JEE Chemistry Coordination Compounds MCQs Set 12

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. CN⁻ is a strong field ligand, so the Ni²⁺ electrons pair up and the complex is square planar with 0 unpaired electrons
2. Cl⁻ is weak, so the complex is tetrahedral with 2 unpaired electrons
3. R correctly explains A.

Teacher's Note:
1. Ligand field strength decides the geometry and magnetism of Ni²⁺ complexes.
2. CN⁻ is at the strong end of the spectrochemical series.

Answer: (c) A is true, but R is false.

Explanation:
1. The cis isomer of an octahedral [M(AA)₂b₂] complex is chiral and exists as two enantiomers, which form a racemic mixture: A is true
2. The complex is octahedral, not square planar, so R is false.

Teacher's Note:
1. Square planar complexes with two bidentate ligands cannot show this isomerism.
2. Racemic mixture means equal amounts of d and l forms.

Answer: (d) A is false, but R is true.

Explanation:
1. A square planar complex lies in a plane, which is itself a plane of symmetry, so it cannot be chiral. Hence A is false
2. The mirror image of Ma₂b₂ can be superimposed on the original, so R is true.

Teacher's Note:
1. Planar molecules are never optically active.
2. Square planar Ma₂b₂ has only cis and trans isomers.

Answer: (c) A is true, but R is false.

Explanation:
1. AgI is yellow because of charge transfer from I⁻ to Ag⁺, while AgF is colourless: A is true
2. Ag⁺ is d¹⁰ in both cases and has no unpaired electron, so R is false.

Teacher's Note:
1. The easily polarised I⁻ gives a low energy charge transfer band in the visible region.
2. d¹⁰ ions cannot show d–d transitions.

Answer: (c) A is true, but R is false.

Explanation:
1. F⁻ is a weak field ligand, so Δ₀ is smaller than the pairing energy and Co³⁺ stays high spin (4 unpaired electrons): A is true
2. The reason says F⁻ is strong, which is false.

Teacher's Note:
1. F⁻ is a weak field ligand and usually gives high spin complexes.
2. CN⁻ and CO are strong field ligands.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Fe(C₅H₅)₂ has the iron bonded to the π electrons of the cyclopentadienyl rings: A is true
2. The iron lies between two parallel rings, so it is a sandwich compound: R is true
3. R describes the shape but is not the explanation of π-bonding.

Teacher's Note:
1. The two rings donate six π electrons each, so Fe gets 18 electrons.
2. Sandwich compounds are always π-bonded.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Chromate is intensely yellow: A is true
2. Cr is +6: R is true
3. Cr(VI) is d⁰, so there is no d–d transition and the colour comes from charge transfer, so R is not the explanation.

Teacher's Note:
1. Colour in d⁰ species comes from ligand to metal charge transfer.
2. MnO₄⁻ is another example.

Answer: (c) A is true, but R is false.

Explanation:
1. [Fe(CN)₆]⁴⁻ is low spin d⁶ with 0 unpaired electrons, and [Fe(CN)₆]³⁻ is low spin d⁵ with 1 unpaired electron: A is true
2. The higher charge on Fe³⁺ gives a larger splitting in the ferricyanide ion, so R is false.

Teacher's Note:
1. Splitting increases with the oxidation state of the metal.
2. The magnetism depends on the d electron count, not on which Δ is larger.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. KCN reduces Cu(II) to Cu(I) and forms very stable [Cu(CN)₄]³⁻, which gives too few Cu⁺ ions to precipitate Cu₂S
2. [Cd(CN)₄]²⁻ is less stable, so enough Cd²⁺ is available to give yellow CdS: A is true
3. R is true and explains A.

Teacher's Note:
1. This is the classical method to separate Cd²⁺ from Cu²⁺.
2. A larger stability constant means fewer free metal ions.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Pink [Co(H₂O)₆]²⁺ changes to blue tetrahedral [CoCl₄]²⁻ in excess chloride: A is true
2. R names the cause of the colour change: R is true and explains A.

Teacher's Note:
1. Octahedral to tetrahedral change alters Δ and hence the colour.
2. Adding water reverses the change.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Glycinate is a bidentate anionic ligand: three gly⁻ neutralise the Co³⁺ charge and provide six donor atoms
2. A complex in which one ligand satisfies both the primary and the secondary valency is called an inner-metallic (inner-complex) compound: A and R are true and R explains A.

Teacher's Note:
1. Inner complexes are neutral chelates such as Ni(DMG)₂.
2. Primary valency is the oxidation state and secondary valency is the coordination number.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Δ_t ≈ 4/9 Δ_o, which is smaller than the pairing energy in nearly all cases, so tetrahedral complexes are almost always high spin
2. A and R are both true and R explains A.

Teacher's Note:
1. Four ligands give weaker splitting than six.
2. Low spin tetrahedral complexes are very rare.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In hydrazine the two N atoms are directly bonded, so chelation would give a three-membered ring with M, N and N
2. Such a ring is highly strained and unstable, so hydrazine acts only as a monodentate or bridging ligand
3. A and R are both true and R explains A.

Teacher's Note:
1. Stable chelate rings are five- or six-membered.
2. Ethylenediamine forms five-membered rings.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. NO₂⁻ > NH₃ > H₂O in ligand field strength; a stronger field absorbs at a shorter wavelength, so A is true
2. Complex stability and colour both depend on the ligand field, so R is true
3. R does not explain the wavelength order (which is explained by the splitting energy), so it is not the correct explanation.

Teacher's Note:
1. Wavelength of absorption is governed by Δ₀.
2. Stability is a thermodynamic property and colour is an energy gap.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Cr(I) with NO⁺ is d⁵
2. The strong field ligands CN⁻ and NO⁺ pair the electrons and two d orbitals are vacated for d²sp³, leaving 1 unpaired electron, so μ = 1.73 BM: A is true
3. R describes exactly this, so R explains A.

Teacher's Note:
1. d⁵ low spin has one unpaired electron.
2. μ = √3 = 1.73 BM.

Answer: (d) A is false, but R is true.

Explanation:
1. Zeise's salt K[PtCl₃(C₂H₄)] has three Cl⁻ and one ethene, so the coordination number of Pt is 4: A is false
2. Ethene bonds through its π electrons from one site and occupies one position, so it is monodentate: R is true.

Teacher's Note:
1. Ethene is counted as one ligand because it occupies one coordination site.
2. Pt(II) with four ligands is square planar.

Answer: (c) A is true, but R is false.

Explanation:
1. The brown ring test uses Fe²⁺ (FeSO₄), which reduces nitrate to NO: A is true
2. Nitrate is converted into NO, not NO₂, so R is false.

Teacher's Note:
1. Brown ring: [Fe(H₂O)₅NO]²⁺.
2. Fe²⁺ is oxidised while NO₃⁻ is reduced.

Answer: (c) A is true, but R is false.

Explanation:
1. Co³⁺ and Co²⁺ aqua ions have different d configurations and splitting, so the colour changes: A is true
2. Δ₀ is smaller for Co²⁺ (lower charge), and the crystal field stabilisation energy decreases on reduction, so R is false.

Teacher's Note:
1. Splitting energy falls with a lower oxidation state.
2. CFSE is larger for the 3+ ion.

Answer: (d) A is false, but R is true.

Explanation:
1. H₂O is a weaker ligand than NH₃, so Δ₀ changes and the wavelength of absorption changes: A is false
2. [Co(NH₃)₆]³⁺ is a low spin complex, so R is also false.

Teacher's Note:
1. Replace a ligand and the splitting and the colour change.
2. Co(III) ammines are low spin.

Answer: (d) A is false, but R is true.

Explanation:
1. Triethylenediamine is a neutral molecule, not an anion, so A is false
2. Propylenediamine can be 1,2-diaminopropane or 1,3-diaminopropane; complexes containing these different ligands are ligand isomers, so R is true.

Teacher's Note:
1. Ligand isomerism arises when the free ligands themselves are isomers.
2. Ethylenediamine (en) is neutral and bidentate.

Answer: (d) A is false, but R is true.

Explanation:
1. [Cr(NH₃)₅(NCS)][ZnCl₄] has a complex cation and a complex anion, so its isomers are coordination isomers (exchange of ligands between Cr and Zn) or linkage isomers, not ionisation isomers: A is false
2. Cr³⁺ is d³, so it is paramagnetic and d–d transitions occur: R is true.

Teacher's Note:
1. Ionisation isomers exchange a ligand with a free counter ion.
2. Coloured d³ complexes have d–d transitions.

Answer: (d) A is false, but R is true.

Explanation:
1. The two compounds contain different metals (Cu, Hg and Co) in different amounts, so they have different molecular formulae and cannot be isomers: A is false
2. SCN⁻ can bind through S or N, so it is ambidentate: R is true.

Teacher's Note:
1. Isomers must have the same molecular formula.
2. Ambidentate ligands cause linkage isomerism only when the same formula is kept.