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

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