JEE Chemistry Coordination Compounds MCQs Set 28

Answer: (a) both are paramagnetic

Explanation:
1. Fe³⁺ is d⁵, so neither complex can be diamagnetic.
2. \( \ce{[Fe(CN)6]^{3-}} \) is low spin with 1 unpaired electron and \( \ce{[FeF6]^{3-}} \) is high spin with 5.
3. Answer: (a)

Teacher's Note:
1. Odd number of \( d \) electrons always leaves unpaired electrons.
2. Both are paramagnetic.

Answer: (c) Co³⁺

Explanation:
1. Co³⁺ accepts electron pairs from the ligands, so it is the Lewis acid.
2. NH₃ donates pairs and is the Lewis base.
3. Answer: (c)

Teacher's Note:
1. Central metal ion: electron pair acceptor.
2. Ligand: electron pair donor.

Answer: (c) \( \ce{[V(NH3)6]^{3+}} \)

Explanation:
1. Absorption in the visible region needs d–d transitions, which require partially filled \( d \) orbitals.
2. Sc³⁺, Ti⁴⁺ are d⁰ and Zn²⁺ is d¹⁰; only V³⁺ (d²) can absorb.
3. Answer: (c)

Teacher's Note:
1. d⁰ and d¹⁰ ions are colourless.
2. Colour comes from d–d transitions.

Answer: (d) \( \ce{[Ni(NH3)6]^{2+}} \)

Explanation:
1. \( \ce{[Ni(NH3)6]^{2+}} \): Ni²⁺ is d⁸, so no \( 3d \) orbital can be vacated for d²sp³ hybridisation and sp³d² (outer orbital) is used.
2. The other three are low spin d⁶ or d⁵ inner orbital complexes.
3. Answer: (d)

Teacher's Note:
1. d⁸ octahedral complexes are always outer orbital.
2. Inner orbital means d²sp³.

Answer: (d) Tetraethyl lead

Explanation:
1. Zeise's salt, ferrocene and dibenzene chromium have π bonds between the metal and ethene or arenes.
2. Tetraethyl lead, Pb(C₂H₅)4, has ordinary σ Pb–C bonds.
3. Answer: (d)

Teacher's Note:
1. π complexes involve π electrons of alkenes or arenes.
2. Zeise's salt: K[PtCl₃(C₂H₄)].

Answer: (a) \( \ce{[Mn(CO)6]^{+}} \)

Explanation:
1. Back-donation from metal to CO π* weakens the C–O bond.
2. The cationic complex \( \ce{[Mn(CO)6]^{+}} \) has the least electron density on the metal, so the least back-bonding and the strongest C–O bond.
3. Answer: (a)

Teacher's Note:
1. More negative charge on metal increases back-bonding.
2. Back-bonding lowers C–O bond order.

Answer: (a) Al(OC₂H₅)₃

Explanation:
1. In aluminium ethoxide the Al is bonded to oxygen, so it is not organometallic.
2. The others contain M–C bonds: Mg–C, Pt–C (π) and Ni–C.
3. Answer: (a)

Teacher's Note:
1. Organometallics have at least one M–C bond.
2. Alkoxides are not organometallic.

Answer: (d) less availability of Ag⁺ ions, as Cu cannot displace Ag from \( \ce{[Ag(CN)2]^{-}} \) ion

Explanation:
1. In \( \ce{[Ag(CN)2]^{-}} \) the concentration of free Ag⁺ is very small, so copper cannot displace silver and a smooth, adherent plate is formed by electrolysis.
2. With AgNO₃ copper would displace Ag from solution and give a poor coating.
3. Answer: (d)

Teacher's Note:
1. Complexation lowers free metal ion concentration.
2. A controlled deposit gives a smooth plating.

Answer: (b) [Co(NH₃)₃](NO₂)3

Explanation:
1. [Co(NH₃)₃(NO₂)₃] is a neutral complex with no ions, so it does not conduct.
2. The other three give 3, 5 and 2 ions respectively.
3. Answer: (b)

Teacher's Note:
1. Neutral complex: non-electrolyte.
2. Count ions per formula unit.

Answer: (b) \( \ce{[Co(ox)3]^{3-}} \)

Explanation:
1. \( \ce{[Co(ox)3]^{3-}} \): Co³⁺ is low-spin d⁶, with oxalate acting as a field strong enough to pair electrons, so it is diamagnetic.
2. \( \ce{[Fe(CN)6]^{3-}} \) has one, \( \ce{[FeF6]^{3-}} \) five and \( \ce{[CoF6]^{3-}} \) four unpaired electrons.
3. Answer: (b)

Teacher's Note:
1. Diamagnetic: all electrons paired.
2. Fluoride is a weak field ligand.

Answer: (c) 1.73 BM

Explanation:
1. Taking NO as NO⁺: \( x + 1 + 0 - 4 = -2 \), so Cr is \( +1 \) (d⁵).
2. With strong field ligands it is low spin with one unpaired electron.
3. \( \mu = \sqrt{1 \times 3} = 1.73 \) BM.
4. Answer: (c)

Teacher's Note:
1. Strong field ligands (CN⁻, NO) give low spin.
2. One unpaired electron gives 1.73 BM.

Answer: (b) \( \ce{[Co(CN)6]^{3-}} \)

Explanation:
1. Δ_o increases with ligand field strength and metal charge.
2. CN⁻ is the strongest field ligand here and the metal is Co³⁺, so \( \ce{[Co(CN)6]^{3-}} \) has the largest Δ_o.
3. Answer: (b)

Teacher's Note:
1. Spectrochemical series: H₂O < C₂O₄²⁻ < NH₃ < CN⁻.
2. Higher charge on metal also increases Δ_o.

Answer: (b) Y

Explanation:
1. \( X \) = \( \ce{[Ni(en)(NH3)4]^{2+}} \) has no stereoisomers.
2. \( Y \) = \( \ce{[Ni(en)2(NH3)2]^{2+}} \) is of the type [M(AA)2b2], with cis (optically active) and trans forms, so it has both kinds of isomerism.
3. \( Z \) = \( \ce{[Ni(en)3]^{2+}} \) shows optical isomerism only.
4. Answer: (b)

Teacher's Note:
1. [M(AA)2b2] has both geometrical and optical isomers.
2. Tris-chelate: optical only.

Answer: (b) Na₂[CdCl₄]

Explanation:
1. Cd²⁺ is d¹⁰, so there are no d–d transitions and \( \ce{[CdCl4]^{2-}} \) is colourless.
2. The Cu(II), Fe(II) and Fe(III) complexes are coloured.
3. Answer: (b)

Teacher's Note:
1. d¹⁰ compounds are colourless.
2. Zn²⁺ and Cd²⁺ salts are white.

Answer: (c) [CrCl₂(H₂O)₄]Cl.2H₂O

Explanation:
1. One third of chlorine precipitating means only one Cl⁻ is ionisable.
2. [CrCl₂(H₂O)₄]Cl·2H₂O has one Cl⁻ outside the sphere, whereas the last option would precipitate two of three.
3. Answer: (c)

Teacher's Note:
1. One free Cl⁻ gives 1/3 precipitation.
2. Coordination number of Cr(III) is 6.

Answer: (c) \( \ce{[Fe(C2O4)3]^{3-}} \)

Explanation:
1. Chelating ligands give extra stability through the chelate effect.
2. Oxalate is a bidentate chelate forming three five-membered rings, so \( \ce{[Fe(C2O4)3]^{3-}} \) is the most stable.
3. Answer: (c)

Teacher's Note:
1. Chelate effect increases stability.
2. Monodentate Cl⁻ and H₂O form weaker complexes.

Answer: (b) CN⁻ > en > NCS⁻ > Cl⁻

Explanation:
1. The spectrochemical series in decreasing field strength is CN⁻ > en > NCS⁻ > Cl⁻.
2. Answer: (b)

Teacher's Note:
1. Weak field ligands: halides.
2. Strong field ligands: CN⁻, CO.

Answer: (a) [PtCl₂(NH₃)₂]

Explanation:
1. Pt²⁺ (\( 5d \) d⁸) forms square planar complexes such as cisplatin.
2. \( \ce{[NiCl4]^{2-}} \), MnO₄⁻ and CrO₄²⁻ are tetrahedral.
3. Answer: (a)

Teacher's Note:
1. Pt(II) is always square planar.
2. Tetrahedral: oxo anions of \( 3d \) metals.

Answer: (a) CH₄

Explanation:
1. A ligand must have a lone pair (or π electrons) to donate.
2. CH₄ has only bonding pairs, so it cannot act as a ligand.
3. Answer: (a)

Teacher's Note:
1. CO, CN⁻ and Br⁻ have lone pairs.
2. No lone pair means no ligand.