Chapter 9: Coordination Compounds Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 9 – Coordination Compounds (Intext Questions)

Intext Question 9.1

Write the formulas for the following coordination compounds:

Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄]
(iii) [Cr(en)₃]Cl₃
(iv) [Pt(NH₃)(Br)(Cl)(NO₂)]⁻
(v) [PtCl₂(en)₂](NO₃)₂
(vi) Fe₄[Fe(CN)₆]₃

Intext Question 9.2

Write IUPAC names of the following coordination compounds:

Answer:
(a) Hexaamminecobalt(III) chloride
(b) Pentaamminechloridocobalt(III) chloride
(c) Potassium hexacyanidoferrate(III)
(d) Potassium trioxalato ferrate(III)
(e) Potassium tetrachloridopalladate(II)
(f) Diamminechlorido(methylamine)platinum(II) chloride

Intext Question 9.3

Indicate the types of isomerism exhibited by the following complexes:

Answer:
(i) Geometrical isomerism (cis and trans forms)
(ii) Optical isomerism (due to three chiral bidentate ligands)
(iii) Linkage isomerism (NO₂⁻ can bind through N or O atom)
(iv) Geometrical isomerism (cis and trans forms)

Intext Question 9.4

Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.

Answer: In [Co(NH₃)₅Cl]SO₄, the chloride is coordinated, and sulfate is free, which can be precipitated with Ba²⁺ ions. In [Co(NH₃)₅SO₄]Cl, sulfate is coordinated and Cl⁻ is free, which can be precipitated by Ag⁺ ions, proving ionisation isomerism.

Intext Question 9.5

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

Answer: In [Ni(CN)₄]²⁻, CN⁻ is a strong field ligand causing pairing of electrons in the 3d orbitals, resulting in dsp² hybridisation and square planar geometry. In [NiCl₄]²⁻, Cl⁻ is a weak field ligand, no pairing occurs, leading to sp³ hybridisation and a tetrahedral geometry with unpaired electrons.

Intext Question 9.6

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

Answer: Cl⁻ is a weak field ligand in [NiCl₄]²⁻, so no electron pairing occurs, resulting in paramagnetism. CO is a strong field ligand in [Ni(CO)₄], causing electron pairing and making the complex diamagnetic.

Intext Question 9.7

[Fe(H₂O)₆]³⁺ is strongly paramagnetic, but [Fe(CN)₆]³⁻ is weakly paramagnetic. Explain.

Answer: H₂O is a weak field ligand, so [Fe(H₂O)₆]³⁺ has 5 unpaired electrons (high spin). CN⁻ is a strong field ligand, causing pairing in [Fe(CN)₆]³⁻, leaving only 1 unpaired electron (low spin).

Intext Question 9.8

Explain why [Co(NH₃)₆]²⁺ is an inner orbital complex, while [Ni(NH₃)₆]²⁺ is an outer orbital complex.

Answer: In [Co(NH₃)₆]²⁺, strong field NH₃ ligands cause electron pairing, leading to inner d²sp³ hybridisation (low spin). In [Ni(NH₃)₆]²⁺, NH₃ is weaker for Ni²⁺, so no pairing, and sp³d² hybridisation (outer orbital, high spin).

Intext Question 9.9

Predict the number of unpaired electrons in [Pt(CN)₄]²⁻.

Answer: Pt²⁺ = 5d⁸. Strong field CN⁻ causes full pairing of electrons. Thus, [Pt(CN)₄]²⁻ has 0 unpaired electrons.

Intext Question 9.10

Explain why hexaaquamanganese(II) has five unpaired electrons, but hexacyanomanganate(II) has only one.

Answer: In [Mn(H₂O)₆]²⁺, H₂O is a weak field ligand, no pairing → 5 unpaired electrons (high spin).
In [Mn(CN)₆]⁴⁻, CN⁻ is a strong field ligand, causing pairing → 1 unpaired electron (low spin).

Intext Question 9.11

Calculate the overall dissociation equilibrium constant for [Cu(NH₃)₄]²⁺, given β₄ = 2.1×10¹³.

Answer: Dissociation constant, K_d = 1 / β₄ = 1 / (2.1 × 10¹³) ≈ 4.76 × 10⁻¹⁴.

NCERT Solutions for Class 12 Chemistry Chapter 9 – Coordination Compounds (Exercise Questions)

Q9.1 Explain the bonding in coordination compounds in terms of Werner’s postulates.

Answer:
Werner proposed that metals show two types of valencies:
1. Primary valency (oxidation state) – ionisable.
2. Secondary valency (coordination number) – non-ionisable, satisfied by ligands.
Primary valencies are shown outside the brackets and secondary within. Secondary valencies are directional, leading to specific geometries (e.g., octahedral, tetrahedral).

Q9.2 Why does FeSO₄ with (NH₄)₂SO₄ give Fe²⁺ test but CuSO₄ with NH₃ does not give Cu²⁺ test?

Answer:
FeSO₄ + (NH₄)₂SO₄ forms a double salt that dissociates completely into Fe²⁺ ions.
CuSO₄ + NH₃ forms [Cu(NH₃)₄]SO₄, a coordination complex where Cu²⁺ is not freely available, so Cu²⁺ test is not observed.

Q9.3 Explain with two examples each:

Answer:
(i) Coordination entity: [Co(NH₃)₆]³⁺, [Fe(CN)₆]³⁻
(ii) Ligand: NH₃, Cl⁻
(iii) Coordination number: 6 in [Co(NH₃)₆]³⁺, 4 in [Ni(CO)₄]
(iv) Coordination polyhedron: Octahedral ([Co(NH₃)₆]³⁺), Tetrahedral ([Ni(CO)₄])
(v) Homoleptic: [Ni(CO)₄]
(vi) Heteroleptic: [CoCl₂(en)₂]⁺

Q9.4 Define unidentate, didentate, and ambidentate ligands. Give examples.

Answer:
– Unidentate: Cl⁻, NH₃
– Didentate: Ethylenediamine (en), Oxalate ion (C₂O₄²⁻)
– Ambidentate: NO₂⁻ (N or O donor), SCN⁻ (S or N donor)

Q9.5 Determine the oxidation number of the metal in the following complexes:

Answer:
(i) +3 (Co)
(ii) +3 (Co)
(iii) +2 (Pt)
(iv) +3 (Fe)
(v) +3 (Cr)

Q9.6 Write formulae for given compounds:

Answer:
(a) [Zn(OH)₄]²⁻
(b) [Pt(NH₃)₆]⁴⁺
(c) K₂[PdCl₄]
(d) [CuBr₄]²⁻
(e) [Co(NH₃)₆]₂(SO₄)₃
(f) K₂[Ni(CN)₄]
(g) K₃[Cr(C₂O₄)₃]
(h) [Co(NH₃)₅(ONO)]²⁺
(i) [Pt(NH₃)₂Cl₂]
(j) [Co(NH₃)₅(NO₂)]²⁺

Q9.7 Write IUPAC names:

Answer:
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine)platinum(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)

Q9.8 What types of isomerism are shown by coordination compounds?

Answer:
– Ionisation isomerism: [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄
– Linkage isomerism: [Co(NH₃)₅NO₂]²⁺ and [Co(NH₃)₅ONO]²⁺
– Geometrical isomerism: [Pt(NH₃)₂Cl₂] (cis/trans)
– Optical isomerism: [Co(en)₃]³⁺ (d- and l- forms)

Q9.9 How many geometrical isomers for given complexes?

Answer:
(i) No geometrical isomer for [Cr(C₂O₄)₃]³⁻
(ii) Two (fac and mer) for [CoCl₃(NH₃)₃]

Q9.10 Draw optical isomers of given complexes.

Answer:
[Cr(C₂O₄)₃]³⁻, [PtCl₂(en)₂]²⁺, [Cr(NH₃)₂Cl₂(en)]⁺ all exhibit optical isomerism (d- and l- forms).

Q9.11 Draw all isomers (geometrical and optical).

Answer:
[CoCl₂(en)₂]⁺ and [Cr(NH₃)₂Cl₂(en)]⁺ exhibit both geometrical (cis/trans) and optical isomerism.

Q9.12 Draw all geometrical isomers of [Pt(NH₃)(Br)(Cl)(Py)]. How many are optically active?

Answer:
Three geometrical isomers exist. None are optically active.

Q9.13 Explain observations when CuSO₄ is treated with KF and KCl.

Answer:
– KF forms green precipitate of [CuF₄]²⁻.
– KCl forms green solution of [CuCl₄]²⁻.

Q9.14 Coordination entity when excess KCN added to CuSO₄?

Answer:
[Cu(CN)₄]³⁻ complex forms.
No CuS precipitate forms with H₂S because Cu is locked in the stable complex.

Q9.15 Discuss nature of bonding in:

Answer:
– [Fe(CN)₆]⁴⁻: d²sp³ (inner orbital), diamagnetic, octahedral.
– [FeF₆]³⁻: sp³d² (outer orbital), paramagnetic, octahedral.
– [Co(C₂O₄)₃]³⁻: d²sp³, diamagnetic, octahedral.
– [CoF₆]³⁻: sp³d², paramagnetic, octahedral.

Q9.16 Diagram: Splitting of d-orbitals in octahedral field.

Answer:
– t₂g (lower): d_xy, d_yz, d_zx
– e_g (higher): d_x²–y², d_z²
Δ₀ = splitting energy between these sets.

Q9.17 What is spectrochemical series?

Answer:
An order of ligands based on their ability to split d-orbitals.
Weak field: H₂O, F⁻
Strong field: CN⁻, CO

Q9.18 What is crystal field splitting energy?

Answer:
The energy difference Δ₀ between t₂g and e_g orbitals. Determines whether the complex is high spin (small Δ₀) or low spin (large Δ₀).

Q9.19 [Cr(NH₃)₆]³⁺ is paramagnetic but [Ni(CN)₄]²⁻ is diamagnetic. Why?

Answer:
[Cr(NH₃)₆]³⁺ has unpaired electrons (paramagnetic).
[Ni(CN)₄]²⁻ has all electrons paired (diamagnetic).

Q9.20 Why [Ni(H₂O)₆]²⁺ is green but [Ni(CN)₄]²⁻ is colourless?

Answer:
[Ni(H₂O)₆]²⁺ has unpaired electrons allowing d–d transitions → green colour.
[Ni(CN)₄]²⁻ has paired electrons → no d–d transitions → colourless.

Q9.21 Different colours of [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ solutions.

Answer:
Different crystal field splitting Δ₀ values due to different ligands (CN⁻ vs H₂O) cause absorption of different wavelengths → different colours.

Q9.22 Nature of bonding in metal carbonyls.

Answer:
σ-bond from CO to metal (lone pair).
π-backbonding from metal d-orbitals to CO’s π* orbitals (synergic bonding).

Q9.23 Central metal oxidation state, d-orbital configuration, and coordination number.

Answer:
(i) K₃[Co(C₂O₄)₃]: +3, d⁶, 6
(ii) cis-[Cr(en)₂Cl₂]Cl: +3, d³, 6
(iii) (NH₄)₂[CoF₄]: +2, d⁷, 4
(iv) [Mn(H₂O)₆]SO₄: +2, d⁵, 6

Q9.24 IUPAC name, oxidation state, configuration, coordination number, stereochemistry, and magnetic moment.

Answer:
(i) Potassium diaquadioxalatochromate(III) trihydrate: +3, d³, 6, octahedral, 3.87 BM
(ii) Pentaamminechloridocobalt(III) chloride: +3, d⁶, 6, octahedral, diamagnetic
(iii) Trichloridotripyridinechromium(III): +3, d³, 6, octahedral, 3.87 BM

Q9.25 What is meant by stability of a coordination compound in solution? State the factors that govern stability of complexes.

Answer:
The stability of a coordination compound refers to its resistance towards dissociation into its components in solution. Stability is measured by its formation constant (β) — higher the β value, greater the stability.

Factors affecting stability:
– Charge on the metal ion: Higher charge favours stronger binding.
– Size of the metal ion: Smaller ions form more stable complexes.
– Nature of ligand: Strong field ligands and chelating ligands (multidentate) enhance stability.
– Basicity of the ligand: More basic ligands bond more strongly with the metal.
– Chelate effect: Complexes with chelating ligands (like en, EDTA) are exceptionally stable due to ring formation.

Q9.26 What is meant by the chelate effect? Give an example.

Answer:
The chelate effect refers to the enhanced stability of coordination compounds containing chelating (bidentate or polydentate) ligands compared to complexes with similar monodentate ligands.

Example:
– [Cu(en)₂]²⁺ is more stable than [Cu(NH₃)₄]²⁺ because en (ethane-1,2-diamine) forms two bonds per ligand, creating stable five-membered rings.

This enhanced stability arises due to the entropy increase when a chelating ligand binds, replacing multiple monodentate ligands.

Q9.27 Discuss briefly the role of coordination compounds in biological systems, analytical chemistry, medicinal chemistry, and metallurgy.

Answer:
(a) Biological Systems: Coordination compounds are vital, e.g.,
– Hemoglobin (Fe²⁺ complex) transports oxygen.
– Chlorophyll (Mg²⁺ complex) is essential for photosynthesis.
– Vitamin B₁₂ (contains Co³⁺) is important for metabolism.

(b) Analytical Chemistry: Complexes like EDTA are used in complexometric titrations to determine metal ion concentrations.

(c) Medicinal Chemistry: Cisplatin [Pt(NH₃)₂Cl₂] is used in chemotherapy for cancer treatment. EDTA is used to treat lead poisoning.

(d) Metallurgy: Gold and silver are extracted by forming soluble cyanide complexes (e.g., [Ag(CN)₂]⁻) in the cyanide process.

Q9.28 How many ions are produced from the complex [Co(NH₃)₆]Cl₂ in solution?

Answer:
The complex dissociates as:
[Co(NH₃)₆]Cl₂ → [Co(NH₃)₆]²⁺ + 2Cl⁻

Thus, a total of 3 ions are produced: one complex cation and two chloride ions.

Correct Answer: 3 ions

Q9.29 Which of the following compounds will exhibit the highest magnetic moment?

Given:
(i) [Cr(H₂O)₆]³⁺ → d³ → 3 unpaired electrons
(ii) [Fe(H₂O)₆]²⁺ → d⁶ → 4 unpaired electrons
(iii) [Zn(H₂O)₆]²⁺ → d¹⁰ → 0 unpaired electrons

Answer:
[Fe(H₂O)₆]²⁺ has 4 unpaired electrons, hence the highest magnetic moment.
Magnetic moment (μ) = √n(n+2) BM
For Fe²⁺: μ = √4(4+2) = √24 ≈ 4.9 BM (maximum)

Q9.30 The oxidation number of cobalt in K[Co(CO)₄] is:

Answer:
Let oxidation number of Co be x.
Given: K⁺ = +1, CO (carbonyl ligand) = neutral.

Thus, x + 0 = –1 (overall charge on complex)
x = –1

Correct Answer: (iii) –1