CLASS 12 · CHEMISTRY · CHAPTER 5

Coordination Compounds

Simple answers to ten in-text question groups, all twenty-eight numbered exercises, and the three extra questions printed in the current NCERT chapter.

41 worked answersComplete chapter answers

Read the NCERT questions ↗← All chapters

41 worked answers · short steps first, explanations when you need them

In-text Questions

Textbook pages 120–136

Question 5.1

Write formulas for: (i) tetraamminediaquacobalt(III) chloride; (ii) potassium tetracyanidonickelate(II); (iii) tris(ethane-1,2-diamine)chromium(III) chloride; (iv) amminebromidochloridonitrito-N-platinate(II); (v) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate; (vi) iron(III) hexacyanidoferrate(II).

Steps only

Key charge-balanced formulas: [Co(NH₃)₄(H₂O)₂]Cl₃, K₂[Ni(CN)₄], [Cr(en)₃]Cl₃, [PtCl₂(en)₂](NO₃)₂, Fe₄[Fe(CN)₆]₃.

Answer

(i) [Co(NH₃)₄(H₂O)₂]Cl₃;
(ii) K₂[Ni(CN)₄];
(iii) [Cr(en)₃]Cl₃;
(iv) [Pt(NH₃)(Br)(Cl)(NO₂)]⁻ (as the platinate

(II) complex ion); (v) [PtCl₂(en)₂](NO₃)₂; (vi) Fe₄[Fe(CN)₆]₃.

Question 5.2

Give IUPAC names for: (i) [Co(NH₃)₆]Cl₃; (ii) [Co(NH₃)₅Cl]Cl₂; (iii) K₃[Fe(CN)₆]; (iv) K₃[Fe(C₂O₄)₃]; (v) K₂[PdCl₄]; (vi) [Pt(NH₃)₂Cl(CH₃NH₂)]Cl.

Steps only

(i) Hexaamminecobalt(III) chloride; (ii) pentaamminechloridocobalt(III) chloride; (iii) potassium hexacyanidoferrate(III); (iv) potassium tris(oxalato)ferrate(III); (v) potassium tetrachloridopalladate(II); (vi) diamminechlorido(methanamine)platinum(II) chloride.

Answer

Question 5.3

State possible isomerism for (i) K[Cr(H₂O)₂(ox)₂]; (ii) [Co(en)₃]Cl₃; (iii) [Co(NH₃)₅(NO₂)](NO₃)₂; (iv) [Pt(NH₃)(H₂O)Cl₂].

Steps only

(i) cis/trans geometrical isomers; the cis form is optically active. (ii) Two optical isomers. (iii) Linkage isomers (NO₂ through N or O) and ionisation isomers; the text notes ten total forms when the relevant geometrical arrangements are counted. (iv) cis and trans geometrical isomers.

Answer

(i) cis/trans geometrical isomers; the cis form is optically active.
(ii) Two optical isomers.
(iii) Linkage isomers (NO₂ through N or O) and ionisation isomers; the text notes ten total forms when the relevant geometrical arrangements are counted.
(iv) cis and trans geometrical isomers.

Question 5.4

How can you show that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅(SO₄)]Cl are ionisation isomers?

Steps only

The first releases SO₄²⁻ and gives BaSO₄ with Ba²⁺, but no AgCl with Ag⁺. The second releases Cl⁻ and gives AgCl with Ag⁺, but no BaSO₄ with Ba²⁺.

Answer

The first releases SO₄²⁻ and gives BaSO₄ with Ba²⁺, but no AgCl with Ag⁺. The second releases Cl⁻ and gives AgCl with Ag⁺, but no BaSO₄ with Ba²⁺.

Question 5.5

Explain using valence bond theory why [Ni(CN)₄]²⁻ is square-planar and diamagnetic, while [NiCl₄]²⁻ is tetrahedral and paramagnetic.

Steps only

[Ni(CN)₄]²⁻: dsp², square planar, diamagnetic. [NiCl₄]²⁻: sp³, tetrahedral, two unpaired electrons, paramagnetic.

Answer

[Ni(CN)₄]²⁻: dsp², square planar, diamagnetic. [NiCl₄]²⁻: sp³, tetrahedral, two unpaired electrons, paramagnetic.

Question 5.6

Why is [NiCl₄]²⁻ paramagnetic but [Ni(CO)₄] diamagnetic although both are tetrahedral?

Steps only

[NiCl₄]²⁻ has two unpaired electrons; Ni(CO)₄ has none.

Answer

[NiCl₄]²⁻ has two unpaired electrons; Ni(CO)₄ has none.

Question 5.7

Why is [Fe(H₂O)₆]³⁺ strongly paramagnetic but [Fe(CN)₆]³⁻ weakly paramagnetic?

Steps only

The aqua complex has five unpaired electrons; the cyanide complex has one.

Answer

The aqua complex has five unpaired electrons; the cyanide complex has one.

Question 5.8

Explain why [Co(NH₃)₆]³⁺ is inner-orbital but [Ni(NH₃)₆]²⁺ is outer-orbital.

Steps only

[Co(NH₃)₆]³⁺: d²sp³ inner-orbital. [Ni(NH₃)₆]²⁺: sp³d² outer-orbital.

Answer

[Co(NH₃)₆]³⁺: d²sp³ inner-orbital. [Ni(NH₃)₆]²⁺: sp³d² outer-orbital.

Question 5.9

Predict unpaired electrons in square-planar [Pt(CN)₄]²⁻.

Steps only

There are 0 unpaired electrons; the complex is diamagnetic.

Answer

There are 0 unpaired electrons; the complex is diamagnetic.

Question 5.10

Why does [Mn(H₂O)₆]²⁺ have five unpaired electrons while [Mn(CN)₆]⁴⁻ has one?

Steps only

The aqua complex is high-spin d⁵ with 5 unpaired electrons; the cyanide complex is low-spin d⁵ with 1.

Answer

The aqua complex is high-spin d⁵ with 5 unpaired electrons; the cyanide complex is low-spin d⁵ with 1.

Chapter 5 Exercises and Extra Questions

Textbook pages 138–140

Question 5.1

Explain bonding in coordination compounds using Werner’s postulates.

Steps only

Primary valence is ionisable; secondary valence is the fixed coordination number and determines the geometry.

Answer

Primary valence is ionisable; secondary valence is the fixed coordination number and determines the geometry.

Question 5.2

Why does FeSO₄ mixed with (NH₄)₂SO₄ give an Fe²⁺ test, while CuSO₄ mixed with four equivalents of aqueous NH₃ does not give a Cu²⁺ test?

Steps only

Mohr’s salt releases Fe²⁺ ions. Ammonia binds Cu²⁺ as [Cu(NH₃)₄]²⁺, so free Cu²⁺ is not available for its usual test.

Answer

Mohr’s salt releases Fe²⁺ ions. Ammonia binds Cu²⁺ as [Cu(NH₃)₄]²⁺, so free Cu²⁺ is not available for its usual test.

Question 5.3

Give examples of coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Steps only

Entity: [Co(NH₃)₆]³⁺, [Fe(CN)₆]⁴⁻. Ligands: NH₃, CN⁻. Coordination numbers: 6 in [Co(NH₃)₆]³⁺; 4 in [Ni(CO)₄]. Polyhedra: octahedral [Co(NH₃)₆]³⁺; tetrahedral [Ni(CO)₄]. Homoleptic: [Co(NH₃)₆]³⁺, [Ni(CO)₄]. Heteroleptic: [Co(NH₃)₄Cl₂]⁺, [Pt(NH₃)₂Cl₂].

Answer

Question 5.4

Define unidentate, didentate and ambidentate ligands and give two examples of each.

Steps only

Unidentate: H₂O, NH₃. Didentate: en, oxalate (C₂O₄²⁻). Ambidentate: NO₂⁻ (N or O donor), SCN⁻ (S or N donor).

Answer

Unidentate: H₂O, NH₃. Didentate: en, oxalate (C₂O₄²⁻). Ambidentate: NO₂⁻ (N or O donor), SCN⁻ (S or N donor).

Question 5.5

Find metal oxidation states in (i) [Co(H₂O)(CN)(en)₂]²⁺ (ii) [CoBr₂(en)₂]⁺ (iii) [PtCl₄]²⁻ (iv) K₃[Fe(CN)₆] (v) [Cr(NH₃)₃Cl₃].

Steps only

(i) Co=+3; (ii) Co=+3; (iii) Pt=+2; (iv) Fe=+3; (v) Cr=+3.

Answer

(i) Co=+3;
(ii) Co=+3;
(iii) Pt=+2;
(iv) Fe=+3;
(v) Cr=+3.

Question 5.6

Write formulas for the ten named coordination compounds in exercise 5.6.

Steps only

(i) [Zn(OH)₄]²⁻ (zincate ion); (ii) K₂[PdCl₄]; (iii) [Pt(NH₃)₂Cl₂]; (iv) K₂[Ni(CN)₄]; (v) [Co(NH₃)₅(ONO)]²⁺ with one suitable counter-anion required by the full compound; (vi) [Co(NH₃)₆]₂(SO₄)₃; (vii) K₃[Cr(C₂O₄)₃]; (viii) [Pt(NH₃)₆]⁴⁺; (ix) [CuBr₄]²⁻; (x) [Co(NH₃)₅(NO₂)]²⁺.

Answer

(i) [Zn(OH)₄]²⁻ (zincate ion);
(ii) K₂[PdCl₄];
(iii) [Pt(NH₃)₂Cl₂];
(iv) K₂[Ni(CN)₄];
(v) [Co(NH₃)₅(ONO)]²⁺ with one suitable counter-anion required by the full compound;
(vi) [Co(NH₃)₆]₂(SO₄)₃;
(vii) K₃[Cr(C₂O₄)₃];
(viii) [Pt(NH₃)₆]⁴⁺;
(ix) [CuBr₄]²⁻;
(x) [Co(NH₃)₅(NO₂)]²⁺.

Question 5.7

Name the nine compounds listed in exercise 5.7.

Steps only

(i) Hexaamminecobalt(III) chloride; (ii) amminemethylaminechloridoplatinum(II) chloride; (iii) hexaaquatitanium(III) ion; (iv) tetraamminechloridonitrito-N-cobalt(III) chloride; (v) hexaaquamanganese(II) ion; (vi) tetrachloridonickelate(II) ion; (vii) hexaamminenickel(II) chloride; (viii) tris(ethane-1,2-diamine)cobalt(III) ion; (ix) tetracarbonylnickel(0).

Answer

Question 5.8

List types of coordination-compound isomerism with one example of each.

Steps only

Structural: ionisation [Co(NH₃)₅Br]SO₄/[Co(NH₃)₅SO₄]Br; linkage [Co(NH₃)₅(NO₂)]²⁺ (N- or O-bound); coordination [Co(NH₃)₆][Cr(CN)₆]/[Cr(NH₃)₆][Co(CN)₆]; solvate [Cr(H₂O)₆]Cl₃/[Cr(H₂O)₅Cl]Cl₂·H₂O. Stereoisomerism: geometrical cis/trans [Pt(NH₃)₂Cl₂], and optical [Co(en)₃]³⁺.

Answer

Question 5.9

How many geometrical isomers are possible for (i) [Cr(ox)₃]³⁻ and (ii) [Co(NH₃)₃Cl₃]?

Steps only

(i) One geometrical arrangement (it has a pair of optical forms). (ii) Two: fac and mer.

Answer

(i) One geometrical arrangement (it has a pair of optical forms).
(ii) Two: fac and mer.

Question 5.10

Describe the optical isomers for (i) [Cr(ox)₃]³⁻ (ii) [PtCl₂(en)₂]²⁺ (iii) [Cr(NH₃)₂Cl₂(en)]⁺.

Steps only

(i) Two mirror forms, Δ and Λ. (ii) The cis form has two optical forms; trans is achiral. (iii) Optical isomerism occurs for the cis arrangement with the two NH₃ and two Cl ligands placed so no mirror plane remains; its mirror image is the second form.

Answer

(i) Two mirror forms, Δ and Λ.
(ii) The cis form has two optical forms; trans is achiral.
(iii) Optical isomerism occurs for the cis arrangement with the two NH₃ and two Cl ligands placed so no mirror plane remains; its mirror image is the second form.

Question 5.11

List geometrical and optical forms of (i) [CoCl₂(en)₂]⁺ (ii) [Co(NH₃)Cl(en)₂]²⁺ (iii) [Co(NH₃)₂Cl₂(en)]⁺.

Steps only

(i) cis and trans; cis exists as a pair of optical isomers, trans is achiral. (ii) cis and trans arrangements of NH₃/Cl; the cis form is optically active as a pair. (iii) fac and mer geometrical arrangements; the unsymmetrical fac arrangement has an optical pair, while the mer form is achiral.

Answer

(i) cis and trans; cis exists as a pair of optical isomers, trans is achiral.
(ii) cis and trans arrangements of NH₃/Cl; the cis form is optically active as a pair.
(iii) fac and mer geometrical arrangements; the unsymmetrical fac arrangement has an optical pair, while the mer form is achiral.

Question 5.12

List geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and say how many are optically active.

Steps only

Three geometrical isomers; none is optically active.

Answer

Three geometrical isomers; none is optically active.

Question 5.13

Why does aqueous CuSO₄ give (i) a green precipitate with KF and (ii) a bright-green solution with KCl?

Steps only

(i) Green CuF₂ precipitate forms. (ii) Soluble chloro-complex ions form, giving a bright-green solution.

Answer

(i) Green CuF₂ precipitate forms.
(ii) Soluble chloro-complex ions form, giving a bright-green solution.

Question 5.14

What complex forms when excess KCN is added to CuSO₄, and why does passing H₂S not precipitate copper sulphide?

Steps only

The coordination entity is [Cu(CN)₄]²⁻ (often represented as a stable Cu(I)-cyanide complex under excess cyanide). H₂S does not precipitate CuS because copper is held in a stable cyanide complex and free copper-ion concentration is very low.

Answer

Question 5.15

Describe VBT bonding for (i) [Fe(CN)₆]⁴⁻ (ii) [FeF₆]³⁻ (iii) [Co(ox)₃]³⁻ (iv) [CoF₆]³⁻.

Steps only

(i) Fe²⁺ d⁶, CN⁻ strong: d²sp³, octahedral, low spin, diamagnetic. (ii) Fe³⁺ d⁵, F⁻ weak: sp³d², octahedral, five unpaired. (iii) Co³⁺ d⁶, oxalate gives an inner-orbital d²sp³ complex, diamagnetic. (iv) Co³⁺ d⁶, F⁻ weak: sp³d², four unpaired, paramagnetic.

Answer

(i) Fe²⁺ d⁶, CN⁻ strong: d²sp³, octahedral, low spin, diamagnetic.
(ii) Fe³⁺ d⁵, F⁻ weak: sp³d², octahedral, five unpaired.
(iii) Co³⁺ d⁶, oxalate gives an inner-orbital d²sp³ complex, diamagnetic.
(iv) Co³⁺ d⁶, F⁻ weak: sp³d², four unpaired, paramagnetic.

Question 5.16

Show the d-orbital splitting in an octahedral field.

Steps only

The lower t₂g set is dxy, dxz, dyz; the higher e_g set is dz², dx²−y². Their separation is Δ₀; relative to the barycentre, t₂g is −0.4Δ₀ and e_g is +0.6Δ₀.

Answer

The lower t₂g set is dxy, dxz, dyz; the higher e_g set is dz², dx²−y². Their separation is Δ₀; relative to the barycentre, t₂g is −0.4Δ₀ and e_g is +0.6Δ₀.

Question 5.17

What is the spectrochemical series? Distinguish weak- and strong-field ligands.

Steps only

A common order is I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. Weak-field ligands give small splitting; strong-field ligands give large splitting.

Answer

A common order is I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. Weak-field ligands give small splitting; strong-field ligands give large splitting.

Question 5.18

Define Δ₀ and explain how it controls electron arrangement.

Steps only

Small Δ₀ relative to pairing energy gives high spin; large Δ₀ gives low spin.

Answer

Small Δ₀ relative to pairing energy gives high spin; large Δ₀ gives low spin.

Question 5.19

Why is [Cr(NH₃)₆]³⁺ paramagnetic while [Ni(CN)₄]²⁻ is diamagnetic?

Steps only

[Cr(NH₃)₆]³⁺ has 3 unpaired electrons; [Ni(CN)₄]²⁻ has 0.

Answer

[Cr(NH₃)₆]³⁺ has 3 unpaired electrons; [Ni(CN)₄]²⁻ has 0.

Question 5.20

Why is [Ni(H₂O)₆]²⁺ green but [Ni(CN)₄]²⁻ colourless?

Steps only

The aqua complex absorbs visible light and appears green; the cyanide complex appears colourless because its absorption is not in the visible region.

Answer

The aqua complex absorbs visible light and appears green; the cyanide complex appears colourless because its absorption is not in the visible region.

Question 5.21

Why do [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ have different colours?

Steps only

Different ligands cause different crystal-field splitting, so the complexes absorb different visible wavelengths and have different colours.

Answer

Different ligands cause different crystal-field splitting, so the complexes absorb different visible wavelengths and have different colours.

Question 5.22

Describe bonding in metal carbonyls.

Steps only

Metal–CO bonding includes CO→metal σ donation and metal→CO π back-donation; the synergic effect strengthens the bond.

Answer

Metal–CO bonding includes CO→metal σ donation and metal→CO π back-donation; the synergic effect strengthens the bond.

Question 5.23

Find oxidation state, d count and coordination number in (i) K₃[Co(ox)₃] (ii) cis-[CrCl₂(en)₂]Cl (iii) (NH₄)₂[CoF₄] (iv) [Mn(H₂O)₆]SO₄.

Steps only

(i) Co +3, d⁶, CN 6. (ii) Cr +3, d³, CN 6. (iii) Co +2, d⁷, CN 4. (iv) Mn +2, d⁵, CN 6.

Answer

(i) Co +3, d⁶, CN 6.
(ii) Cr +3, d³, CN 6.
(iii) Co +2, d⁷, CN 4.
(iv) Mn +2, d⁵, CN 6.

Question 5.24

For each listed complex give name, oxidation state, d count, coordination number, geometry and magnetic moment.

Steps only

(i) K[Cr(H₂O)₂(ox)₂]·3H₂O: potassium diaquabis(oxalato)chromate(III) trihydrate; Cr³⁺ d³, CN 6, octahedral, 3 unpaired, μ≈3.87 BM. (ii) [Co(NH₃)₅Cl]Cl₂: pentaamminechloridocobalt(III) chloride; Co³⁺ d⁶, CN 6, octahedral low-spin, 0 unpaired, μ=0. (iii) [CrCl₃(py)₃]: trichloridotris(pyridine)chromium(III); Cr³⁺ d³, CN 6, octahedral, 3 unpaired, μ≈3.87 BM. (iv) Cs[FeCl₄]: caesium tetrachloridoferrate(III); Fe³⁺ d⁵, CN 4, tetrahedral, 5 unpaired, μ≈5.92 BM. (v) K₄[Mn(CN)₆]: potassium hexacyanidomanganate(II); Mn²⁺ d⁵, CN 6, octahedral low-spin, 1 unpaired as given in NCERT data, μ≈1.73 BM.

Answer

Question 5.25

Explain the violet colour of [Ti(H₂O)₆]³⁺ using crystal-field theory.

Steps only

The complex absorbs blue-green light for a t₂g→e_g transition and appears violet, the complementary colour.

Answer

The complex absorbs blue-green light for a t₂g→e_g transition and appears violet, the complementary colour.

Question 5.26

What is the chelate effect? Give an example.

Steps only

The increased stability of a complex with a multidentate ligand is the chelate effect; for example, [Ni(en)₃]²⁺.

Answer

The increased stability of a complex with a multidentate ligand is the chelate effect; for example, [Ni(en)₃]²⁺.

Question 5.27

Give one role/example of coordination compounds in biology, medicine, analysis and metallurgy.

Steps only

Biology: haemoglobin carries oxygen. Medicine: cisplatin is an anticancer drug. Analysis: EDTA measures water hardness. Metallurgy: cyanide complexes help extract gold.

Answer

Biology: haemoglobin carries oxygen. Medicine: cisplatin is an anticancer drug. Analysis: EDTA measures water hardness. Metallurgy: cyanide complexes help extract gold.

Question 5.28

How many ions are produced when Co(NH₃)₆Cl₂ dissolves?

Steps only

Three ions: one [Co(NH₃)₆]²⁺ and two Cl⁻ ions.

Answer

Three ions: one [Co(NH₃)₆]²⁺ and two Cl⁻ ions.

Question 5.29

Which has the highest magnetic moment: [Cr(H₂O)₆]³⁺, [Fe(H₂O)₆]²⁺ or [Zn(H₂O)₆]²⁺?

Steps only

[Fe(H₂O)₆]²⁺ has the highest moment: 4 unpaired electrons (about 4.90 BM spin-only).

Answer

[Fe(H₂O)₆]²⁺ has the highest moment: 4 unpaired electrons (about 4.90 BM spin-only).

Question 5.30

Which is expected to be most stable: [Fe(H₂O)₆]³⁺, [Fe(NH₃)₆]³⁺, [Fe(ox)₃]³⁻ or [FeCl₆]³⁻?

Steps only

[Fe(ox)₃]³⁻ is expected to be most stable because oxalate chelates the iron ion.

Answer

[Fe(ox)₃]³⁻ is expected to be most stable because oxalate chelates the iron ion.

Question 5.31

Order absorption wavelengths for [Ni(NO₂)₆]⁴⁻, [Ni(NH₃)₆]²⁺ and [Ni(H₂O)₆]²⁺.

Steps only

Increasing wavelength: [Ni(NO₂)₆]⁴⁻ < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺.

Answer

Increasing wavelength: [Ni(NO₂)₆]⁴⁻ < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺.