex ion is octahedral. It is diamagnetic as it does not contain any unpaired electron. Also (n-1)d orbitals are involved in hybridization. So it is a inner orbital or low spin complex ion.

(ii) [FeF₆]^3- : The oxidation state of Fe in this complex is +3.
Fe (Z = 26) = [Ar]3d⁶4s²

Fe⁺³ (Z = 23) = [Ar] 3d⁵ 4s⁰

F^– is a weak ligand. So pairing does not happens.
Here 3d orbitals are not available to take part in bonding.
Hence nd orbitals are involved in hybridisation and sp³d² hybridisation takes place.
[FeF₆]^3- :

Because of the presence of five unpaired electrons, the complex is paramagnetic.
Also outer d orbitals are involved in hybridization, so it is an outer orbital or high spin complex ion.

iii) [Co(C₂O₄)₃]^3- : The oxidation state of Co in this complex is +3
Co (Z = 27) = [Ar]3d⁷4s²

Co⁺³ = [Ar] 3d⁶4s⁰

C₂O₄^-2 is a strong ligand .
So pairing happens and hence d²sp³ hybridization takes place.
Now 6 empty d²sp³ hybrid orbitals overlap with 6 ligand orbitals ( C₂O₄^-2) to form complex ion.
[Co(C₂O₄)₃]^3- :

Since, all the electrons are paired, this complex is diamagnetic. It is an inner orbital complex because of the involvement of (n-1)d orbital for bonding.

(iv) [CoF₆]^3- : The oxidation state of Co in this complex is +3.
Co (Z = 27) : [Ar] 3d⁷4s²,

Co⁺³ (Z = 24) [Ar] 3d⁶4s⁰

F^– is a weak field ligand. So pairing does not happen
Thus 3d orbitals are not available to take part in bonding.
Hence nd orbitals are involved in hybridisation and sp³d² hybridisation takes place.
[CoF₆]^3- :

Because of the presence of four unpaired electrons, the complex is paramagnetic.
Since outer d orbitals take part in bonding it is an outer orbital complex ion or high spin complex ion.

Question 3.
Sketch the splitting of d-orbitals in an octahedral crystal field. A:
Answer:

Question 4.
Explain the applications of coordination compounds in different Fields.
Answer:
1) Coordination compounds are used in the estimation of qualitative and quantitative analysis of compounds. For this purpose chelating reagents like EDTA, DMG (dimethyl glyoxime) α-nitroso-β-naphthol are used.

2) Ca⁺², Mg⁺² ions present in hard water are estimated by chelating agent sodium salt of EDTA.

3) Silver and gold are extracted from ore by complex formation and metal displacement method.
Ex: AgCl + 2NaCN → Na[Ag(CN)₂] + NaCl
2Na[Ag(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Ag↓

4) Certain metals like Ni, Fe are purified by complex formation and subsequent decomposition of complex. In Mond’s process TMi’ is purified in this way.

5) Coordination complexes are very important in biological systems.

1. Chlorophyll present in plants is a complex of Mg⁺² ion
2. Vitamin B12 (Cyanocobalamine) is a complex of CO⁺³ ion.
3. In Hemoglobin, Heme is the complex of Fe⁺² ion
4. Enzymes like carboxypeptidase and carbonic anhydrase act as catalyst in biochemical reactions.

6) Coordination compounds are used as catalysts in industries. Wilkinson catalyst in a rhodium complex [(Ph₃P)₃RhCl] is used for the hydrogenation of alkenes.

7) In photography, in the process of developing [Ag(S₂O₃)₂]^3- is formed when unreduced AgBr film dissolves in hypo solution.

8) Recently complexes are used in the medicinal chemistry for treatment of diseases.
Ex: EDTA is used in the treatment of lead poisoning.

Question 5.
a) Write four postulates of Werner’s theory of coordination compounds.
b) Explain cis-trans isomerism in [Pt(NH₃)₂Cl₂] complex.
Answer:
Werner’s theory explains the structure, bonding, and isomerism of complex compounds.
a) Postulates of Werner’s theory:

1. In coordination compounds metals show two types of linkages (valences)- primary and secondary.
2. The primary valences are normally ionisable and are satisfied by negative ions.
3. The secondary valences are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the coordination number and is fixed for a metal.
4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements corresponding to different coordination numbers.

b) Cis-trans isomerism in [Pt(NH₃)₂Cl₂] complex:
Geometrical or Cis-trans isomerism:
Cis Isomerism: When two ligands of same type occupy adjacent positions in coordination sphere of central metal atom, then it is called cis-isomer.

Trans Isomerism: If the two ligands occupy far apart opposite to each other in coordination sphere, then it is called trans isomer.

The given complex [Pt(NH₃)₂Cl₂] is in the square complex of the form [MX₂L₂] and its Cis isomer and trans isomer are shown in the figure.

Objective Questions

Question 1.
In coordination compounds, ligands act as:
1. Electron donors
2. Electron acceptors
3. Proton donors
4. Neutral molecules only
Answer:
1. Electron donors

Question 2.
The coordination number of Co in [CO(NH₃)₆] Cl₃ is:
1. 3
2. 4
3. 6
4. 2
Answer:
3. 6

Question 3.
Which among the following is a neutral ligand?
1. CN^–
2. Cl^–
3. NH₃
4. OH^–
Answer:
3. NH₃

Question 4.
The oxidation state of Fe in K₄[Fe(CN)₆] is:
1. +1
2. +2
3. +3
4. +4
Answer:
2. +2

Question 5.
Which ligand is bidentate?
1. NH₃
2. Cl^–
3. en
4. H₂O
Answer:
3. en

Question 6.
EDTA is:
1. Monodentate
2. Bidentate
3. Tetradentate
4. Hexadentate
Answer:
4. Hexadentate

Question 7.
Which ligand is ambidentate?
1. NH₃
2. H₂O
3. NO₂^–
4. EDTA
Answer:
3. NO₂^–

Question 8.
The coordination entity in (CU(NH₃)₄]SO₄ is:
1. SO₄^2-
2. Cu²⁺
3. [Cu(NH₃)₄]²⁺
4. NH₃
Answer:
3. [Cu(NH₃)₄]²⁺

Question 9.
The IUPAC name of [CO(NH₃)₅Cl]Cl₂ is:
1. Pentaamminechloridocobalt(III) chloride
2. Pentaamrninechlorocobalt(II) chloride
3. Chloropentaamminecobalt(III) chloride
4. Pentaamminecobalt chloride
Answer:
1. Pentaamminechloridocobalt(III) chloride

Question 10.
When 0.1 mol CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to
1) 1:3 electrolyte
2) 1:2 electrolyte
3) 1:1 electrolyte
4) 3:1 electrolyte
Answer:
4) 3:1 electrolyte

Question 11.
Werner’s theory explains:
1. Hybridisation
2. Radioactivity
3. Coordination compounds
4. Electrolysis
Answer:
3. Coordination compounds

Question 12.
Primary valency corresponds to:
1. Coordination number
2. Oxidation state
3. Number of ligands
4. Shape
Answer:
2. Oxidation state

Question 13.
Secondary valency corresponds to: .
1. Oxidation number
2. Charge on ion
3. Coordination number
4. Atomic number
Answer:
3. Coordination number

Question 14.
Which of the following is a chelating ligand?
1. NH₃
2. H₂O
3. en
4. Cl^–
Answer:
3. en

Question 15.
Which of the following can show linkage isomerism?
1. NH₃
2. H₂O
3. NO₂^–
4. Cl^–
Answer:
3. NO₂^–

Question 16.
The compounds [Co(SO₄)(NH₃)₅]Br and [CO(SO₄)(NH₃)₅]Cl represent
1) linkage isomerism
2) ionisation isomerism
3) coordination isomerism .
4) no isomerism
Answer:
4) no isomerism

Question 17.
The oxidation state of Pt in [PtCl₆]^2- is:
1. +2
2. +3
3. +4
4. +6
Answer:
3. +4

Question 18.
The shape of [Co(NH₃)₆]³⁺ is:
1. Tetrahedral
2. Square planar
3. Octahedral
4. Linear
Answer:
3. Octahedral

Question 19.
Hemoglobin contains:
1. Cu
2. Mg
3. Fe
4. Co
Answer:
3. Fe

Question 20.
Chlorophyll contains:
1. Cu
2. Mg
3. Fe
4. Zn
Answer:
2. Mg

Question 21.
Which complex shows ionisation isomerism?
1. [Co(NH₃)₅SO₄]Br
2. [Ni(CO)₄]
3. [Zn(NH₃)₄]⁺
4. [Ag(NH₃)₃]⁺
Answer:
1. [Co(NH₃)₅SO₄]Br

Question 22.
Geometrical isomerism is shown by:
1. [Co(NH₃)₆]³⁺
2. [Pt(NH₃)₂ Cl₂]
3. [Zn(NH₃)₄]²⁺
4. [Ag(NH₃)₂]⁺
Answer:
2. [Pt(NH₃)₂ Cl₂]

Question 23.
The shape of [Ni(CO)₄] is:
1. Square planar
2. Tetrahedral
3. Octahedral
4. Linear
Answer:
2. Tetrahedral

Question 24.
The hybridisation in [Ni(CN)₄]^2- is:
1. sp₃
2. dsp₂
3. sp₂
4. d₂sp₃
Answer:
2. dsp₂

Question 25.
Which complex is diamagnetic?
1. [NiCl₄]^2-
2. [FeCl₄]^–
3. [Ni(CN)₄]^2-
4. [COF₆]^3-
Answer:
3. [Ni(CN)₄]^2-

Question 26.
Which ligand is a strong field ligand?
1. F^–
2. H₂O
3. Cl^–
4. CN^–
Answer:
4. CN^–

Question 27.
Which ligand is a weak field ligand?
1. CN^–
2. CO
3. F^–
4. en
Answer:
3. F^–

Question 28.
Colour in coordination compounds is mainly due to:
1. Proton transfer
2. d-d transitions
3. Nuclear charge
4. Radioactivity
Answer:
2. d-d transitions

Question 29.
The effective atomic number concept was proposed by:
1. Werner
2. Sidgwick
3. Pauling
4. Bohr
Answer:
2. Sidgwick

Question 30.
Which complex is used in electroplating?
1. [Ag(CN)₂]^–
2. Chlorophyll
3. Hemoglobin
4. Vitamin B₁₂
Answer:
1. [Ag(CN)₂]^–

Question 31.
The correct IUPAG name of [Pt(NH₃)₂Cl₂] is
1) Diamminedichloridoplatinum (II)
2) Diamminedichloridoplatinum (IV)
3) Diamminedichloridoplatinum (0)
4) Dichloridodiammineplatinum (IV)
Answer:
1) Diamminedichloridoplatinum (II)

Question 32.
Which complex has square planar geometry?
1. [Ni(CO)₄]
2. [NiCl₄]^2-
3. [Ni(CN)₄]^2-
4. [COF₆]^3-
Answer:
3. [Ni(CN)₄]^2-

Question 33.
The hybridisation of [CoF₆ ]^3- is:
1. dsp²
2. d²sp³
3. sp³
4. sp²
Answer:
2. d²sp³

Question 34.
Which complex is low spin?
1.[COF₆]^3-
2.[FeF₄]^3-
3. [Fe(CN)₂]^4-
4. [NiCl₄]^2-
Answer
3. [Fe(CN)₂]^4-

Question 35.
Which of the following is an organometallic compound?
1. K₄[Fe(CN)₄]
2. [Ni(CO)₄]
3. [CO(NH₃)₆] Cl₃
4. [CU(NH₃)₄]SO₄
Answer:
2. [Ni(CO)₄]

Question 36.
The ligand CO is:
1. Weak field ligand
2. Strong field ligand
3. Ambidentate ligand
4. Bidentate ligand
Answer:
2. Strong field ligand

Question 37.
Which among the following is paramagnetic?
1. [Ni(CN)₄]>]^2-
2. [Zn(NH₃)₄]>]²⁺
3. [COF₆]>]^3-
4. [PtCl₄]>]^2-
Answer:
3. [COF₆]>]^3-

Question 38.
Ethylene diaminetetraacetate (EDTA) ion is
1) tridentate ligand with three “N” donor atoms
2) hexadentate ligand with four “O” and two “N” donor atoms
3) unidentate ligand
4) bidentate ligand with two “N” donor atoms
Answer:
2) hexadentate ligand with four “O” and two “N” donor atoms

Question 39.
The complex ion present in Prussian blue is:
1. [Fe(CN)₆ ]^4-
2. [CU(NH₃)₄]²⁺
3. [Ni(CO)₄]
4. [CoCl₄]^2-
Answer:
1. [Fe(CN)₆ ]^4-

Question 40.
Which one is NOT a coordination compound?
1. K₄[Fe(CN)₆] ]
2. [Cu(NH₃)₄]SO₄
3. NaCl
4. [CO(NH₃)₆] Cl₃
Answer:
3. NaCl

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 2 Electrochemistry Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 2nd Lesson Electrochemistry Questions and Answers

I. Multiple Choice Questions

Question 1.
During electrolytic refining of metals the refined metal is obtained at (on) the [ ]
1) anode
2) cathode
3) surface of the electrolyte
4) Both 1 & 2
Answer:
2) cathode
In electrolytic refining, the impure metal is made anode and pure metal is deposited on cathode.
Oxidation at anode → metal ions go into solution → reduction at cathode → pure metal deposits.

Question 2.
A device that converts the energy of combustion of H₂ and CH₄ directly into electrical energy is known as [ ]
1) Electrolytic cell
2) Concentration cell
3) Leclanche cell
4) Fuel cell
Answer:
4) Fuel cell
A fuel cell is a device that converts combustion energy of fuel directly into electrical energy.

Question 3.
The electrode potential of a standard hydrogen electrode is
1) 0.0V
2) 1.0V
3) 2.0V
4) 3.0V
Answer:
1) 0.0V
Standard Hydrogen Electrode (SHE) is used as a reference electrode.
By convention, its potential is taken as zero for all calculations.

Question 4.
The standard cell potential of a Daniel cell is [ ]
1) 1.6 V
2) 1.9 V
3) 0.6 V
4) 1.1V
Answer:
4) 1.1V
Daniel Cell: Zn | Zn²⁺ || Cu²⁺| Cu
E° = \(\mathrm{E}_{\text {Cathode }}^0\) – \(\mathrm{E}_{\text {Anode }}^0\) = 0.34 – (-0.76) = 1.10V

Question 5.
Units of specific conductance are [ ]
1) S m
2) S m^-1
3) S^-1 m^-1
4) Ohm m
Answer:
2) S m^-1
Specific conductance (k) = conductance × cell constant. ⇒ S × \(\frac{1}{\mathrm{~m}}\)
SI unit = Siemens per meter. S/m

Question 6.
Consider the following cell reaction [ ]
Mg(s) + 2Ag⁺ (0.0001M) → Mg⁺² (0.1M) + 2 Ag(s)
of E°_cell is 3.17 V, the approximate value of Ecell will be
1) 2.96V
2) 3.52V
3) 4.11V
4) 8.61V
Answer:
1) 2.96V
The cell is represented as Mg | Mg²⁺ (0.130M) || Ag⁺(0.0001M)|Ag

Question 7.
The molar conductivity of a solution increases with [ ]
1) decrease in temperature
2) decrease in concentration
3) increase in concentration
4) decrease in ionic mobility
Answer:
2) decrease in concentration
Molar conductivity (\(\Lambda_{\mathrm{m}}\)) depends on mobility of ions.
On dilution → ions are far apart → less interaction → higher mobility.
So \(\Lambda_{\mathrm{m}}\)m increases as concentration decreases. \(\Lambda_{\mathrm{m}}\) = \(\frac{\mathrm{k} \times 1000}{\mathrm{M}}\) \(\Lambda_{\mathrm{m}}\) ∝ \(\frac{1}{\mathrm{M}}\)

Question 8.
In a dry cell which of the following is the electrolyte?
1) NH₄Cl + ZnCl₂
2) H₂SO₄ + NaOH
3) NaOH+KOH
4) NaOH+HCI
Answer:
1) NH₄Cl + ZnCl₂
Dry cell uses a paste electrolyte, not liquid.
Composition: NH₄Cl + ZnCl₂ + MnO₂ (depolariser)

Question 9.
Which of the following is correct about cathode in an electro chemical cell? [ ]
1) Oxidation occurs
2) Electrons loss occur
3) Reduction occurs
4) No reaction occurs
Answer:
3) Reduction occurs
At cathode:
Reduction = gain of electrons always occurs (in both electrolytic & galvanic cells).

Question 10.
During the electrolysis of aqueous NaCl solution using platinum electrodes, the product obtained at the cathode is
1) Cl₂
2) H₂
3) O₂
4) Na
Answer:
2) H₂
Ionisation: 2NaCl → 2Na⁺ + 2Cl
At anode: 2Cl^– → Cl₂ + 2e^–
At cathode: 2H₂O + 2e^– → 2OH^– + H₂↑
2Na⁺ + 2OH^– → 2NaOH

Question 11.
If a current of 0.5 amperes flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
1) 1.50 × 1021
2) 3.75 × 1023
3) 2.25 × 1022
4) 1.10 × 1023
Answer:
3) 2.25 × 1022
Given i = 0.5 amp, t = 2 × 60 × 60s
Q = it = 0.5 × 7200 = 3600C
96000C = 1 mole electrons = 6.022 × 10²³
3600C = ____ No.of electorns
∴ No.of electrons = \(\frac{3600 \times 6.022 \times 10^{23}}{96500}\) = 2.25 × 10²² electrons

Question 12.
A lead storage battery is mainly used in [ ]
1) watches
2) automobiles
3) inverters
4) Both 2 & 3
Answer:
4) Both 2 & 3
Lead storage battery is used where large current + rechargeability is needed.
Common uses: automobiles & inverters.

Question 13.
The standard electrode potential for Daniel cell is 1.1 V. The standard Gibbs energy for the following reaction is (IF = 96500 C mol^-1) []
Zn(s) + Cu²⁺(aq) → Zn²⁺ (aq) + Cu (s)
1) -212.3 kJ/mol
2) -444.4 kJ/mol
3) -101.3 kJ/mol
4) -323.8 kJ/mol
Answer:
1) -212.3 kJ/mol
Zn(s) + Cu²⁺(aq) → Zn²⁺ (aq) + Cu (s)
From above equation n = 2, F = 96500C, \(\mathrm{E}_{\text {Cell }}^0\) = 1.1V
\(\Delta \mathrm{G}_{\mathrm{F}}^0\) = -nF\(\mathrm{E}_{\text {Cell }}^0\)
Δ\(\mathrm{G}_{\mathrm{F}}^0\) = -2 × 96500 × 1.1 = -21200J.mole^-1 = -212.3 KJ.mole^-1

Question 14.
The electrical conductance of a metallic conductor is due to [ ]
1) flow of free electrons only
2) flow of ions only
3) Both by flow of electrons and ions
4) Neither by flow of electrons nor by ions
Answer:
1) flow of free electrons only
In metals, current is carried by free electrons, not ions.
No ionic movement in solid metals. Flow of free electrons only

Question 15.
Which of the following metals listed in the options is least reactive?
1) Fe
2) Mg
3) Au
4) Cu
Answer:
3) Au
Least reactive metal = most noble metal.
Reactivity order: Mg > Fe> Cu> Au Gold is least reactive.

II. Fill in the Blanks

Question 1.
1 Faraday value is equal to _____ C mol-1
Answer:
96487 (or) 96500

Question 2.
Corrosion of iron is mainly due to ____ and ____
Answer:
Corrosion of iron is mainly due to water and air

Question 3.
The units of molar conductivity are _____
Answer:
Scm² mol^-1 (or) S m²mol^-1

Question 4.
Lead storage battery is an example of ____ cell.
Answer:
secondary

Question 5.
In Daniell cell the oxidation half reaction occurs at ____
Answer:
anode

III. One Word Answer Questions

Question 1.
Write an expression showing the relation between E°_cell and Gibbs energy change.
Answer:
ΔG° = -nFE°_(cell) is the relation between E°_(cell) and Gibbs energy change.
Here n is the no. of electrons involved in oxidation (or) reduction, E°cell = standard EMF

Question 2.
Give an example of an inert electrode.
Answer:
Gold / Platinum / Graphite are the examples of an inert electrode.

Question 3.
What is the charge of an electron in coulombs?
Answer:
Charge of an electron in coulombs is -1.602 × 10^-19 coulombs

Question 4.
In a galvanic cell chemical energy is converted to electrical energy by which reaction?
Answer:
In a galvanic cell chemical energy is converted to electrical energy by Spontaneous redox reaction.

Question 5.
What happens to the colour of a solution when a zinc wire is dipped in CuSO₄ solution?
Answer:
Blue colour of CuSO₄ disappears when a zinc wire is dipped in CuSO₄ solution.

IV. Very Short Answer Questions

Question 1.
How is a galvanic cell represented on paper as per IUPAC convention? Give one example.
Answer:
As per IUPAC convention, a galvanic cell is represented as follows.
For the Daniel Cell: Zn/Zn⁺² (C₁)|| Cu⁺²(C₂)/Cu
The electrode on which oxidation takes place is written on the left hand side and the other electrode on which reduction takes place is written on the right hand side.
Oxidation electrode and reduction electrode are separated by double line which represents salt bridge. C₁ and C₂ represent concentrations of Zn⁺² and Cu⁺² ions respectively.

Question 2.
What is standard hydrogen electrode?
Answer:

• Standard hydrogen electrode is a reference electrode
• H₂ gas adsorbed on platinum at one atmosphere placed in contact with 1MHCl concentration in solution is treated as standard hydrogen electrode (SHE).
• Potential of this electrode is taken as zero.
• By using the standard hydrogen electrode (SHE) we can measure the potential of single electrode.

Question 3.
Write the Nernst equation for the EMF of the cell Ni(s) / Ni²⁺(aq) // Ag⁺(aq) / Ag(s)
Answer:

Question 4.
What is cell constant of a conductivity cell?
Answer:
Cell constant is the ratio of distance between electrodes and area of cross section. It is denoted by l/a. Its unit is Cm^-1 or m^-1.

Question 5.
Define molar conductivity \(\Lambda_m\) and how is it related to conductivity (k)?
Answer:
Molar conductivity is defined as the conductance of the solution containing one gram-mole of the electrolyte, when the electrodes are unit distance apart.
Molar conductivity \(\Lambda_m\) = \(\frac{\mathrm{k} \times 1000}{\mathrm{M}}\). Here k is conductivity, ‘M’ is molarity.

Question 6.
State Faraday’s first law of electrolysis.
Answer:
Faraday’s first law of electrolysis:
“The mass (m) of the substance deposited or liberated at the electrode is directly proportional to the quantity of electric charge (Q) passing through the electrolyte”.
Thus, m ∝ Q

Question 7.
State Faraday’s second law of electrolysis.
Answer:
Faraday’s second law of electrolysis:
“When same quantity of electric charge is passed through different electrolytes, connected in series the amount of different substances deposited at the electrodes is directly proportional to their equivalent masses”.
Mathematically, \(\frac{\mathrm{m}_1}{\mathrm{~m}_2}\) = \(\frac{E_1}{E_2}\)

Question 8.
What is a fuel cell? How is it different from a conventional galvanic cell?
Answer:
A fuel cell is a device that converts combustion energy of fuel directly into electrical energy. Conventional galvanic cell converts chemical energy of the spontaneous redox reaction into electrical energy.
Fuel cells relatively generate less pollution when compared to conventional galvanic cells.

Question 9.
What is metallic corrosion? Give one example.
Answer:
Metallic Corrosion: It is the process of slow eating up of metals by gases and water vapours present in atmosphere due to the formation of certain compounds like oxides, sulphides, carbonates, etc
Ex: Rusting of Iron (Fe converts into its oxide)

Question 10.
Give the electro-chemical reaction that represents the corrosion or rusting of iron.
Answer:
Corrosion reaction:
Anode: 2Fe → 2Fe⁺² + 4e^–
Cathode: O₂ + 4H⁺ + 4e^– → 2H₂O
Overall reaction: 2Fe + O₂ + 4H⁺ → 2Fe⁺² + 2H₂O

V. Short Answer Questions

Question 1.
What are galvanic cells? Explain the working of a galvanic cell with a neat sketch taking Daniel cell as example.
Answer:
Galvanic cells convert chemical energy into electrical energy.
They involve in spontaneous redox reactions.

Construction:
Daniel cell consists of two half cells

1. anode half cell and
2. cathode half cell.

The two half cells are connected by a salt bridge containing a saturated solution of KNO₃ in agar-agar gel. The anode half cell consists of zinc plate dipped in ZnSO₄ solution and the cathode half cell consists of a copper plate dipped in CuSO₄ solution.

Working: When the two half cells are connected externally through a voltmeter, current flows out from the cell due to potential difference.
Electrode reactions:
At anode : Zn → Zn⁺² + 2e ̄ (oxidation)
At cathode: Cu⁺² + 2e^– → Cu (reduction)
Net cell reaction: Zn_(s) + \(\mathrm{Cu}^{+2}{ }_{(\mathrm{aq})}\) → \(\mathrm{Zn}^{+2}(\mathrm{aq})\) + Cu_(s)

Question 2.
State and explain Nernst equation with the help of a metallic electrode and a non-metallic electrode.
Answer:
Nernst Equation: The equation that gives the dependence of the electrode potential on the concentration of ions with which electrode is reversible is known as Nernst equation.
Nernst Equation: Electrode potential E = E° – \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}}\)log\(\frac{\text { [Products] }}{\text { [Reactants] }}\)
Where E° = Standard reduction potential of the cell at 25°C.
n = Number of electrons involved in cell reaction.
R = Gas constant R (8.314 J K^-1 mol^-1);
T = Temperature = 273 + 25 = 298K;
F = Faraday constant = 96,487 ≈ 96500 c mol^-1;

(a) For Metal Electrode:

(b) For Non-metal Electrode:

Question 3.
On what factors the electrical conductance of an aqueous solution of electrolyte depends?
Answer:
The conductance of the solution of an electrolyte depends on the following factors

1. Nature of the electrolyte: Strong electrolytes ionize almost completely in the solution and hence conduct electricity to a large extent. Whereas weak electrolytes ionise to a small extent and hence conduct electricity to a small extent.
2. Size of the ions: Greater the size of the ions or greater the solvation of the ions, lesser is the conductance.
3. Nature of the solvent and its viscosity: Electrolytes ionise more in a polar solvent. Hence, greater the polarity of the solvent, greater is the ionization and hence greater is the electrical conductance. Also greater the viscosity of a solvent, lesser is the conductance.
4. Concentration of the solution: Higher the concentration of the solution, lesser is the conductance. This is because in a weak electrolyte, the ionization is less whereas in a strong electrolyte, the interionic attractions are higher at higher concentration.
   With dilution, incase of weak électrolytes ionization increases and incase of strong electrolytes, interionic attraction decreases and hence conductance increases.
5. Temperature: On increasing the temperature, the dissociation of electrolyte increases and hence the conductance increases.

Question 4.
State and explain Kohlrausch’s law of independent migration of ions.
Answer:
Kohlrausch law of independent migration of ions states that “limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte, at infinite dilution.”

Thus, if \(\lambda_{\mathrm{Na}^{+}}^0\) and \(\lambda^0 \mathrm{Cl}^{-}\) are the limiting molar conductivities of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation
\(\Lambda_{(\mathrm{NaCl})}^0\) = \(\lambda_{\mathrm{Na}^{+}}^0\) + \(\lambda_{\mathrm{C} l^{-}}^0\)

In general, if an electrolyte on dissociation gives V₊ cations and V_– anion then its limiting molar conductivity is given by
\(\Lambda_{\mathrm{m}}^{\circ}\) = \(V_{+} \lambda_{+}^0\) + \(\mathrm{V}_{-} \lambda_{-}^0\)
Here, \(\lambda_{+}^0\) and \(\lambda_{-}^0\) are the limiting molar conductivities of the cation and anion respectively.

Question 5.
What are the products obtained at the cathode and the anode during the electrolysis of the following when platinum electrodes are used in the electrolysis
(a) Molten KCl
(b) Aq. CuSO₄ solution
(c) Aq.K₂SO₄ solution
Answer:

Question 6.
What are fuel cells? How are they different from galvanic cells? Give the construction of H₂-O₂ fuel cell.
Answer:
A fuel cell is a device that converts the energy from combustion of fuel into electrical energy. The fuel cell like any other electrochemical cell has two electrodes and an electrolyte.
Oxidation of fuel occurs at the anode: Fuel → oxidation product + ne^–

The oxidant gets reduced at the cathode: Oxidation + ne^– → Reduction product. Fuel cells are different from conventional galvanic cells in the following aspects:

a) In a fuel cell, reactants are fed into the cell and products are removed from the cell. These do not form integral part of cell as in a conventional galvanic cell.
b) Chemical energy is not stored in a fuel cell, whereas in a galvanic cell chemical energy is converted into electrical energy.

H₂-O₂ fuel cell: This fuel cell consists of porous carbon electrodes suspended in concentrated NaOH solution.

Catalysts like finely divided platinum or palladium is incorporated into the electrodes for increasing the rate of electrode reactions.
H₂ and O₂ gases are bubbled at the surface of the electrodes.

The electrode reactions:
At cathode: O_2(g) + 2H₂O + 4e ^– → 4\(\mathrm{H}_{\mathrm{aq}}^{-}\)
At anode: 2H_2(g) + 4O\(\mathrm{H}_{\mathrm{aq}}^{-}\) → 4H₂O_(Cl) + 4e^–
Overall reaction: 2H_2(g) + O_2(g) → 2H₂O_(Cl)

Question 7.
What is metallic corrosion? Explain it with respect to iron corrosion.
Answer:

Metallic corrosion: It is the process of slow eating up of metals by gases and water vapours present in atmosphere due to the formation of certain compounds like oxides, sulphides, carbonates,etc.,

Corrosion of Iron is known as Rusting. Rusting is an electrochemical process that occurs when iron is exposed to oxygen and moisture.

Electrochemical Mechanism:

1. Formation of Anodic and Cathodic Regions
2. On the iron surface, tiny regions act as: “Anode ? oxidation occurs “Cathode ? reduction occurs

The corrosion process is electrochemical in nature.
On the iron surface, iron becomes oxidized in the presence of water.

On the metal surface, oxidation takes places at anode and reduction takes place at cathode.

At anode: Fe_(s) → \(\mathrm{Fe}_{\mathrm{aq}}^{+2}\) + 2e^– (Iron atoms lose electrons and go into solution)
At cathode: \(\frac{1}{2} \mathrm{O}_{2(\mathrm{aq})}\) + H₂O_(l) + 2e^– → 2O\(\mathrm{H}_{\mathrm{aq}}^{-}\) (Oxygen reacts with water to form hydroxide ions)
Overall reaction: Fe_(s) + \(\frac{1}{2} \mathrm{O}_{2(\mathrm{aq})}\) + H₂O_(l) → \(\mathrm{Fe}_{\mathrm{aq}}^{+2}\) + 2O\(\mathrm{H}_{\mathrm{aq}}^{-}\)

Formation of Rust: Ferrous ions react with hydroxide ions
\(\mathrm{Fe}_{\mathrm{aq}}^{+2}\) + 2OH^– → Fe(OH)₂
On further oxidation Fe(OH)₂ → Fe(OH)₃
The Ferrous ions finally form hydrated ferric oxide Fe₂O₃.xH₂O (rust).

Question 8.
Explain the variation of molar conductivity with the change in the concentration of the electrolyte. Give reasons.
Answer:
Molar conductivity of electrolytes generally, increases with dilution.

(i) For Strong Electrolytes (NaCl, KCl):
\(\Lambda^0{ }_{\mathrm{m}}\) increases slightly with dilution.
The increase is gradual and linear

Reason:
Strong electrolytes are almost completely dissociated even at higher concentrations
On dilution:
Inter-ionic attractions decrease, Ion mobility increases
Hence, conductivity increases slightly

(ii) For Weak Electrolytes (CH₃COOH, NH₄OH):
\(\Lambda^0{ }_{\mathrm{m}}\) increases sharply with dilution
Reason:
Weak electrolytes are partially ionised
On dilution:
Degree of ionisation increases (Le Chatelier’s principle)
More ions are produced. Hence, conductivity increases rapidly.
Graphical representation of the variation of \(\Lambda_m\) vs \(\sqrt{\mathrm{C}}\)
For a strong electrolyte (KCl) and a weak electrolyte (CH₃COOH) is shown in the graph.
Strong electrolyte – straight line (small increase)
Weak electrolyte – curved line (steep increase)

Question 9.
Explain with a suitable example the relation between the Gibbs energy of chemical reaction (G) and the functioning of the electrochemical cell.
Answer:
Electrical work done in one second in an electrochemical cell is equal to decrease in its Gibbs energy. Electrical work done decrease in free energy.
In the case of an electrochemical cell, the electrical work done is equal to the electrical energy produced. This in turn is equal to the product of the quantity of electricity flowing in the circuit and the EMF of the cell.
Now, for every one mole of electrons transferred in the cell reaction, the quantity of electricity that flows through the cell is one faraday (1F = 96,500 Coulomb).
Hence, if n moles of electrons are transferred, in any cell reaction, the quantity of electricity flowing = nF Faraday
If Ecell represents the EMF of the cell, then electrical work done = nF E_cell
Hence -Δ_rG = nF E_cell
For comparing different cells, standard cell potentials are used which are represented by E⁰_cell.
The corresponding free energy change is called the standard change of the reaction, represented by Δ_rG⁰.
Hence we can write -ΔG⁰ = nF \(\mathrm{E}^0{ }_{\text {cell }}\)
Thus for a cell reaction Zn(s) + \(\mathrm{Cu}^{+2}(\mathrm{aq})\) → \(\mathrm{Zn}^{+2}(\mathrm{aq})\) + Cu_(s)
Δ_rG⁰ = -2F \(\mathrm{E}^0{ }_{\text {cell }}\) [∵ n=2= no.of electrons involved]

Question 10.
Give the construction and working of a standard hydrogen electrode with a neat diagram.
Answer:
Standard hydrogen electrode (SHE):
It is a reference electrode and its reduction potential is arbitrarily assigned as zero volt at all temperatures.

By using ‘SHE’ we can measure the potential of single electrode.
H₂ gas adsorbed on platinum at one atmosphere placed in contact with 1M H⁺ concentration in solution is treated as standard hydrogen electrode (SHE).
Potential of this electrode is taken as zéro.
Hydrogen electrode is represented as Pt, H_2(g) 1 atm / \(\mathrm{H}_{(\mathrm{aq})}^{+}\) (C = 1)

VI. Long Answer Questions

Question 1.
What are electrochemical cells? How are they constructed? Explain the working of the galvanic cells. Write any three differences between electrochemical cells and electrolytic cells.
Answer:
Electrochemical cells(Galvanic cells): They convert chemical energy into electrical energy.
They involve in spontaneous redox reactions. Construction: Daniel cell consists of two half cells
i) anode half cell and
ii) cathode half cell.

The two half cells are connected by a salt bridge containing a saturated solution of KNO₃ in agar-agar gel.
The anode half cell consists of zinc plate dipped in ZnSO₄ solution and the cathode half cell consists of a copper plate dipped in CuSO₄ solution.

Working: When the two half cells are connected externally through a voltmeter, current flows out from the cell due to potential difference.
Electrode reactions:
At anode: Zn → Zn⁺² + 2e^– (oxidation)
At cathode: Cu⁺² + 2e^– → Cu (reduction)
Net cell reaction:Zn_(s) + Cu⁺²_(aq) → Zn⁺²_(aq) + Cu_(s)

Differences between electrochemical cells and electrolytic cells:

Question 2.
Give the different types of batteries and explain the construction and working of each type of battery.
Answer:

Batteries are of two types:
1) Primary Batteries
2) Secondary Batteries

1) Primary Batteries (Non- Rechargeable) :
(a) Dry cell:
Construction: The cell consists of a zinc container which acts as the anode.
The cathode is a carbon rod (graphite) surrounded by powdered manganese dioxide and carbon.
The electrolyte is a moist paste of ammonium chloride
(NH₄Cl) and zinc chloride (ZnCl₂).
Electrode reactions:
Anode: Zn_(s) → Zn⁺² + 2e^–
Cathode: MnO₂ + \(\mathrm{NH}_4^{+}\) + e^– → MnO(OH) + NH₃
The cell exhibits a potential of nearly 1.5 V.

(b) Mercury cell:

It consists of zinc-mercury amalgam as anode and a mixture of HgÒ and carbon as the cathode. The electrolyte is a paste of KOH and ZnO.
Electrode reactions:
Anode: Zn(Hg) + 2OH^– → ZnO_(s) + H₂O + 2e^–
Cathode: HgO+ H₂O + 2e^– → Hg_(l) + 2OH^–
The overall reaction is represented by Zn_(Hg) + HgO_(s) → ZnO_(s) + Hg_(l)
The cell potential is approximately 1.35 V and it remains constant during its life.

2) Secondary Batteries (Rechargeable) :
(c) Lead storage battery:
It consists of a lead anode and a grid of lead packed with lead dioxide (PbO₂) as cathode. A 38% solution of sulphuric acid is used as electrolyte.

The cell reactions :
Anode: Pb_(s) + \(\mathrm{SO}_{4(\mathrm{aq})}^{-2}\) → PbSO_4(s) + 2e^–
Cathode: \(\mathrm{PbO}_{2(\mathrm{~s})}\) + \(\mathrm{SO}_{4(\mathrm{aq})}^{-2}\) + \(4 \mathrm{H}_{\mathrm{aq}}^{+}\) + 2e^– → PbSO_4(s) + 2H₂O_(l)
These reactions occur during discharge, that is during use of the battery.
This is commonly used in automobiles and inverters.

(d) Nickel Cadmium Battery:

A rechargeable nickel-cadmium cell in a jelly roll arrangement and separated by a layer soaked in moist sodium or potassium hydroxide

Construction:
It is a jelly roll arrangement and separated by a layer soaked in moist sodium or potassium hydroxide.
Anode: Cadmium
Cathode: Nickel oxide hydroxide (NiO(OH))
Electrolyte: KOH
Working: Overall reaction:
Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)
These batteries have longer life than the lead storage cell.

Question 3.
Calculate the concentration of silver ions in the cell constructed by using 0.1M concentration of Cu²⁺ and Ag⁺ ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V. [\(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0\) = +0.80V; \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = +0.34V]
Solution:
At anode : Cu → Cu⁺² + 2e^–
At cathode: 2Ag⁺ + 2e → 2Ag
Cell reaction: Cu + 2Ag⁺ → Cu⁺² + 2Ag
\(\mathrm{E}_{\text {cell }}^0\) = \(\mathrm{E}_{\text {cathode }}^0\) – \(\mathrm{E}_{\text {anode }}^0\) = 0.80 – 0.34 = +0.46V
Given E_cell = 0.422 V; [Cu⁺²] = 0.1 M
Applying Nernst equation

Question 4.
Calculate the degree of dissociation (α) of CH₃COOH at 298 K
Given that \(\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^0\) = 11.75cm² mol^-1
Answer:
Given \(\Lambda_{\mathrm{m}}^0\left(\mathrm{CH}_3 \mathrm{COOH}\right)\) = \(\Lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0\) + \(\lambda_{\mathrm{H}^{+}}^0\) = 40.95 + 349.15 = 390.1
Formula: Degree of dissociation α = \(\frac{\Lambda_m^c}{\Lambda_m^0}\) = \(\frac{11.75}{390.1}\) = 0.03012 = 3.012 × 10^-2

Question 5.
a) State and explain Kohlrausch’s law of independent migration of ions along with two applications.
b) Define molar conductivity and limiting molar conductivity.
Answer:
(a) Kohlrausch’s law of independent migration of ions:
Kohlrausch law of independent migration of ions states that “limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte, at infinite dilution.”
Thus, if \(\lambda_{\mathrm{Na}^{+}}^0\) and \(\lambda_{\mathrm{C} l^{-}}^0\) are the limiting molar conductivities of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation.
\(\Lambda_{(\mathrm{NaCl})}^{\circ}\) = \(\lambda_{\mathrm{Na}^0}^0\) + \(\lambda_{\mathrm{Cl}^{-}}^0\)
In general, if an electrolyte on dissociation gives V₊ cations and V_– anion then its limiting molar conductivity is given by \(\Lambda_{\mathrm{m}}^{\mathrm{o}}\) = \(V_{+} \lambda_{+}^0\) + \(V_{-} \lambda_{-}^0\)
Here, \(\lambda_{+}^0\) and \(\lambda_{-}^0\) are the limiting molar conductivities of the cation and anion respectively.

Applications of Kohlarusch’s law:

(i) Calculation of molar conductivities of weak electrolyte at infinite dilution:
For example molar conductivity of acetic acid (CH₃COOH) at infinite dilution can be obtained from molar conductivities at infinite dilution of strong electrolytes like CH₃COONa, HCl and NaCl as shown below:

(ii) Determination of degree of dissociation of weak electrolytes:
Degree of dissociation (α) = \(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\mathrm{o}}}\)

(iii) Determination of dissociation constant (K) of weak electrolytes: K = \(\frac{C \alpha^2}{1-\alpha}\)

b) Molar conductivity and limiting molar conductivity:
Molar conductivity is defined as the conductance of the solution containing one gram- mole of the electrolyte when the electrodes are unit distance apart having sufficient area of cross section to hold the electrolyte.
\(\Lambda_{\mathrm{m}}\) = \(\frac{\mathrm{k} \times 1000}{\mathrm{M}}\)
Where k is conductivity, ‘M’ is molarity and \(\Lambda_{\mathrm{m}}\) is molar conductivity
Limiting molar conductivity \(\left(\Lambda_{\mathrm{m}}^0\right)\) is the molar conductivity and infinite dilution.
It is defined as the molar conductivity of any electrolyte when the concentration of electrolyte approaches zero (i.e., at infinite dilution)
Mathematically, \(\Lambda_{\mathrm{m}}^0\) = \(\lim _{\mathrm{c} \rightarrow 0} \boldsymbol{\Lambda}_{\mathrm{m}}\) Here, \(\boldsymbol{\Lambda}_{\mathbf{m}}\) is molar conductivity at concentration C.

Textual Solved Problems

Question 1.
Represent the cell in which the following reaction takes place Mg(s) + 2Ag⁺(0.0001 M) → Mg2⁺(0.130M) + 2Ag(s)
Calculate its E_(cell) if E^–_(cell)= 3.17 V.
Answer:
The cell is represented as Mg | Mg²⁺ (0.130M) || Ag⁺(0.0001M)|Ag
E_(cell) = E^–_(cell) – \(\frac{\mathrm{RT}}{2 \mathrm{~F}} \ln \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}\) = 3.17 V – \(\frac{0.059 \mathrm{~V}}{2}\)log\(\frac{0.130}{(0.0001)^2}\) = 3.17V – 0.21V = 2.96 V.

Question 2.
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) \(\mathrm{E}_{\text {(cell) }}^0\) = 0.46 V
Answer:
From the above equation n = 2, \(\mathrm{E}_{\text {(cell) }}^0\) = 0.46 V
\(\mathrm{E}_{(\text {cell })}^0\) = \(\frac{0.059 \mathrm{~V}}{2}\)logK_c = 0.46
0.46 V = \(\frac{0.059}{2}\)log K_c ⇒ logK_c = \(\frac{0.46 \times 2}{0.059}\) = 15.6 ⇒ K_c = _e15.6 = 3.92 × 10¹⁵.

Question 3.
The standard emf of Daniel cell is 1.1V. Calculate the standard Gibbs energy for the cell reaction:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Answer:
Zn(s) + Cu²⁺(aq) → Zn²⁺ (aq) + Cu (s)
From above equation n = 2, F = 96500C, \(\mathrm{E}_{\text {Cell }}^0\) = 1.1 V.
∆\(G_F^g\) = -nF\(\mathrm{E}_{\text {Cell }}^0\)
∆\(\mathrm{G}_{\mathrm{F}}^{\mathrm{O}}\) = -2 × 96500 × 1.1 = -21200J.mole^-1 = -212.3 KJ.mole^-1

Question 4.
Calculate \(\Lambda_{\mathrm{m}}^0\) for CaCl₂ and MgSO₄ from the given data Ca²⁺ = 119.08 cm² mol^-1. Cl = 76.38 cm² mol^-1 Mg²⁺ = 106.0 S cm² mol^-1 \(\mathrm{SO}_4^{2-}\) = 160.0S cm² mol^-1
Solution:
We know from Kohlrausch law that

Question 5.
\(\Lambda_m^0\) for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm²mol^-1 respectively. Calculate \(\Lambda^0\) for HAc.
Answer:

Question 6.
A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Answer:
Given time t = 10 min. = 600s, current i = 1.5 amperes
charge = current × time = 1.5 × 600 = 900C
According to the reaction: Cu²⁺(aq) + 2e^– = Cu(s)
We require 2F or 2 × 96500 C deposit 1 mole or 63 g of Cu.
For 900C, the mass of Cu deposited = (63 ×900 )/(2 × 96500 ) = 0.2938 g

Objective Questions

Question 1.
The quantity of charge required to obtain one mole of aluminium from Al₂O₃ is ____
1) IF
2) 6F
3) 3F
4) 2F
Answer:
3) 3F

Question 2.
The weight of silver (at.wt= 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be
1) 5.4 g
2) 10.8 g
3) 54.0 g
4) 108.0 g
Answer:
4) 108.0 g

Question 3.
The number of Faradays (F) required to produce 20g of Calcium from molten CaCl₂ (Atomic mass of Ca = 40gmol^-1) is
1) 1
2) 2
3) 3
4) 4
Answer:
1) 1

Question 4.
When 0.1 mol of \(\mathrm{MnO}_4^{2-}\) is oxidised the quantity of electricity required to completely oxidise Mn\(\mathrm{O}_4^{2-}\) to Mn\(\mathrm{O}_4^{-}\) is
1) 96500C
2) 2 × 96500C
3) 9650C
4) 96.50C
Answer:
3) 9650C

Question 5.
Which of the following statement is not correct about an inert electrode in a cell?
1) It does not participate in the cell reaction.
2) It provides surface either for oxidation or for reduction reaction.
3) It provides surface for conduction of electrons.
4) It provides surface for redox reaction.
Answer:
4) It provides surface for redox reaction.

Question 6.
The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ____
1) Cell potential
2) Cell emf
3) Potential difference
4) Cell voltage
Answer:
2) Cell emf

Question 7.
Which cell will measure standard electrode potential of copper electrode?
1) Pt(S)|H₂ (g,0. 1 bar)| H⁺(aq, 1 M) || Cu²⁺(aq., 1 M)|Cu
2) Pt(s)|H₂(g, 1 bar)H⁺(aq.,1 M)||Cu²⁺(aq., 2M)|Cu
3) Pt(s)|H₂(g,1 bar)H₊ (aq, 1M)||Cu²⁺(aq,1M)Cu
4) Pt(s)|H₂ (g, 1 bar 1MjCu(aq., 1 M)IC u
Answer:
3) Pt(s)|H₂(g,1 bar)H₊ (aq, 1M)||Cu²⁺(aq,1M)Cu

Question 8.
An electrochemical cell can behave like an electrolytic cell when___
1) E_cell=0
2) E_cell > E_ext
3) E_ext > E_cell
4) E_ext = E_ext
Answer:
3) E_ext > E_cell

Question 9.
Using the data given below find out the strongest reducing agent.

1) Cl^–
2) Cr
3) Cr³⁺
4) Mn²⁺
Answer:
2) Cr

Question 10.
Standard electrode potential of three metals X, Y, and Z are -1.2 V, + 0.5 V and -3.0 V respectively. The reducing power of these metals will be
1) X > Y > Z
2) Y > Z > X
3) Y > X > Z
4) Z > X > Y
Answer:
4) Z > X > Y

Question 11.
For the cell reaction:
2Fe³⁺ (aq) + 2l^– (aq) → 2Fe²⁺ (aq) + I₂(aq)
\(\mathrm{E}_{\text {cell }}^0\) = 0.24V at 298 K. The standard Gibbs’ energy (∆_rG°) of the cell reaction is (given that F = 96500 Cmol^-1)
1) 23.16 kJmol^-1
2) -46.32 kJ mol^-1
3) -23. 16kJ mol^-1
4) 46.32 kJ mol^-1
Answer:
2) -46.32 kJ mol^-1

Question 12.
For the reduction of silver ions with copper metal, the standard cell potential is 0.46 V at 25°C. The value of standard Gibbs energy, ∆G° will be
1) -89.0 kJ
2) -89.0 J
3) -44.5 kJ
4) -98.0 kJ
Answer:
1) -89.0 kJ

Question 13.
If the \(\mathrm{E}_{\text {Cell }}^O\) for a given reaction has a negative value, then which of the following gives the correct relationship for the values of ∆G° and K_eq?
1) ∆G° > 0; K_eq < 1 2) ∆G° > 0; K_eq > 1
3) ∆G° < 0; K_eq > 1
4) ∆G°< 0; K_eg < 1 Answer: 1) ∆G° > 0; K_eq < 1

Question 14.
For a cell involving one electron \(E_{\text {Cell }}^0\) = 0.59V at 298K, the equilibrium constant for the cell reaction is given that \(\frac{2.303 \mathrm{RT}}{\mathrm{~F}}\) 0.059V at T = 298K
1) 1.0 × 10³⁰
2) 1.0 × 10²
3) 1.0 × 10⁵
4) 1.0 × 10¹⁰
Answer:
4) 1.0 × 10¹⁰

Question 15.
The cell constant of a conductivity cell ___
1) changes with change of electrolyte.
2) changes with change of concentration of electrolyte.
3) changes with temperature of electrolyte.
4) remains constant for a cell.
Answer:
4) remains constant for a cell.

Question 16.
\(\Lambda_{\mathrm{m}\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\mathrm{O}}\) is equal to ____

Answer:

Question 17.
The molar conductance of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm²mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is
1) 540.48 Scm²mol^-1
2) 201.28 Scm²mol^-1
3) 390.71 Scm²mol^-1
4) 698.28 Scm²mol^-1
Answer:
3) 390.71 Scm²mol^-1

Question 18.
The molar conductivity of 0.007M acetic acid is 20Scm²mol^-1. What is the dissociation constant of acetic acid?
\(\Lambda_{\mathrm{H}}^{\circ}\) = 350 S cm²mol^-1‚ \(\Lambda_{\mathrm{CH}, \mathrm{COO}}^0\)
1) 2.50 × 10^-5 mol L^-1
2) 1.75 × 10^-4 mol L^-1
3) 2.50 × 10^-4 mol L^-1
4) 1.75 × 10^-5 mol L^-1
Answer:
4) 1.75 × 10^-5 mol L^-1

Question 19.
While charging the lead storage battery
1) PbSO₄ anode is reduced to Pb
2) PbSO₄ cathode is reduced to Pb
3) PbSO₄ cathode is oxidised to Pb
4) PbSO₄ anode is oxidised to PbO₄
Answer:
1) PbSO₄ anode is reduced to Pb

AP Intermediate 2nd Year Physics Textbook Solutions

Physics Inter 2nd Year

• Chapter 1 Electric Charges and Fields Questions and Answers
• Chapter 2 Electrostatic Potential and Capacitance Questions and Answers
• Chapter 3 Current Electricity Questions and Answers
• Chapter 4 Moving Charges and Magnetism Questions and Answers
• Chapter 5 Magnetism and Matter Questions and Answers
• Chapter 6 Electromagnetic Induction Questions and Answers
• Chapter 7 Alternating Currents Questions and Answers
• Chapter 8 Electromagnetic Waves Questions and Answers
• Chapter 9 Ray Optics and Optical Instruments Questions and Answers
• Chapter 10 Wave Optics Questions and Answers
• Chapter 11 Dual Nature of Radiation and Matter Questions and Answers
• Chapter 12 Atoms Questions and Answers
• Chapter 13 Nuclei Questions and Answers
• Chapter 14 Semiconductor Devices Questions and Answers

AP Inter 2nd Year Physics Weightage BluePrint 2026-2027

AP Inter 2nd Year Study Material

Regular practice with AP Inter 2nd Year Chemistry Study Material Chapter 6 Haloalkanes and Haloarenes Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Chemistry 6th Lesson Haloalkanes and Haloarenes Questions and Answers

I. Multiple Choice Questions

Question 1.
What is the general formula for haloalkanes? (X=haIogen atom, n = 1, 2, 3…)
1) C_nH_2nX
2) C_nH_2n+1X
3) C_nH_2n-1X
4) C_nH_2n-3X
Answer:
2) C_nH_2n+1X
Haloalkanes are derived from alkanes by replacing one H with X.
Alkane formula = C_nH_2n+2⟶ replace one H ⟶ ?C_nH_2n+1X

Question 2.
Which of the following reagents is used for replacement of -OH group in an alcohol by Cl group?
1) SO₂Cl₂
2) AqueousHCl
3) PCl₃, SOCl₂
4) Nacl
Answer:
3) PCl₃, SOCl₂
Converting alcohol (-OH) ⟶ alkyl chloride (-Cl) needs chlorinating agents.
Best reagents: PCl₃, SOCl₂
3R-OH +PCl₃⟶ 3R-Cl + H₃PO₃
R-CH + SOCl₂ ⟶ R-Cl + SC₂ + HCl

Question 3.
Identify the compound Y in the following reaction sequence

Answer:

Reaction sequence: Aniline ⟶ diazonium salt ⟶ Sandmeyer reaction (Cu₂Cl₂/ HCl).
Gives chlorobenzene.

Question 4.
A primary alkyl halide would prefer to undergo
1) S_N¹ reaction
2) S_N² reaction
3) α-elimination
4) racemization
Answer:
2) S_N² reaction
Primary alkyl halides favor backside attack (less steric hindrance).So S_N² dominates.

Question 5.
Which of the following is the suitable medium for preparing Grignard reagents?
1) Dry acetone
2) Dry ether
3) Concentrated HCl
4) Alcoholic KOH
Answer:
2) Dry ether
Grignard reagents are destroyed by water/protic solvents.
Need dry, aprotic medium → Dry ether

Question 6.
Compared to haloalkanes, the reactivity of haloarenes towards nucleophilic substitution reactions is
1) low
2) high
3) very high
4) equal
Answer:
1) low
Haloarenes are less reactive due to:resonance stabilization partial double bond character So substitution is difficult.

Question 7.
Chlorobenzene is formed by the reaction of chlorine with benzene in the presence of anhydrous AlCl₃. Which of the following species attacks the benzene ring in this reaction?
1) Cl^–
2) Cl⁺
3) AlCl₃
4) [AlCl₄]^-1
Answer:
2) Cl⁺
Electrophilic substitution → attacking species must be electrophile,
Cl⁺ is the actual attacking species.

Question 8.
Arrange the following compounds in increasing order of their boiling points.

1) (b) < (a) < (c)
2) (a) < (b) < (c)
3) (c) < (a) < (b)
4) (c) < (b) < (a)
Answer:
3) (c) < (a) < (b)
Boiling point depends on: molecular mass branching (more branching → lower BP)
Arrange accordingly. Answer: (c) < (a) < (b)

Question 9.
Which of the following alky! halide will undergo S_N¹ reaction most readily?
1) (CH₃)₃C—F
2) (CH₃)₃C—Cl
3) (CH₃)₃C—Br
4) (CH₃)₃C—I
Answer:
4) (CH₃)₃C—I
S_N¹ favors tertiary carbocation stability. (CH₃)₃C-I forms most stable carbocation + best leaving group.

Question 10.
Which reagent will you use for the following reaction?
CH₃CH₂CH₂CH₃ ⟶ CH₃CH₂CH₂CH₂CI + CH₃CH₃CHCICH₃
1) Cl₂/UV light
2) NaCl + H₂SO₄
3) Cl₂ gas in dark
4) Cl₂ gas in the presence of iron in dark
Answer:
1) Cl₂/UV light
Reaction shows free radical substitution → needs UV light
Chlorination of alkanes occurs under UV.

Question 11.
Which of the following represents the correct name of the compound, CH₃CHCl₂?
1) Ethylidene chloride
2) Ethylene dichloride
3) 1,2-Dichloroethane
4) Vic-dichloride
Answer:
1) Ethylidene chloride
CH₃-CHCl₂ has both Cl atoms on the same carbon (geminal dihalide),
Common name: Ethylidene chloride (not ethylene dichloride),

Question 12.
In alkyl halides, carbon-halogen bond is polarized due to
1) greater electronegativity of halogen atom than carbon
2) greater electronegativity of carbon than halogen atom
3) partial positive charge on halogen atom and a partial negative charge on carbon
4) tendency of halogen atom to donate electrons
Answer:
1) greater electronegativity of halogen atom than carbon
Halogens are more electronegative than carbon.
So they pull electron density → C becomes δ⁺ and X becomes δ^–.
Greater electronegativity of halogen atom than carbon

Question 13.
Chlorobenzene is
1) less reactive than benzyl chloride
2) more reactive than ethyl bromide
3) nearly as reactive as methyl chloride
4) more reactive than isopropyl chloride
Answer:
1) less reactive than benzyl chloride
Chlorobenzene is less reactive due to: resonance stabilization, partial double bond character of C-Cl.
So it reacts slower than alkyl halides. Less reactive than benzyl chloride