CBSE Class 12 Chemistry Coordination Compounds MCQs Set 18

Answer: (c) dsp², square planar

Explanation:
1. Ni²⁺ is 3d⁸. CN⁻ is a strong field ligand and pairs the 3d electrons, leaving one 3d orbital empty.
2. One 3d, one 4s and two 4p orbitals hybridise: dsp².
3. The shape is square planar and the ion is diamagnetic.

Teacher's Note:
1. Strong field ligands with Ni(II) give square planar complexes.
2. Weak field ligands such as Cl⁻ give tetrahedral ones.

Answer: (a) 2

Explanation:
1. Ni²⁺ is 3d⁸, with two unpaired electrons.
2. Cl⁻ is a weak field ligand and cannot pair them, so the ion is sp³ hybridised and tetrahedral.
3. It keeps two unpaired electrons.

Teacher's Note:
1. A weak field ligand leaves the electron arrangement of the free ion unchanged.

Answer: (d) [Co(NH₃)₆]³⁺

Explanation:
1. An inner orbital complex uses 3d orbitals (d²sp³).
2. NH₃ is a strong field ligand and pairs the 3d⁶ electrons of Co³⁺, so [Co(NH₃)₆]³⁺ is inner orbital.
3. F⁻ is a weak field ligand and Ni²⁺ (3d⁸) has no pair of free 3d orbitals to offer, so those use 4d orbitals (outer orbital).

Teacher's Note:
1. Inner orbital means (n−1)d orbitals are used in hybridisation.
2. Inner orbital complexes are low spin.

Answer: (b) 1.73 BM

Explanation:
1. Fe³⁺ is 3d⁵. CN⁻ is a strong field ligand, so the electrons pair to leave one unpaired electron.
2. μ = \( \sqrt{1(1+2)} \) = 1.73 BM.

Teacher's Note:
1. Find n first, then apply the formula μ = √n(n+2) BM.

Answer: (d) sp³d² hybridised and paramagnetic

Explanation:
1. F⁻ is a weak field ligand and cannot pair the 3d⁶ electrons of Co³⁺.
2. Outer 4d orbitals are used, so the hybridisation is sp³d².
3. Four unpaired electrons remain, so the complex is paramagnetic.

Teacher's Note:
1. Weak field ligand: outer orbital, high spin, paramagnetic.

Answer: (b) tetrahedral and diamagnetic

Explanation:
1. Nickel in [Ni(CO)₄] is in the zero oxidation state: 3d⁸4s².
2. CO is a strong field ligand and the electrons pair up as 3d¹⁰, so there are no unpaired electrons.
3. The 4s and three 4p orbitals hybridise (sp³), giving a tetrahedral shape.

Teacher's Note:
1. Zero oxidation state with CO gives a full 3d¹⁰ arrangement.

Answer: (a) d(x²−y²) and d(z²)

Explanation:
1. The ligands approach along the axes.
2. d(x²−y²) and d(z²) point along the axes, so they are repelled most and form the e_g set at higher energy.
3. dxy, dyz and dzx lie between the axes and form the lower t_2g set.

Teacher's Note:
1. e_g is two orbitals along the axes; t_2g is three orbitals between the axes.

Answer: (c) Δt = 4/9 Δ₀

Explanation:
1. A tetrahedral complex has only four ligands, and none points directly at the d orbitals.
2. As a result the splitting is only about 4/9 of the octahedral value.

Teacher's Note:
1. Remember: Δt is roughly less than half of Δ₀.
2. This is why tetrahedral complexes are high spin.

Answer: (b) CN⁻

Explanation:
1. The spectrochemical series runs I⁻ < H₂O < NH₃ < CN⁻.
2. CN⁻ is the strongest field ligand of the four.

Teacher's Note:
1. Weak field: I⁻, Br⁻, Cl⁻, F⁻. Strong field: NH₃, en, CN⁻, CO.

Answer: (d) t₂g³ e_g¹

Explanation:
1. When Δ₀ is less than the pairing energy P, it costs less energy to put the fourth electron in e_g than to pair it in t₂g.
2. So the arrangement is t₂g³ e_g¹, with four unpaired electrons (high spin).

Teacher's Note:
1. Weak field means Δ₀ < P and high spin.

Answer: (a) t₂g⁶ e_g⁰ and is diamagnetic

Explanation:
1. Fe²⁺ is d⁶. CN⁻ is a strong field ligand, so Δ₀ > P and all six electrons pair in t₂g.
2. The arrangement is t₂g⁶ e_g⁰ with no unpaired electrons.

Teacher's Note:
1. Strong field with d⁶ gives the fully paired low spin arrangement.

Answer: (c) the single d electron absorbs visible light and moves from t₂g to e_g

Explanation:
1. Ti³⁺ is d¹ with its electron in the t₂g level.
2. It absorbs blue-green light to jump to e_g, and the transmitted colour is purple.

Teacher's Note:
1. Colour of a complex = complement of the colour it absorbs.

Answer: (b) Zn²⁺ has a completely filled d subshell, so no d-d transition is possible

Explanation:
1. Zn²⁺ is 3d¹⁰.
2. With every d orbital full, an electron cannot be excited to a higher d level, so no visible light is absorbed.

Teacher's Note:
1. d⁰ and d¹⁰ ions give colourless complexes.

Answer: (d) linkage isomerism

Explanation:
1. NO₂⁻ is ambidentate and binds through N in the first and through O in the second.
2. Isomers that differ in the donor atom of one ligand show linkage isomerism.

Teacher's Note:
1. Ambidentate ligands lead to linkage isomerism.

Answer: (c) cis and trans forms

Explanation:
1. In [MA₂B₂] the two identical ligands can be adjacent (cis) or opposite (trans).
2. So [Pt(NH₃)₂Cl₂] has two geometrical isomers.

Teacher's Note:
1. Fac and mer apply to octahedral MA₃B₃ complexes.

Answer: (a) cis-[CoCl₂(en)₂]⁺

Explanation:
1. cis-[CoCl₂(en)₂]⁺ has no plane or centre of symmetry, so its mirror image cannot be superimposed.
2. The trans form has a plane of symmetry. Square planar and simple tetrahedral complexes such as the last two are not chiral.

Teacher's Note:
1. Optical isomers are non-superimposable mirror images (enantiomers).

Answer: (b) ionisation isomerism

Explanation:
1. The two formulae are the same, but SO₄²⁻ and Br⁻ swap places between the coordination sphere and the counter ion.
2. The first gives SO₄²⁻ ions (white precipitate with BaCl₂) and the second gives Br⁻ ions (precipitate with AgNO₃).

Teacher's Note:
1. Ionisation isomers give different ions in solution.

Answer: (c) solvate (hydrate) isomers

Explanation:
1. They have the same formula CrCl₃.6H₂O.
2. The number of water molecules inside the coordination sphere differs, which is solvate isomerism.

Teacher's Note:
1. Hydrate isomers differ in how many water molecules are ligands.

Answer: (d) dimethylglyoxime

Explanation:
1. Dimethylglyoxime (DMG) reacts with Ni²⁺ in ammoniacal solution to give a bright red chelate precipitate.
2. The precipitate is weighed to estimate nickel.

Teacher's Note:
1. DMG is didentate and forms five-membered chelate rings.

Answer: (a) [Ni(CO)₄]

Explanation:
1. Impure nickel reacts with carbon monoxide at about 330-350 K to form volatile [Ni(CO)₄].
2. On stronger heating this tetracarbonyl decomposes and deposits pure nickel, while CO is recovered.

Teacher's Note:
1. Volatile nickel tetracarbonyl allows separation from non-volatile impurities.