JEE Chemistry Coordination Compounds MCQs Set 10

Answer: (c) [Cr(NH₃)₆]³⁺

Explanation:
1. Zn²⁺ is d¹⁰ with 0 unpaired electrons
2. Ti³⁺ is d¹ with 1 unpaired electron
3. Cr³⁺ is d³ with 3 unpaired electrons
4. Co³⁺ is d⁶ and in the strong ammonia field it is low spin with 0 unpaired electrons
5. Cr³⁺ complex is the most paramagnetic.

Teacher's Note:
1. Count unpaired electrons in each metal ion.
2. NH₃ is a strong enough ligand to make Co³⁺ complexes low spin.

Answer: (a) [Co(NH₃)₅Cl]²⁺

Explanation:
1. Longer wavelength means smaller Δ₀ (E = hc/λ)
2. Cl⁻ is the weakest field ligand in this set (H₂O < NH₃ < en, and Cl⁻ is below H₂O)
3. So [Co(NH₃)₅Cl]²⁺ has the smallest Δ₀ and absorbs the longest wavelength.

Teacher's Note:
1. Weaker field ligand means smaller splitting, so absorption at a longer wavelength.
2. Spectrochemical order: Cl⁻ < H₂O < NH₃ < en.

Answer: (c) [PtCl₄]²⁻

Explanation:
1. [Cu(NH₃)₄]²⁺ and [Cu(H₂O)₄]²⁺: Cu²⁺ is d⁹, 1 unpaired electron
2. [NiCl₄]²⁻: tetrahedral Ni²⁺, 2 unpaired electrons
3. [PtCl₄]²⁻: Pt²⁺ is d⁸, square planar dsp², all electrons paired.

Teacher's Note:
1. Pt²⁺ always forms square planar, diamagnetic d⁸ complexes.
2. Odd electron ions such as Cu²⁺ are always paramagnetic.

Answer: (c) [Ni(NH₃)₆]²⁺

Explanation:
1. Cr³⁺ and Co³⁺ ammines are inner orbital (d²sp³) complexes
2. [Ni(NH₃)₆]²⁺: Ni²⁺ is d⁸ with no vacant 3d orbital pair, so it uses outer 4d orbitals (sp³d²) and keeps 2 unpaired electrons
3. [Zn(NH₃)₆]²⁺ is outer orbital too, but d¹⁰ is diamagnetic.

Teacher's Note:
1. Outer orbital complexes use 4s, 4p and 4d orbitals (sp³d²).
2. d⁸ with six ligands cannot free two 3d orbitals, so it is outer orbital.

Answer: (a) The red complex has tetrahedral geometry

Explanation:
1. The red complex is bis(dimethylglyoximato)nickel(II)
2. Ni²⁺ is dsp² hybridised and the complex is square planar and diamagnetic, so the statement of tetrahedral geometry is false
3. DMG is bidentate and the complex has two O–H···O hydrogen bonds.

Teacher's Note:
1. This is the standard test for Ni²⁺.
2. Ni(DMG)₂ is square planar, not tetrahedral.

Answer: (d) \(-\frac{12}{5}\)Δ₀ + 3P

Explanation:
1. Low spin d⁶ is t₂g⁶ e_g⁰
2. CFSE = 6 × (−0.4Δ₀) = −2.4Δ₀ = −12/5 Δ₀
3. All six electrons are paired in three t₂g orbitals, so three pairing energies are included: 3P
4. Energy = −12/5 Δ₀ + 3P.

Teacher's Note:
1. Count the pairs in the t₂g set: six electrons make three pairs.
2. −2.4 can also be written as −12/5.

Answer: (b) [Cu(NH₃)₄]²⁺

Explanation:
1. 1.73 BM corresponds to one unpaired electron (√3)
2. [CoCl₆]⁴⁻ has 3 unpaired electrons
3. [Cu(NH₃)₄]²⁺: Cu²⁺ d⁹ has 1 unpaired electron
4. [Ni(CN)₄]²⁻ has 0 and TiCl₄ (Ti⁴⁺, d⁰) has 0.

Teacher's Note:
1. μ = √(n(n+2)) gives 1.73 BM for n = 1.
2. d⁹ ions always have one unpaired electron.

Answer: (b) cis-[PtCl₂(NH₃)₂]

Explanation:
1. Cisplatin, cis-[PtCl₂(NH₃)₂], binds to DNA and stops the division of cancer cells.

Teacher's Note:
1. Only the cis isomer of the platinum complex is active.
2. It is used in the treatment of several cancers.

Answer: (d) [Co(NO₂)Br(NH₃)₄]Cl

Explanation:
1. Linkage isomerism needs an ambidentate ligand such as NO₂⁻
2. Ionisation isomerism needs different anions, one in the sphere and one outside, that can exchange
3. [Co(NO₂)Br(NH₃)₄]Cl has NO₂⁻ (ambidentate) and also Br⁻ inside with Cl⁻ outside, so both occur
4. In options 1 and 2 the counter ion is the same NO₂⁻ as the ligand, so exchange gives only linkage isomers; [Co(NH₃)₆]Cl₂ has neither.

Teacher's Note:
1. Check for an ambidentate ligand and for two different anions.
2. Exchange of identical groups gives no new isomer.

Answer: (d) CoCl₃·3NH₃

Explanation:
1. CoCl₃·3NH₃ is [Co(NH₃)₃Cl₃], with all three Cl⁻ inside the coordination sphere
2. It gives no free Cl⁻ and so no precipitate with AgNO₃.

Teacher's Note:
1. Only ionisable chloride gives AgCl with silver nitrate.
2. Coordination number 6 means 3 + 3 here, so all chlorides are bonded to cobalt.

Answer: (d) [Co(CN)₆]³⁻ has no unpaired electrons and will be in a low-spin configuration

Explanation:
1. Co³⁺ is d⁶
2. CN⁻ is a strong field ligand, so the electrons pair up as t₂g⁶ e_g⁰
3. There are no unpaired electrons and the complex is low spin.

Teacher's Note:
1. Strong field ligand means low spin.
2. t₂g⁶ is completely filled, so the complex is diamagnetic.

Answer: (b) hexacyanidoferrate(III) ion

Explanation:
1. Six cyanide ligands: hexacyanido
2. Iron in an anionic complex takes the name ferrate
3. Oxidation state: x + 6(−1) = −3, so x = +3.

Teacher's Note:
1. Anionic complexes end in -ate and use the Latin name for iron (ferrate).
2. IUPAC names cyanide ligand as cyanido.

Answer: (c) dsp²

Explanation:
1. Ni²⁺ is 3d⁸; CN⁻ is a strong field ligand and pairs the electrons
2. One 3d orbital is vacated, so dsp² hybridisation occurs
3. The geometry is square planar.

Teacher's Note:
1. Strong field ligands with d⁸ give dsp² square planar complexes.
2. The complex is diamagnetic.

Answer: (c) 9

Explanation:
1. en and oxalate are bidentate: 2 × 2 + 2 = 6, so the coordination number is 6
2. Charge of the complex cation = +1: x + 0 + (−2) = +1, so x = +3
3. Sum = 6 + 3 = 9.

Teacher's Note:
1. Count donor atoms, not ligands, for the coordination number.
2. The outer Cl⁻ gives the charge on the complex ion.

Answer: (a) 3

Explanation:
1. The cation is [Co(en)₂Cl₂]⁺ of the type [M(AA)₂b₂]
2. Trans form: 1 (achiral)
3. Cis form: 2 optical isomers
4. Total = 3.

Teacher's Note:
1. [M(AA)₂b₂] has 3 stereoisomers in all.
2. The cis isomer is chiral.

Answer: (d) d–d transition in Cr³⁺

Explanation:
1. Cr³⁺ replaces some Al³⁺ in the octahedral holes of the Al₂O₃ lattice
2. Cr³⁺ is d³ and has partly filled d orbitals
3. Absorption of visible light by d–d transition of Cr³⁺ gives the red colour.

Teacher's Note:
1. Al³⁺ has no d electrons and cannot show d–d transitions.
2. The same ion in beryl (emerald) gives green.

Answer: (d) oxalate

Explanation:
1. t₂g⁶ e_g⁰ is the low spin d⁶ configuration, which needs a strong enough field
2. Cl⁻, F⁻ and S-bonded SCN⁻ are weak field ligands, which leave Co³⁺ high spin
3. Oxalate is the strongest of the four, and [Co(C₂O₄)₃]³⁻ is low spin and diamagnetic.

Teacher's Note:
1. Position in the spectrochemical series decides the spin state.
2. Co³⁺ is low spin with all ligands except F⁻ and a few others.

Answer: (c) [Fe(CO)₄]²⁻

Explanation:
1. Back bonding lengthens the C–O bond
2. Among the isoelectronic carbonyls Ni(CO)₄, [Co(CO)₄]⁻ and [Fe(CO)₄]²⁻, the greater negative charge gives more back donation
3. [Fe(CO)₄]²⁻ has the largest negative charge, so the longest C–O bond. [Mn(CO)₆]⁺ has the shortest.

Teacher's Note:
1. Isoelectronic species: compare the charge on the metal.
2. More negative charge on the metal means more π back bonding.

Answer: (d) CN⁻ > C₆H₅⁻ > Br⁻ > NH₃

Explanation:
1. The trans effect series is CN⁻, CO, C₂H₄ > PR₃, H⁻ > CH₃⁻, C₆H₅⁻, SCN⁻, I⁻ > Br⁻, Cl⁻ > py, NH₃ > OH⁻, H₂O
2. So the order is CN⁻ > C₆H₅⁻ > Br⁻ > NH₃.

Teacher's Note:
1. π-acceptor and strong σ-donor ligands have the largest trans effect.
2. The trans effect is used to plan the synthesis of cis and trans platinum complexes.

Answer: (d) d⁸

Explanation:
1. Strong Jahn–Teller distortion occurs when the e_g set is unevenly occupied (high spin d⁴, d⁹, low spin d⁷)
2. High spin d⁷ has uneven t₂g occupation and shows only weak distortion
3. High spin d⁸ is t₂g⁶ e_g², a symmetrical arrangement, so no distortion.

Teacher's Note:
1. Distortion occurs for degenerate orbitals with unequal electron occupancy.
2. d⁸ octahedral is symmetrical and does not distort.

Answer: (a) Na₃[Fe(C₂O₄)₃]

Explanation:
1. Chelating ligands form much more stable complexes than monodentate ligands (chelate effect)
2. Oxalate in [Fe(C₂O₄)₃]³⁻ forms three five-membered chelate rings with Fe³⁺
3. The other complexes have only monodentate ligands, so the tris(oxalato)ferrate(III) is the most stable.

Teacher's Note:
1. Each chelate ring adds extra stability because of the gain in entropy.
2. Ring size of five or six is most favourable.

Answer: (c) π-bond containing ligands are classical

Explanation:
1. CO and NH₃ (neutral) are stronger than many anions, so 1 is true
2. Chelating ligands bind more strongly, so 2 is true
3. Ligands that contain π bonds (such as C₂H₄, CO) can accept electrons into π* orbitals; the π-bonded ligands like alkenes are non-classical, so 3 is incorrect
4. A less electronegative donor donates its pair more readily, so 4 is true.

Teacher's Note:
1. Non-classical ligands accept as well as donate electron density.
2. Classical ligands are simple σ donors.

Answer: (b) Fe₂(SO₄)₃ is used in this process

Explanation:
1. Nitrate is reduced to NO by Fe²⁺ in acid, and Fe²⁺ is oxidised to Fe³⁺
2. Freshly prepared FeSO₄ is used, not Fe₂(SO₄)₃
3. The brown ring is [Fe(H₂O)₅NO]²⁺.

Teacher's Note:
1. Use FeSO₄ solution and conc. H₂SO₄ down the side of the test tube.
2. The ring forms at the junction of the two layers.