JEE Chemistry Coordination Compounds MCQs Set 07

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Multiple Choice Questions

Question 1: Which of the following systems has the maximum number of unpaired electrons?
(a) d5 (octahedral, low spin)
(b) d8 (tetrahedral)
(c) d6 (octahedral, low spin)
(d) d3 (octahedral)
Show Answer & Explanation

Answer: (d) d3 (octahedral)

Explanation:
1. d5 low spin: 1; d8 tetrahedral: 2; d6 low spin: 0; d3: 3
2. d3 has the most.

Teacher's Note:
1. Low spin reduces unpaired electrons.
2. d3 has no choice of spin state.

Question 2: In an octahedral crystal field, the t2g orbitals are
(a) raised in energy by 0.4Δo
(b) lowered in energy by 0.4Δo
(c) raised in energy by 0.6Δo
(d) lowered in energy by 0.6Δo
Show Answer & Explanation

Answer: (b) lowered in energy by 0.4Δo

Explanation:
1. The t2g set is lowered by 0.4Δo and the eg set is raised by 0.6Δo relative to the barycentre.

Teacher's Note:
1. The weighted average energy stays unchanged: 3(−0.4) + 2(0.6) = 0.
2. CFSE = −0.4 n(t2g) + 0.6 n(eg).

Question 3: If Δo < P, the correct electronic configuration for a d4 system will be
(a) t2g4 eg0
(b) t2g3 eg1
(c) t2g0 eg4
(d) t2g2 eg2
Show Answer & Explanation

Answer: (b) t2g3 eg1

Explanation:
1. Δo < P means electron pairing costs more than promotion, so electrons spread out: high spin
2. d4 high spin: t2g3 eg1.

Teacher's Note:
1. Weak field means small Δo.
2. High spin d4 has four unpaired electrons.

Question 4: Match the complex ions A [CrF6]4−, B [MnF6]4−, C [Cr(CN)6]4−, D [Mn(CN)6]4− with the number of unpaired electrons: (i) one, (ii) two, (iii) three, (iv) four, (v) five. The correct match is
(a) A–iv, B–v, C–ii, D–i
(b) A–ii, B–i, C–iii, D–v
(c) A–iv, B–i, C–ii, D–v
(d) A–iv, B–v, C–iii, D–ii
Show Answer & Explanation

Answer: (a) A–iv, B–v, C–ii, D–i

Explanation:
1. A: Cr2+ d4 weak field: 4
2. B: Mn2+ d5 weak field: 5
3. C: Cr2+ d4 strong field: 2
4. D: Mn2+ d5 strong field: 1.

Teacher's Note:
1. Find the d count first.
2. Weak ligand F−, strong ligand CN−.

Question 5: Consider the formation reactions: (i) Fe2+ + 6NH3 ⇌ [Fe(NH3)6]2+ (Kf) and (ii) Fe2+ + 3en ⇌ [Fe(en)3]2+ (Kf′). Which is true?
(a) Kf > Kf′
(b) Kf < Kf′
(c) Kf = Kf′
(d) they cannot be compared
Show Answer & Explanation

Answer: (b) Kf < Kf′

Explanation:
1. en is a chelating ligand: the chelate effect (greater entropy gain) makes [Fe(en)3]2+ more stable
2. Hence Kf′ is larger.

Teacher's Note:
1. More particles are released when 3 en replace 6 NH3 or water.
2. Chelates are more stable than non-chelates.

Question 6: The most stable complex among the following is
(a) [Co(H2O)6]3+
(b) [Co(NH3)2(en)2]3+
(c) [Co(en)3]3+
(d) [Co(NH3)4(en)]3+
Show Answer & Explanation

Answer: (c) [Co(en)3]3+

Explanation:
1. The complex with the maximum number of chelate rings (three) is the most stable: [Co(en)3]3+.

Teacher's Note:
1. Number of rings: 3, 2, 1, 0.
2. Chelation increases stability.

Question 7: The theoretical spin-only magnetic moment when Fe(SCN)3 reacts with a solution containing F− ions to give a complex is
(a) 2.83 BM
(b) 3.87 BM
(c) 5.92 BM
(d) 1.73 BM
Show Answer & Explanation

Answer: (c) 5.92 BM

Explanation:
1. Fe(SCN)3 + F− forms [FeF6]3−
2. Fe3+ d5 high spin: five unpaired electrons
3. μ = √35 = 5.92 BM.

Teacher's Note:
1. The blood-red thiocyanate colour disappears (used to mask Fe3+).
2. F− is a weak field ligand.

Question 8: Which high spin complex has the largest crystal field stabilisation energy (CFSE)?
(a) [Mn(H2O)6]2+
(b) [Cr(H2O)6]2+
(c) [Mn(H2O)6]3+
(d) [Cr(H2O)6]3+
Show Answer & Explanation

Answer: (d) [Cr(H2O)6]3+

Explanation:
1. High spin CFSE (in Δo): Mn2+ d5: 0; Cr2+ d4: −0.6; Mn3+ d4: −0.6; Cr3+ d3: −1.2
2. Also Δo is larger for the +3 ion, so [Cr(H2O)6]3+ has the largest.

Teacher's Note:
1. d3 has the most stabilisation among these.
2. Higher charge increases Δo.

Question 9: Match the complex ions (P) [Mn(H2O)6]2+, (Q) [Cr(H2O)6]2+, (R) [Fe(H2O)6]2+, (S) [Cr(H2O)6]3+ with the CFSE: 1) 0.6Δo, 2) 0.4Δo, 3) 0, 4) 1.2Δo. The correct code is
(a) P–3, Q–1, R–2, S–4
(b) P–1, Q–2, R–3, S–4
(c) P–4, Q–3, R–2, S–1
(d) P–3, Q–1, R–4, S–2
Show Answer & Explanation

Answer: (a) P–3, Q–1, R–2, S–4

Explanation:
1. P: d5 high spin: 0
2. Q: d4 high spin t2g3 eg1: −1.2 + 0.6 = −0.6Δo
3. R: d6 high spin t2g4 eg2: −1.6 + 1.2 = −0.4Δo
4. S: d3: −1.2Δo.

Teacher's Note:
1. Water is weak field, so all are high spin.
2. CFSE = −0.4 n(t2g) + 0.6 n(eg).

Question 10: The square planar complex cisplatin [Pt(NH3)2Cl2] shows
(a) geometrical isomerism
(b) optical isomerism
(c) linkage isomerism
(d) none
Show Answer & Explanation

Answer: (a) geometrical isomerism

Explanation:
1. M a2 b2 square planar has cis and trans forms.

Teacher's Note:
1. Only the cis form is the anticancer drug.
2. Square planar complexes are optically inactive.

Question 11: The two compounds sulphatopentaamminecobalt(III) bromide and sulphatopentaamminecobalt(III) chloride represent
(a) linkage isomerism
(b) ionisation isomerism
(c) coordination isomerism
(d) no isomerism
Show Answer & Explanation

Answer: (d) no isomerism

Explanation:
1. They have different anions (Br− and Cl−), so they are different compounds with different formulae, not isomers.

Teacher's Note:
1. Isomers must have the same molecular formula.
2. Ionisation isomers swap a ligand and a counter ion.

Question 12: Which of the following complexes cannot exhibit geometrical isomerism?
(a) [Pt(NH3)2Cl(NO2)]
(b) [Pt(gly)2]
(c) [Cu(en)2]2+
(d) [Pt(H2O)(NH3)BrCl]
Show Answer & Explanation

Answer: (c) [Cu(en)2]2+

Explanation:
1. [Cu(en)2]2+ is square planar M(AA)2: only one arrangement
2. The others (Ma2bc, M(AB)2, Mabcd) show geometrical isomerism.

Teacher's Note:
1. Symmetrical chelates give no cis–trans isomerism.
2. Glycinate is an unsymmetrical bidentate ligand.

Question 13: Which of the following compounds will exhibit linkage isomerism?
(a) [Pt(NH3)2Cl2]
(b) [Co(NH3)5NO2]Cl2
(c) [Co(NH3)4Cl2]Cl
(d) [Co(en)2Cl2]Cl
Show Answer & Explanation

Answer: (b) [Co(NH3)5NO2]Cl2

Explanation:
1. Linkage isomerism needs an ambidentate ligand such as NO2−
2. Only [Co(NH3)5(NO2)]Cl2 contains one.

Teacher's Note:
1. Nitro (N) and nitrito (O) are the linkage forms.
2. The others contain only chloride and NH3.

Question 14: Which of the following complexes will show geometrical isomerism?
(a) [Pt(NH3)2Cl2]
(b) Ni(CO)4
(c) Na3[Ni(CN)4]
(d) K[Ag(CN)2]
Show Answer & Explanation

Answer: (a) [Pt(NH3)2Cl2]

Explanation:
1. Square planar M a2 b2 gives cis and trans
2. Tetrahedral Ni(CO)4 and linear [Ag(CN)2]− have none.

Teacher's Note:
1. Tetrahedral Ma2b2 has no geometrical isomers.
2. Geometrical isomerism appears in square planar and octahedral complexes.

Question 15: Which of the following complexes will show optical isomerism?
(a) [Cr(NH3)6]2+
(b) [Ni(H2O)6]2+
(c) [Pt(NH3)3Br]NO3
(d) [Cr(en)3]Cl3
Show Answer & Explanation

Answer: (d) [Cr(en)3]Cl3

Explanation:
1. [Cr(en)3]3+ has three bidentate ligands arranged like a propeller: non-superimposable mirror images.

Teacher's Note:
1. Octahedral M(AA)3 complexes are chiral.
2. The others have a plane of symmetry.

Question 16: [Cr(H2O)6]Cl3 and [Cr(H2O)4Cl2]Cl·2H2O are examples of
(a) linkage isomerism
(b) hydrate isomerism
(c) ligand isomerism
(d) none of these
Show Answer & Explanation

Answer: (b) hydrate isomerism

Explanation:
1. The two compounds differ in the number of water molecules inside the coordination sphere: hydrate (solvate) isomerism.

Teacher's Note:
1. They have the same formula CrCl3·6H2O.
2. They differ in colour: violet versus green.

Question 17: A planar complex of the type [Mabcd] gives
(a) two optical isomers
(b) two geometrical isomers
(c) three optical isomers
(d) three geometrical isomers
Show Answer & Explanation

Answer: (d) three geometrical isomers

Explanation:
1. Square planar Mabcd: fix a; b, c or d can be trans to a, giving three geometrical isomers.

Teacher's Note:
1. Planar complexes are not chiral.
2. Count by choosing the ligand trans to a.

Question 18: The theoretical number of geometrical isomers expected for the octahedral complex [Mabcdef] is
(a) 0
(b) 30
(c) 15
(d) 9
Show Answer & Explanation

Answer: (c) 15

Explanation:
1. Six different ligands: 15 geometrical isomers (30 stereoisomers including 15 enantiomeric pairs).

Teacher's Note:
1. Fix a; choose its trans partner (5 ways) and arrange the remaining four in 3 ways = 15.
2. Each of the 15 is chiral.

Question 19: Which of the following complexes produces a precipitate with AgNO3 and exists in two geometrical isomeric forms?
(a) PtCl2·4NH3
(b) PtCl2·3NH3
(c) PtCl4·4NH3
(d) PtCl4·2NH3
Show Answer & Explanation

Answer: (c) PtCl4·4NH3

Explanation:
1. PtCl4·4NH3 = [Pt(NH3)4Cl2]Cl2: ionic chloride gives AgCl and the octahedral M a4 b2 exists as cis and trans
2. [Pt(NH3)4]Cl2 and [Pt(NH3)3Cl]Cl have no geometrical isomers; [Pt(NH3)2Cl4] has no ionic Cl.

Teacher's Note:
1. Both conditions must hold.
2. Octahedral Ma4b2 gives cis and trans.

Question 20: cis-[Pt(en)2Cl2]Br2·2H2O and trans-[Pd(en)2Br2]Cl2·2H2O are
(a) ionisation isomers
(b) geometrical isomers
(c) hydrate isomers
(d) none of these
Show Answer & Explanation

Answer: (d) none of these

Explanation:
1. The two compounds contain different metals (Pt and Pd), so they are not isomers of each other.

Teacher's Note:
1. Isomers must have identical molecular formulae.
2. Metals differ.

Question 21: Which of the following octahedral complexes has the maximum number of stereoisomers?
(a) [M(CN)6]±
(b) [M(en)(CN)4]±
(c) [M(en)2(CN)2]±
(d) [M(en)3]±
Show Answer & Explanation

Answer: (c) [M(en)2(CN)2]±

Explanation:
1. [M(CN)6]: 1; [M(en)(CN)4]: 1; [M(en)2(CN)2]: trans + cis pair = 3; [M(en)3]: 2 enantiomers
2. The third has the maximum.

Teacher's Note:
1. M(AA)2b2: cis (chiral) plus trans.
2. Count enantiomers separately.

Question 22: Which of the following complexes shows no geometrical isomerism?
(a) [Co(NH3)4Cl2]+
(b) [Co(NH3)3(NO2)3]
(c) [Cr(en)3]3+
(d) [Pt(gly)2]
Show Answer & Explanation

Answer: (c) [Cr(en)3]3+

Explanation:
1. [Cr(en)3]3+ is M(AA)3: only optical isomers
2. The other complexes have cis–trans or fac–mer forms.

Teacher's Note:
1. Tris-chelate octahedra have no geometrical isomers.
2. The others have two different positions possible.

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