JEE Chemistry Coordination Compounds MCQs Set 32

Answer: (a) 10, 1

Explanation:
1. Co₂(CO)₈ has six terminal CO and two bridging CO groups.
2. Terminal CO give 6 metal–carbon bonds and each bridging CO makes 2, giving 4 more, so 10 in all.
3. There is one Co–Co bond.
4. Answer: (a)

Teacher's Note:
1. Bridging CO bonds to both metals.
2. Co–Co bond: one.

Answer: (a) 3d¹_xy, 3d¹_yz, 3d¹_zx

Explanation:
1. Cr³⁺ is d³; \( \mu = \sqrt{3 \times 5} = 3.87 \) BM, matching the given value.
2. In an octahedral field the three electrons occupy the lower t_2g set (d_xy, d_yz, d_zx) singly, with parallel spins.
3. Answer: (a)

Teacher's Note:
1. d³ octahedral is t_2g³.
2. The e_g set contains d_z² and d_x²−y².

Answer: (d) chlorobis(ethylenediamine)nitrito-O-cobalt(III) ion

Explanation:
1. Ligands are named alphabetically: chlorido, bis(ethylenediamine), nitrito-O.
2. The ion is a cation, so the metal is named cobalt(III) with no ‘-ate’ ending.
3. Option \( (a) \) wrongly uses cobaltate, and \( (b) \) and \( (c) \) have wrong ligand prefixes and order.
4. Answer: (d)

Teacher's Note:
1. Cationic complex: no -ate ending.
2. ONO⁻ is nitrito-O.

Answer: (a) CN⁻ > CH₃⁻ > NO₂⁻ > Br⁻

Explanation:
1. In the trans effect series strong π-acceptors and σ-donors come first: CN⁻ > CH₃⁻ > NO₂⁻ > Br⁻.
2. Answer: (a)

Teacher's Note:
1. Trans effect: CN⁻, CO, C₂H₄ are strongest.
2. Halides are weak trans directors.

Answer: (a) one Co–Co bond, six terminal CO and two bridging CO

Explanation:
1. The bridged structure of Co₂(CO)₈ has a Co–Co bond, 6 terminal CO and 2 bridging CO.
2. This gives each cobalt an 18-electron configuration.
3. Answer: (a)

Teacher's Note:
1. Count electrons: Co \( (9) + 8 \) from 4 CO \( + 1 \) from Co–Co bond.
2. Bridging CO sits between two metals.

Answer: (a) greater than x

Explanation:
1. In Zeise's salt, ethene donates π electrons to Pt and Pt back-donates electrons into the π* orbital of ethene.
2. This lowers the C–C bond order, so the bond lengthens.
3. Answer: (a)

Teacher's Note:
1. Back-bonding fills π*.
2. Weaker bond means longer bond.

Answer: (a) 22σ and 15π

Explanation:
1. Fe₂(CO)₉ has six terminal CO (each C≡O has \( 1\sigma + 2\pi \), plus 1 Fe–C σ) and three bridging CO (each \( C=O \) has \( 1\sigma + 1\pi \), plus 2 Fe–C σ).
2. \( \sigma = 6 + 6 + 3 + 6 + 1 \) (Fe–Fe) = 22; \( \pi = 12 + 3 = 15 \).
3. Answer: (a)

Teacher's Note:
1. Terminal CO: triple bond; bridging CO: double bond.
2. Count Fe–C σ bonds carefully.

Answer: (a) 4 water molecules

Explanation:
1. In CuSO₄.5H₂O, written [Cu(H₂O)₄]SO₄.H₂O, four water molecules are bonded to Cu²⁺.
2. The fifth water molecule is hydrogen bonded to sulphate.
3. Answer: (a)

Teacher's Note:
1. Four H₂O are coordinated.
2. The fifth is lattice water.

Answer: (a) raised by (2/5)Δ_t

Explanation:
1. In a tetrahedral field the \( e \) set is lowered by (3/5)Δ_t and the t₂ set is raised by (2/5)Δ_t relative to the barycentre.
2. The weighted sum \( 2(-3/5) + 3(2/5) = 0 \) confirms this.
3. Answer: (a)

Teacher's Note:
1. Tetrahedral: \( e \) below, t₂ above.
2. Barycentre rule: total shift is zero.

Answer: (c) x = 2, y = 2

Explanation:
1. A square planar \( \ce{[PtX4]^{2-}} \) complex needs four ligands, so \( x + y = 4 \).
2. \( \ce{[PtCl2Br2]^{2-}} \) is of the type [Ma₂b₂] and exists as cis and trans isomers.
3. \( \ce{[PtClBr3]^{2-}} \) and \( \ce{[PtCl3Br]^{2-}} \) are of the type [Mab₃] with no isomers.
4. Answer: (c)

Teacher's Note:
1. Square planar Ma₂b₂: two isomers.
2. Ma₃b: one.

Answer: (a) \( \ce{[CoCl6]^{4-}} \)

Explanation:
1. The metal ion is the same, so Δ_o depends on the ligand and on the charge of the complex.
2. Cl⁻ is a weak field ligand while CN⁻, en and NH₃ are stronger, so \( \ce{[CoCl6]^{4-}} \) has the smallest splitting.
3. Answer: (a)

Teacher's Note:
1. Spectrochemical series: Cl⁻ < NH₃ < en < CN⁻.
2. Weak ligand means small Δ_o.

Answer: (c) Red complex has a tetrahedral geometry

Explanation:
1. The red Ni(dmg)₂ complex is square planar (dsp²), diamagnetic, with two O–H···O hydrogen bonds.
2. Dimethylglyoxime binds through two nitrogen atoms, so it is bidentate.
3. The statement that the geometry is tetrahedral is false.
4. Answer: (c)

Teacher's Note:
1. Ni(dmg)₂ is square planar.
2. Used for the test and estimation of nickel.

Answer: (a) thiosulphato

Explanation:
1. Oxalato, glycinato and ethylenediamine are bidentate and form rings with the metal.
2. Thiosulphato is bonded through a single sulphur atom, so it is monodentate and not a chelating ligand.
3. Answer: (a)

Teacher's Note:
1. Chelating ligands have two or more donor atoms that bind one metal.
2. Thiosulphate is a monodentate ligand.

Answer: (c) \( \ce{[HgI4]^{2-}} \)

Explanation:
1. HgI₂ dissolves in excess KI to form the soluble complex K₂[HgI₄].
2. HgI₂ + 2I⁻ → \( \ce{[HgI4]^{2-}} \); this is the basis of Nessler's reagent.
3. Answer: (c)

Teacher's Note:
1. Hg²⁺ forms tetrahedral \( \ce{[HgX4]^{2-}} \).
2. Nessler's reagent is alkaline K₂[HgI₄].

Answer: (d) measuring the solid state magnetic moment

Explanation:
1. Fe(II) (low spin d⁶) with NO⁺ gives a diamagnetic ion, whereas Fe(III) (d⁵) with neutral NO would leave unpaired electrons and be paramagnetic.
2. The magnetic moment therefore distinguishes the two descriptions; concentrations and decomposition do not.
3. Answer: (d)

Teacher's Note:
1. Different electron count gives different magnetism.
2. Magnetic measurement reveals unpaired electrons.