NCERT Solutions for Class 12 Chemistry Chapter 5 – Coordination Compounds (English Medium)

🧿 Get NCERT Solutions for Class 12 Chemistry Chapter 5 – Coordination Compounds with accurate answers and step-by-step explanations in a neat, exam-ready format. This chapter covers Werner’s theory, IUPAC naming, coordination number & ligands, isomerism (geometrical, optical, linkage), bonding (VBT & CFT), magnetic properties, stability, and applications of coordination compounds. Perfect for CBSE boards and NEET/JEE practice. ✅

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NCERT Solutions for Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes

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

Answer

👉 According to Werner’s theory, bonding in coordination compounds is explained using two kinds of valencies of the central metal atom/ion (CMA):

📌 Primary valency (oxidation state) and 📌 Secondary valency (coordination number) .

👉 These two valencies decide which ions are inside/outside the coordination sphere and also decide the shape (geometry) of the complex.

Explanation (Step by Step)

1) Two types of valencies (linkages) 🤔

✅ (i) Primary valency = Oxidation number (O.N.)

📌 Nature: Non-directional and ionisable

👉 It is satisfied by negative ions (anions) .

👉 These ions usually stay outside the coordination sphere and dissociate in solution.

✅ (ii) Secondary valency = Coordination number (C.N.)

📌 Nature: Directional and non-ionisable

👉 It is satisfied by ligands (neutral molecules like NH₃, H₂O or anions like Cl⁻) that are directly attached to the metal.

👉 These ligands stay inside the coordination sphere .

2) Fixed number + fixed geometry 📌

👉 Werner proposed that the groups/ions attached by secondary valency arrange themselves in a definite spatial pattern (geometry).

Examples (common):

• C.N. = 6 → octahedral
• C.N. = 4 → tetrahedral or square planar

3) Example to understand bonding clearly ✅

Consider: [CoCl₂(NH₃)₄]Cl

📌 Here:

• Inside bracket (coordination sphere): [CoCl₂(NH₃)₄]⁺

• Outside bracket (ionisable ion): Cl⁻

👉 Primary valency (O.N.) of Co = +3

Reason:
Let oxidation state of Co = x
x + 2(−1) + 4(0) = +1 ⟹ x − 2 = +1 ⟹ x = +3

✅ So, primary valency is satisfied by 3 Cl⁻ in total .

But only 2 Cl⁻ are inside (non-ionisable) , and 1 Cl⁻ is outside (ionisable) .

👉 Secondary valency (C.N.) = 2 (Cl⁻) + 4 (NH₃) = 6

📌 So the complex has octahedral geometry .

Key takeaway 🤔

• Outside ion (Cl⁻) shows ionisation (gives precipitate with AgNO₃)
• Inside ligands do not ionise easily because they are directly bonded to the metal.

Did You Know?

✅ Werner’s coordination theory was so important that it earned him the Nobel Prize (1913) .

👉 His biggest contribution was proving that ligands have fixed positions in space, which directly explains isomerism (like cis/trans) in coordination compounds.

Q.2: FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ ion. Explain why.

Answer

👉 FeSO₄ + (NH₄)₂SO₄ forms a double salt (Mohr’s salt) , which dissociates completely in water to give free Fe²⁺ ions , so it gives the usual Fe²⁺ tests.

👉 But CuSO₄ + NH₃ (1:4) forms a complex compound [Cu(NH₃)₄]²⁺ , which does not provide free Cu²⁺ ions easily, so the normal Cu²⁺ tests are not shown.

Explanation (Step by Step)

1) FeSO₄ + (NH₄)₂SO₄ → Double salt (Mohr’s salt) 📌

When mixed in 1:1, they form Mohr’s salt (a double salt):

👉 FeSO₄(aq) + (NH₄)₂SO₄(aq) + 6H₂O(l)
→ FeSO₄·(NH₄)₂SO₄·6H₂O (Mohr’s salt)

🤔 Important: Double salts dissociate completely in water :

👉 FeSO₄·(NH₄)₂SO₄·6H₂O (aq)
⇌ Fe²⁺(aq) + 2NH₄⁺(aq) + 2SO₄²⁻(aq) + 6H₂O(l)

✅ Since Fe²⁺ ions are free in solution, the solution gives the Fe²⁺ test.

2) CuSO₄ + NH₃ → Complex compound 📌

When CuSO₄ is treated with excess NH₃ (about 1:4), Cu²⁺ forms a stable ammine complex:

👉 CuSO₄(aq) + 4NH₃(aq)
→ [Cu(NH₃)₄]SO₄(aq)

or in ionic form:

👉 Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)

🤔 Here, copper is mostly present as complex ion, not as “free” Cu²⁺ .

✅ इसलिए normal Cu²⁺ test (which needs free Cu²⁺) clearly नहीं मिलता.

Did You Know?

✅ Double salt vs complex salt is a classic NCERT concept:

👉 Double salts dissociate into simple ions completely in water (so they show tests of their ions).

👉 Complexes keep the metal ion bound inside the complex ion (so free Mⁿ⁺ tests may disappear or become weak).

Q.3: Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Answer

1) Coordination entity

📌 A coordination entity is a central metal atom/ion bonded to a fixed number of ligands, written inside [ ] as one unit.

👉 Examples:

1. [CoCl₃(NH₃)₃]

2. [Ni(CO)₄]

2) Ligand

📌 A ligand is an ion or molecule that donates a lone pair of electrons to the central metal to form a coordinate bond .

👉 Examples:

1. NH₃ (neutral ligand)

2. Cl⁻ (anionic ligand)

3) Coordination number (C.N.)

📌 Coordination number is the number of donor atoms (or ligands) directly attached to the central metal ion.

👉 Examples:

1. [Co(NH₃)₆]³⁺ → C.N. of Co = 6

2. [PtCl₄]²⁻ → C.N. of Pt = 4

4) Coordination polyhedron

📌 The 3D arrangement (shape) of ligand donor atoms around the metal is called the coordination polyhedron.

👉 Examples:

1. [Co(NH₃)₆]³⁺ → Octahedral

2. [PtCl₄]²⁻ → Square planar

(Also common: tetrahedral, e.g., [Ni(CO)₄] is tetrahedral.)

5) Homoleptic complex

📌 A homoleptic complex has only one kind of ligand attached to the metal.

👉 Examples:

1. [Co(NH₃)₆]³⁺ (only NH₃)

2. [PtCl₄]²⁻ (only Cl⁻)

6) Heteroleptic complex

📌 A heteroleptic complex has more than one kind of ligand attached to the metal.

👉 Examples:

1. [Co(NH₃)₄Cl₂]⁺ (NH₃ and Cl⁻)

2. [Pt(NH₃)₂Cl₂] (NH₃ and Cl⁻)

Explanation (Simple)

🤔 Think of a coordination compound like a “central metal + surrounding helpers (ligands)” .

📌 The number of helpers = coordination number , and the way they sit in space = coordination polyhedron .

✅ If all helpers are the same = homoleptic , if different = heteroleptic .

Did You Know?

✅ The famous anticancer drug cisplatin is a heteroleptic complex: [Pt(NH₃)₂Cl₂] .

👉 Its cis form is active, while the trans form is much less effective.

Q.4: What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.

Answer

1) Unidentate ligand

📌 A unidentate ligand has only one donor atom , so it forms one coordinate bond with the central metal atom/ion (CMA).

👉 Examples:

1. Cl⁻ (donor atom: Cl)

2. NH₃ (donor atom: N)

2) Bidentate ligand

📌 A bidentate ligand has two donor atoms , so it can form two coordinate bonds with the same metal ion (it “grips” the metal like a claw).

👉 Examples:

1. en (ethane-1,2-diamine) = H₂N–CH₂–CH₂–NH₂ (donor atoms: 2 N)

2. C₂O₄²⁻ (oxalate ion) (donor atoms: 2 O)

3) Ambidentate ligand

📌 An ambidentate ligand has two possible donor atoms , but it can donate only one at a time (it can attach through either donor site, not both together).

👉 Examples:

1. NO₂⁻ (can bind via N = nitro or via O = nitrito)

2. SCN⁻ (can bind via S = thiocyanato or via N = isothiocyanato)

Explanation (Simple)

🤔 Think of “dentate” as “teeth” used to hold the metal:

📌 Uni = 1 tooth, Bi = 2 teeth, Ambi = 2 options but uses only 1 at a time.

Did You Know?

✅ Ambidentate ligands can give linkage isomerism .

👉 Example: NO₂⁻ can form M–NO₂ (nitro) or M–ONO (nitrito) , which are different compounds with different properties.

Q.5: Specify the oxidation numbers of the metals in the following coordination entities: (i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₄]²⁻
(iv) K₃[Fe(CN)₆]
(v) [Cr(NH₃)₃Cl₃]

Answer ✅ (Oxidation number of metal)

(i) [Co(H₂O)(CN)(en)₂]²⁺

Let oxidation number of Co = x

📌 Charges: H₂O = 0, CN⁻ = −1, en = 0 (neutral), total charge = +2

👉 x + 0 + (−1) + 0 = +2
✅ x = +3

Co = +3

(ii) [CoBr₂(en)₂]⁺

Let oxidation number of Co = x

📌 Br⁻ = −1 each, en = 0, total charge = +1

👉 x + 2(−1) + 0 = +1
✅ x = +3

Co = +3

(iii) [PtCl₄]²⁻

Let oxidation number of Pt = x

📌 Cl⁻ = −1 each, total charge = −2

👉 x + 4(−1) = −2
👉 x − 4 = −2
✅ x = +2

Pt = +2

(iv) K₃[Fe(CN)₆]

Let oxidation number of Fe = x

📌 K = +1 each ⇒ 3K = +3
CN⁻ = −1 each ⇒ 6 CN = −6
Whole compound is neutral.

👉 (+3) + x + (−6) = 0
✅ x = +3

Fe = +3

(v) [Cr(NH₃)₃Cl₃]

Let oxidation number of Cr = x

📌 NH₃ = 0, Cl⁻ = −1 each, complex is neutral (0)

👉 x + 3(0) + 3(−1) = 0
👉 x − 3 = 0
✅ x = +3

Cr = +3

Explanation (Quick logic)

🤔 Neutral ligands like H₂O, NH₃, en contribute 0 charge.

📌 Anionic ligands like CN⁻, Cl⁻, Br⁻ contribute −1 each.

👉 Add charges of ligands + metal oxidation number = overall charge of the complex.

Did You Know?

✅ en (ethane-1,2-diamine) is a neutral bidentate ligand, so it changes coordination number and stability, but does not change oxidation number directly.

Q.6: Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanidonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt(III)

Answer ✅

(i) Tetrahydroxidozincate(II)

👉 [Zn(OH)₄]²⁻

(ii) Potassium tetrachloridopalladate(II)

👉 K₂[PdCl₄]

(iii) Diamminedichloridoplatinum(II)

👉 [Pt(NH₃)₂Cl₂]

(iv) Potassium tetracyanidonickelate(II)

👉 K₂[Ni(CN)₄]

(v) Pentaamminenitrito-O-cobalt(III)

👉 [Co(NH₃)₅(ONO)]²⁺

📌 Here nitrito-O means NO₂⁻ is bonded through O.

(vi) Hexaamminecobalt(III) sulphate

👉 [Co(NH₃)₆]₂(SO₄)₃

(vii) Potassium tri(oxalato)chromate(III)

👉 K₃[Cr(C₂O₄)₃]

(viii) Hexaammineplatinum(IV)

👉 [Pt(NH₃)₆]⁴⁺

(ix) Tetrabromidocuprate(II)

👉 [CuBr₄]²⁻

(x) Pentaamminenitrito-N-cobalt(III)

👉 [Co(NH₃)₅(NO₂)]²⁺

📌 Here nitrito-N means NO₂⁻ is bonded through N (nitro form).

Explanation (Quick logic)

🤔 ammine (NH₃) is neutral, while hydroxido (OH⁻), chlorido (Cl⁻), bromido (Br⁻), cyanido (CN⁻), oxalato (C₂O₄²⁻), nitrito (NO₂⁻) are anionic.

📌 Add charges to match the metal oxidation state and overall charge, then balance with counter-ions like K⁺, SO₄²⁻, etc.

Did You Know?

✅ Nitrito-O (ONO) and nitrito-N (NO₂) are examples of linkage isomerism.

👉 Same formula, different attachment atom (O vs N), different properties!