JEE Chemistry Coordination Compounds MCQs Set 09

Answer: (c) [Cu(CN)₄]³⁻

Explanation:
1. Cu²⁺ is reduced by CN⁻ to Cu⁺ and cyanogen is evolved: 2Cu²⁺ + 4CN⁻ → 2CuCN + (CN)₂
2. CuCN dissolves in excess KCN: CuCN + 3CN⁻ → [Cu(CN)₄]³⁻
3. The soluble complex is therefore K₃[Cu(CN)₄].

Teacher's Note:
1. CN⁻ is both a reducing agent and a strong ligand here.
2. Cu(I) with CN⁻ forms a stable tetrahedral complex, which is used to separate copper from cadmium.

Answer: (c) linkage isomerism, ionisation isomerism and geometrical isomerism

Explanation:
1. NO₂⁻ is ambidentate (nitro or nitrito), so linkage isomerism is possible
2. NO₂⁻ and Cl⁻ can swap between the sphere and the counter ion, giving [Co(NH₃)₄(NO₂)Cl]NO₂, so ionisation isomerism is possible
3. The two NO₂ groups can be cis or trans, so geometrical isomerism is possible
4. Both the cis and trans forms have a plane of symmetry, so there is no optical isomerism.

Teacher's Note:
1. Optical isomerism in octahedral complexes needs a chiral arrangement such as cis-[M(AA)₂X₂].
2. Check each type of isomerism separately for the given formula.

Answer: (b) [Cr(H₂O)₆]²⁺, [Fe(H₂O)₆]²⁺

Explanation:
1. Spin only moment depends only on the number of unpaired electrons
2. [Cr(H₂O)₆]²⁺: Cr²⁺ is d⁴ with weak field water, 4 unpaired electrons
3. [Fe(H₂O)₆]²⁺: Fe²⁺ is d⁶ with weak field water, 4 unpaired electrons
4. Others: [CoCl₄]²⁻ has 3, [Mn(H₂O)₆]²⁺ has 5. So option 2 is correct.

Teacher's Note:
1. Count unpaired electrons first; μ = √(n(n+2)) BM.
2. Water is a weak field ligand and gives high spin complexes with Cr²⁺ and Fe²⁺.

Answer: (c) 3

Explanation:
1. The complex is of the type [M(AA)₂b₂]
2. The trans form is one isomer (achiral)
3. The cis form is chiral and exists as a pair of enantiomers (d and l)
4. Total = 1 + 2 = 3.

Teacher's Note:
1. [M(AA)₂b₂] gives 1 trans and 2 cis (optical) isomers.
2. Always check the cis form for optical activity.

Answer: (a) NH₃, Cl

Explanation:
1. Cisplatin is cis-[Pt(NH₃)₂Cl₂]
2. Its ligands are ammonia and chloride.

Teacher's Note:
1. Cisplatin is a square planar Pt(II) complex.
2. Only the cis isomer is active; the trans isomer is not.

Answer: (c) V₂O₅ : Haber–Bosch process

Explanation:
1. Wilkinson's catalyst hydrogenates alkenes, so pair 1 is correct
2. The Ziegler–Natta catalyst polymerises alkenes, so pair 2 is correct
3. Nickel is used in hydrogenation, so pair 4 is correct
4. The Haber–Bosch process uses iron (with molybdenum as promoter), and V₂O₅ is the catalyst of the contact process, so pair 3 is the mismatch.

Teacher's Note:
1. V₂O₅ is used in the contact process for H₂SO₄.
2. Fe with Mo promoter is used for ammonia synthesis.

Answer: (a) (i) and (iv)

Explanation:
1. XeF₄: sp³d², 4 bond pairs and 2 lone pairs, square planar
2. SF₄: see-saw
3. [NiCl₄]²⁻: Cl⁻ is a weak ligand, so it is tetrahedral (sp³)
4. [PdCl₄]²⁻: Pd²⁺ is d⁸ (4d), always square planar. Hence (i) and (iv).

Teacher's Note:
1. d⁸ ions of the 4d and 5d series (Pd²⁺, Pt²⁺) are square planar.
2. Ni²⁺ is square planar only with strong field ligands such as CN⁻.

Answer: (b) [Co(en)₂Cl₂]Cl

Explanation:
1. Enantiomorphs (mirror-image isomers) need a chiral complex
2. cis-[Co(en)₂Cl₂]⁺ has no plane or centre of symmetry, so it exists as d and l forms
3. The other three cations or anions have a plane of symmetry or no chiral arrangement.

Teacher's Note:
1. Complexes containing two bidentate ligands in cis positions are chiral.
2. Complexes of the type [M(NH₃)₄X₂] are achiral.

Answer: (b) [Ni(H₂O)₆]²⁺

Explanation:
1. Water is a weak field ligand, so all complexes are high spin
2. Unpaired electrons: Cr²⁺ = 4, Mn²⁺ = 5, Fe²⁺ = 4, Ni²⁺ = 2
3. Ni²⁺ has the fewest, so [Ni(H₂O)₆]²⁺ is the least paramagnetic.

Teacher's Note:
1. Minimum paramagnetism means the fewest unpaired electrons.
2. d⁸ in an octahedral field always has 2 unpaired electrons (t₂g⁶ e_g²).

Answer: (a) [Co(CN)₆]³⁻

Explanation:
1. The metal ion is Co³⁺ in all four, so only the ligand matters
2. Spectrochemical series: H₂O < C₂O₄²⁻ < NH₃ < CN⁻
3. CN⁻ is the strongest field ligand, so Δ₀ is maximum for [Co(CN)₆]³⁻.

Teacher's Note:
1. Δ₀ increases along the spectrochemical series.
2. CN⁻ and CO are π-acceptor ligands and give the largest splitting.

Answer: (d) [Cr(NH₃)₆]³⁺

Explanation:
1. Visible light is absorbed through d–d transitions, which need a partly filled d subshell
2. Zn²⁺ is d¹⁰, Sc³⁺ is d⁰ and Ti⁴⁺ is d⁰, so these are colourless
3. Cr³⁺ is d³, so [Cr(NH₃)₆]³⁺ absorbs visible light (it is yellow).

Teacher's Note:
1. d⁰ and d¹⁰ ions give colourless complexes.
2. Transition metal complexes are coloured because of d–d transitions.

Answer: (d) TiF₆²⁻ and Cu₂Cl₂

Explanation:
1. TiF₆²⁻: Ti⁴⁺ is d⁰, so it is colourless
2. Cu₂Cl₂: Cu⁺ is d¹⁰, so it is colourless
3. CoF₆³⁻ (Co³⁺, d⁶) and NiCl₄²⁻ (Ni²⁺, d⁸) have unpaired d electrons and are coloured.

Teacher's Note:
1. No d–d transition is possible in d⁰ or d¹⁰ ions.
2. Look at the oxidation state of the metal first.

Answer: (c) [Co(NH₃)₃Cl₃]⁰

Explanation:
1. [Co(en)₃]³⁺ and cis-[Co(en)₂Cl₂]⁺ are chiral
2. [Co(NH₃)₃Cl₃] exists only as fac and mer forms, and both have a plane of symmetry, so neither is optically active
3. [Co(en)Cl₂(NH₃)₂]⁺ has an isomer with Cl, Cl, NH₃, NH₃ arranged all cis that is chiral.

Teacher's Note:
1. Complexes of the type [MA₃B₃] show geometrical but not optical isomerism.
2. Optical activity needs the absence of any plane of symmetry.

Answer: (b) [NiCl₄]²⁻

Explanation:
1. [Pt(NH₃)₂Cl₂] is square planar and shows cis and trans forms
2. [Ni(en)₃]²⁺ is chiral, so it shows optical isomerism
3. [Ni(NH₃)₄(H₂O)₂]²⁺ shows cis and trans forms
4. [NiCl₄]²⁻ is tetrahedral with identical ligands and shows no isomerism.

Teacher's Note:
1. A tetrahedral complex with one kind of ligand has no isomers.
2. Square planar [MA₂B₂] and octahedral [MA₄B₂] show geometrical isomerism.

Answer: (b) [Ni(CN)₄]²⁻

Explanation:
1. Ni²⁺ (d⁸), Cr³⁺ (d³) and Fe²⁺ (d⁶) complexes of water or ammonia are coloured by d–d transitions in the visible region
2. CN⁻ is a very strong ligand; in [Ni(CN)₄]²⁻ the large splitting moves the absorption into the ultraviolet region, so the ion is practically colourless.

Teacher's Note:
1. A very large Δ shifts the absorption to a higher energy region (UV).
2. Strong field ligands like CN⁻ often give colourless or pale complexes.

Answer: (c) geometrical isomerism

Explanation:
1. Same formula with different spatial arrangement of Cl⁻ and NH₃ gives cis (violet) and trans (green) forms
2. This is geometrical isomerism.

Teacher's Note:
1. Different colours with the same formula point to geometrical isomers.
2. [MA₄B₂] octahedral complexes have cis and trans forms.

Answer: (a) −0.6 Δ₀

Explanation:
1. High spin d⁴ is t₂g³ e_g¹
2. CFSE = 3(−0.4Δ₀) + 1(+0.6Δ₀) = −1.2Δ₀ + 0.6Δ₀ = −0.6Δ₀.

Teacher's Note:
1. Each t₂g electron contributes −0.4Δ₀ and each e_g electron +0.6Δ₀.
2. Low spin d⁴ would be −1.6Δ₀ + P.

Answer: (a) [Fe(CO)₅]⁻

Explanation:
1. Back donation from metal d orbitals to the π* orbital of CO weakens the C–O bond and lengthens it
2. A negative charge on the metal increases back donation, a positive charge decreases it
3. The species with the most negative metal centre, [Fe(CO)₅]⁻, has the longest C–O bond.

Teacher's Note:
1. More electron density on the metal means stronger M→CO back bonding.
2. Stronger back bonding means a longer C–O bond and a shorter M–C bond.

Answer: (c) 8,000 cm⁻¹

Explanation:
1. For the same metal and ligand, Δt = (4/9)Δ₀
2. Δt = (4/9) × 18000 = 8000 cm⁻¹.

Teacher's Note:
1. Tetrahedral splitting is always smaller than octahedral splitting.
2. Use Δt ≈ 4/9 Δ₀.

Answer: (b) [Ni(CN)₄]²⁻

Explanation:
1. [NiCl₄]²⁻: Ni²⁺ d⁸, tetrahedral, 2 unpaired electrons
2. [Ni(CN)₄]²⁻: strong field CN⁻ pairs the electrons, giving square planar dsp² and no unpaired electrons
3. [CuCl₄]²⁻: Cu²⁺ d⁹, 1 unpaired electron
4. [CoF₆]³⁻: high spin d⁶, 4 unpaired electrons.

Teacher's Note:
1. Strong field ligands force pairing in d⁸ Ni²⁺ to give square planar complexes.
2. Diamagnetic means no unpaired electrons.

Answer: (c) Coordination isomerism

Explanation:
1. Both are salts with a complex cation and a complex anion
2. The metal ions Co³⁺ and Cr³⁺ exchange their ligands between the cation and the anion
3. This is coordination isomerism.

Teacher's Note:
1. Coordination isomerism needs both the cation and the anion to be complex.
2. Ligands are interchanged between the two metal centres.

Answer: (a) 3

Explanation:
1. It is a square planar complex of the type [Mabcd]
2. It has 3 geometrical isomers.

Teacher's Note:
1. Square planar [Mabcd] always has 3 isomers.
2. Fix one ligand and place the others trans to it in turn.

Answer: (b) Mn(CO)₆⁺

Explanation:
1. A positive charge on the metal reduces back donation to CO
2. Less back donation leaves the C–O bond closer to a triple bond, so it is strongest
3. Mn(CO)₆⁺ has the positive charge, so it has the strongest C–O bond.

Teacher's Note:
1. Charge on the metal decides the strength of back bonding.
2. Order of C–O bond strength: Mn(CO)₆⁺ > Cr(CO)₆ > V(CO)₆⁻.