CBSE Class 12 Chemistry Coordination Compounds MCQs Set 17

Answer: (b) 6

Explanation:
1. Ethane-1,2-diamine (en) is didentate, so each en gives two donor atoms: 2 × 2 = 4.
2. Two chloride ions add two more bonds.
3. Coordination number = 4 + 2 = 6.

Teacher's Note:
1. Count donor atoms, not ligands.
2. A didentate ligand uses two coordination positions.

Answer: (d) 0

Explanation:
1. CO is a neutral ligand, so it contributes no charge.
2. The complex is neutral, so x + 4(0) = 0 and x = 0.

Teacher's Note:
1. Neutral ligands such as CO, NH₃ and H₂O do not change the oxidation state.
2. Metal carbonyls have the metal in a zero oxidation state.

Answer: (a) NO₂⁻

Explanation:
1. An ambidentate ligand has two different donor atoms and can bind through either one.
2. NO₂⁻ can bind through nitrogen (nitrito-N) or through oxygen (nitrito-O).
3. Cl⁻, NH₃ and H₂O have only one kind of donor atom.

Teacher's Note:
1. Other common ambidentate ligands are SCN⁻ (S or N) and CN⁻ (C or N).
2. They give rise to linkage isomerism.

Answer: (c) 6

Explanation:
1. EDTA⁴⁻ has two nitrogen atoms and four carboxylate oxygen atoms that can donate lone pairs.
2. Total donor atoms = 2 + 4 = 6, so it is hexadentate.

Teacher's Note:
1. Denticity means the number of donor atoms that bind one metal ion.
2. EDTA forms five chelate rings, one fewer than its denticity.

Answer: (b) potassium hexacyanidoferrate(III)

Explanation:
1. The complex ion [Fe(CN)₆]³⁻ is an anion, so the metal name ends in -ate (Latin stem: ferrate).
2. 3(+1) + x + 6(−1) = 0, so x = +3.
3. The cation is named first, so the name is potassium hexacyanidoferrate(III).

Teacher's Note:
1. Name the cation first and the anion second.
2. Work out the oxidation state from the charge on the complex ion.

Answer: (d) pentaamminechloridocobalt(III) chloride

Explanation:
1. The complex ion is [Co(NH₃)₅Cl]²⁺ because two Cl⁻ are outside.
2. x + 0 − 1 = +2, so Co is +3.
3. Ligands go in alphabetical order (ammine before chlorido), and the cationic complex keeps the plain metal name.

Teacher's Note:
1. Alphabetical order uses the ligand names, ignoring prefixes such as penta.
2. Only an anionic complex takes the ending -ate.

Answer: (c) K₂[PdCl₄]

Explanation:
1. The anion contains Pd²⁺ and four Cl⁻: 2 − 4 = −2, so the ion is [PdCl₄]²⁻.
2. Two K⁺ balance the charge, so the formula is K₂[PdCl₄].

Teacher's Note:
1. Write the charge on the complex ion first, then balance it with counter ions.

Answer: (a) 4

Explanation:
1. [Co(NH₃)₆]Cl₃ dissociates into [Co(NH₃)₆]³⁺ and three Cl⁻.
2. Total = 1 + 3 = 4 ions. The ammonia ligands stay in the complex ion.

Teacher's Note:
1. The coordination sphere stays intact in water.
2. Count the complex ion as a single ion.

Answer: (d) [Co(NH₃)₅Cl]Cl₂

Explanation:
1. Two moles of AgCl mean that only two Cl⁻ are ionisable.
2. The third Cl⁻ is bonded to cobalt inside the sphere, and with 5 NH₃ the coordination number is 6.
3. The formula is [Co(NH₃)₅Cl]Cl₂.

Teacher's Note:
1. Moles of AgCl = number of free chloride ions.
2. The coordination number of Co(III) is 6.

Answer: (b) oxidation state

Explanation:
1. Primary valency is ionisable and is satisfied by negative ions.
2. It is the oxidation state of the metal.
3. Secondary valency is the coordination number.

Teacher's Note:
1. Primary: oxidation state, ionisable, non-directional.
2. Secondary: coordination number, non-ionisable, directional.

Answer: (a) negative ions or neutral molecules

Explanation:
1. The secondary valency is the coordination number, and it is filled by ligands.
2. Ligands can be anions (Cl⁻, CN⁻) or neutral molecules (NH₃, H₂O).
3. It is non-ionisable and directional.

Teacher's Note:
1. Compare with primary valency, which only anions can satisfy.

Answer: (c) [Co(NH₃)₆]³⁺

Explanation:
1. A homoleptic complex has only one kind of donor group.
2. In [Co(NH₃)₆]³⁺ all six ligands are NH₃.
3. The other complexes contain two different kinds of ligands, so they are heteroleptic.

Teacher's Note:
1. Homo means same; hetero means different.
2. Look for more than one type of ligand.

Answer: (d) 3

Explanation:
1. Each didentate en closes one five-membered ring with the metal.
2. Rings = 3 × (2 − 1) = 3.

Teacher's Note:
1. Rings per ligand = denticity − 1.
2. Unidentate ligands make no rings.

Answer: (b) aqua

Explanation:
1. According to IUPAC rules, neutral ligands keep their names, with a few exceptions.
2. H₂O is named aqua, NH₃ is ammine, CO is carbonyl and NO is nitrosyl.

Teacher's Note:
1. Learn the four exceptions: aqua, ammine, carbonyl, nitrosyl.

Answer: (c) −1

Explanation:
1. Oxalate carries −2 each and water is neutral.
2. Charge = +3 + 2(−2) + 2(0) = −1.

Teacher's Note:
1. Add the metal charge to the sum of the ligand charges.

Answer: (a) [Co(NH₃)₃Cl₃]

Explanation:
1. Conductance rises with the number of ions formed.
2. [Co(NH₃)₃Cl₃] is neutral and gives no ions.
3. The others give 2, 3 and 4 ions respectively.

Teacher's Note:
1. A neutral complex does not ionise.

Answer: (d) FeSO₄.(NH₄)₂SO₄.6H₂O

Explanation:
1. A double salt breaks up completely into its simple ions in water.
2. Mohr's salt FeSO₄.(NH₄)₂SO₄.6H₂O gives Fe²⁺, NH₄⁺ and SO₄²⁻ ions.
3. The other three keep their complex ion intact.

Teacher's Note:
1. Double salts lose their identity in solution; complexes do not.

Answer: (b) tetracarbonylnickel(0)

Explanation:
1. CO is the neutral ligand carbonyl, and there are four of them.
2. The complex is neutral, so the metal is in the zero oxidation state, and the name has no -ate ending.

Teacher's Note:
1. Use a Roman numeral 0 for the zero oxidation state.

Answer: (c) potassium dicyanidoargentate(I)

Explanation:
1. The complex ion [Ag(CN)₂]⁻ is an anion, so silver takes its Latin stem and ends in -ate: argentate.
2. x + 2(−1) = −1, so Ag is +1.
3. Name: potassium dicyanidoargentate(I).

Teacher's Note:
1. Latin stems: ferrate (Fe), cuprate (Cu), argentate (Ag), aurate (Au).

Answer: (a) 5

Explanation:
1. K₄[Fe(CN)₆] dissociates into four K⁺ and one [Fe(CN)₆]⁴⁻.
2. Total = 4 + 1 = 5 ions.

Teacher's Note:
1. The complex ion stays whole.