JEE Chemistry Coordination Compounds MCQs Set 06

Chemistry Objective Questions and Answers: Coordination Compounds

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Multiple Choice Questions

Question 1: Which of the following compounds is not coloured?
(a) Na2[CuCl4]
(b) Fe4[Fe(CN)6]3
(c) Na2[CdCl4]
(d) Na2CrO4
Show Answer & Explanation

Answer: (c) Na2[CdCl4]

Explanation:
1. Cd2+ is d10, so no d–d transition is possible and the compound is colourless
2. Cu(II) (d9), Prussian blue (charge transfer) and chromate are coloured.

Teacher's Note:
1. Colour needs unpaired d electrons for d–d transitions or charge transfer.
2. d0 and d10 ions are usually colourless.

Question 2: A person suffering from lead poisoning should be fed with
(a) hypo
(b) cisplatin
(c) [Ca(EDTA)]2−
(d) DMG
Show Answer & Explanation

Answer: (c) [Ca(EDTA)]2−

Explanation:
1. [Ca(EDTA)]2− exchanges Ca2+ for Pb2+ to give the more stable [Pb(EDTA)]2−, which is excreted.

Teacher's Note:
1. This is chelation therapy.
2. Ca is exchanged because Pb–EDTA is more stable.

Question 3: Which of the following is an inner d (low spin) complex?
(a) [Co(H2O)6]3+
(b) [RhF6]3−
(c) [IrF6]3−
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. Co3+(3d) in [Co(H2O)6]3+ pairs its electrons to give d2sp3
2. 4d and 5d metals (Rh, Ir) have large Δo and form low spin inner orbital complexes even with F−.

Teacher's Note:
1. Δo increases down a group.
2. 4d and 5d complexes are nearly always low spin.

Question 4: In a complex ML6n+, Mn+ has five d electrons and L is a weak field ligand. According to crystal field theory, the magnetic behaviour corresponds to how many unpaired electrons?
(a) 0
(b) 5
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. A weak field gives a small Δo, so electrons occupy all five d orbitals singly first
2. t2g3 eg2 gives 5 unpaired electrons.

Teacher's Note:
1. d5 high spin has the maximum number of unpaired electrons.
2. μ = √35 = 5.92 BM.

Question 5: The oxidation state of iron in [Fe(CN)6]3− is
(a) −6
(b) +3
(c) −3
(d) +6
Show Answer & Explanation

Answer: (b) +3

Explanation:
1. x + 6(−1) = −3, so x = +3.

Teacher's Note:
1. Charge on complex = oxidation state of metal + sum of ligand charges.
2. CN− carries −1.

Question 6: The crystal field stabilisation energy is zero for
(a) K2[MnF6]
(b) K3[Fe(CN)6]
(c) K3[FeF6]
(d) K4[Fe(CN)6]
Show Answer & Explanation

Answer: (c) K3[FeF6]

Explanation:
1. [FeF6]3−: Fe3+ d5 high spin, t2g3 eg2: CFSE = 3(−0.4) + 2(0.6) = 0
2. The other complexes have d3 or low spin configurations with non-zero CFSE.

Teacher's Note:
1. d0, d10 and high spin d5 have zero CFSE.
2. F− is a weak field ligand.

Question 7: Match the ligands (A) ethylenediamine, (B) hydrazine, (C) dimethylglyoximato (dmg−), (D) diethylenetriamine (dien) with the properties (P) bidentate, (Q) tridentate, (R) forms hydrazinium ion on protonation. The correct pairing is
(a) A–P, B–R, C–P, D–Q
(b) A–P, B–Q, C–R, D–P
(c) A–Q, B–R, C–P, D–P
(d) A–R, B–P, C–Q, D–Q
Show Answer & Explanation

Answer: (a) A–P, B–R, C–P, D–Q

Explanation:
1. en and dmg− each have two donor atoms: bidentate (P)
2. Hydrazine N2H4 adds a proton to give hydrazinium N2H5+ (R)
3. dien has three N donors: tridentate (Q).

Teacher's Note:
1. Count donor atoms to find denticity.
2. dien = NH2CH2CH2NHCH2CH2NH2.

Question 8: The strongest C–O bond is present in
(a) [Cr(CO)6]+
(b) [Fe(CO)5]
(c) [V(CO)6]−
(d) all have equal strength
Show Answer & Explanation

Answer: (a) [Cr(CO)6]+

Explanation:
1. Strong back bonding lengthens and weakens the C–O bond
2. The cation has the least electron density on the metal and so the weakest back bonding
3. Hence the C–O bond is strongest in the cationic carbonyl.

Teacher's Note:
1. Positive charge on the metal reduces M→CO π back donation.
2. C–O stretching frequency is highest in cationic carbonyls.

Question 9: Which of the following complexes does not exhibit optical isomerism?
(a) tetrahedral complexes
(b) square planar complexes
(c) octahedral complexes
(d) polynuclear complexes
Show Answer & Explanation

Answer: (b) square planar complexes

Explanation:
1. Square planar complexes have a plane of symmetry (the molecular plane), so they are optically inactive
2. Suitably substituted tetrahedral, octahedral and polynuclear complexes can be chiral.

Teacher's Note:
1. Optical isomerism requires the absence of a plane and centre of symmetry.
2. Chelate octahedral complexes such as [Co(en)3]3+ are common examples.

Question 10: The correct geometry of the [NiCl4]2− complex is
(a) tetrahedral
(b) octahedral
(c) square planar
(d) pyramidal
Show Answer & Explanation

Answer: (a) tetrahedral

Explanation:
1. Cl− is a weak field ligand, so Ni2+ uses sp3 hybridisation and the shape is tetrahedral.

Teacher's Note:
1. Weak field ligands do not force pairing for a dsp2 arrangement.
2. [NiCl4]2− has two unpaired electrons.

Question 11: H2N–CH2–CH2–NH2 serves as a
(a) monodentate ligand
(b) chelating ligand
(c) bridging ligand
(d) cationic ligand
Show Answer & Explanation

Answer: (b) chelating ligand

Explanation:
1. Ethylenediamine has two N donors that bind the same metal ion forming a five-membered ring
2. A ligand forming a ring is a chelating ligand.

Teacher's Note:
1. Number of chelate rings = denticity − 1 for each ligand.
2. Five- and six-membered rings are most stable.

Question 12: Which of the complex ions [Ni(CN)4]2−, [Cu(CN)4]2−, [Cu(NH3)4]2+ is square planar?
(a) [Ni(CN)4]2− only
(b) [Cu(CN)4]2− only
(c) [Cu(NH3)4]2+ only
(d) all of them
Show Answer & Explanation

Answer: (d) all of them

Explanation:
1. d8 Ni(II) with a strong ligand (dsp2) and d9 Cu(II) complexes are square planar
2. So all three are square planar.

Teacher's Note:
1. Square planar geometry is typical of d8 and d9 ions.
2. Pt(II), Pd(II) and Au(III) also form square planar complexes.

Question 13: The geometric shape and oxidation number of copper in [Cu(NH3)4(OH2)2]2+ are respectively
(a) tetrahedral, +2
(b) square planar, −2
(c) linear, +3
(d) octahedral, +2
Show Answer & Explanation

Answer: (d) octahedral, +2

Explanation:
1. Six ligands (four NH3, two H2O) give an octahedral shape
2. The ligands are neutral, so Cu is +2.

Teacher's Note:
1. CN = 6 means octahedral.
2. Cu(II) here is tetragonally distorted.

Question 14: For [FeF6]3− and [Fe(CN)6]3−, the magnetic moment of the fluoride complex is expected to be
(a) the same as that of the cyanide complex
(b) larger than that of the cyanide complex because there are more unpaired electrons in the fluoride complex
(c) smaller than that of the cyanide complex because there are more unpaired electrons in the fluoride complex
(d) larger than that of the cyanide complex because there are fewer unpaired electrons in the fluoride complex
Show Answer & Explanation

Answer: (b) larger than that of the cyanide complex because there are more unpaired electrons in the fluoride complex

Explanation:
1. [FeF6]3− is high spin (5 unpaired electrons, 5.92 BM)
2. [Fe(CN)6]3− is low spin (1 unpaired electron, 1.73 BM).

Teacher's Note:
1. F− is a weak field ligand, CN− a strong one.
2. More unpaired electrons means a larger moment.

Question 15: Which of the following contains one unpaired electron in the 4p orbital?
(a) [Cu(NH3)2]+
(b) [Cu(NH3)4]2+
(c) [Cu(CN)4]3−
(d) [Ni(CN)4]2−
Show Answer & Explanation

Answer: (b) [Cu(NH3)4]2+

Explanation:
1. In [Cu(NH3)4]2+, the Cu2+ d9 ion promotes one 3d electron to 4p to make dsp2 hybrids, leaving one unpaired electron in 4p
2. The other ions have no unpaired electron.

Teacher's Note:
1. Cu2+ d9 has one unpaired electron.
2. This is the usual VBT explanation.

Question 16: Which of the complexes (a) K4[Os(CN)6], (b) [Mo(CO)6], (c) [Mn(CN)6]4− are low spin and diamagnetic?
(a) a, b and c
(b) a and b only
(c) a and c only
(d) b and c only
Show Answer & Explanation

Answer: (b) a and b only

Explanation:
1. Os(II) d6 and Mo(0) d6 with strong ligands are low spin t2g6, diamagnetic
2. Mn(II) d5 low spin has one unpaired electron: paramagnetic.

Teacher's Note:
1. Low spin d6 octahedral is diamagnetic.
2. d5 cannot be diamagnetic.

Question 17: How many unpaired electrons are present in the brown ring complex [Fe(H2O)5(NO)]SO4?
(a) 4
(b) 3
(c) 0
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The complex has Fe(I) with NO+ and d7: three unpaired electrons (consistent with Fe2+ NO)
2. μ ≈ 3.87 BM.

Teacher's Note:
1. The brown ring complex is paramagnetic.
2. Three unpaired electrons give √15 BM.

Question 18: Which of the following statements about Fe(CO)5 is correct?
(a) it is paramagnetic and high spin
(b) it is diamagnetic and high spin
(c) it is diamagnetic and low spin
(d) it is paramagnetic and low spin
Show Answer & Explanation

Answer: (c) it is diamagnetic and low spin

Explanation:
1. Fe(0) is d8; with strong field CO the 3d electrons pair, leaving no unpaired electron
2. Hence it is diamagnetic and low spin (dsp3).

Teacher's Note:
1. CO is a strong field π-acid ligand.
2. All metal carbonyls with even electron count are diamagnetic.

Question 19: Which is true for [Ni(en)2]2+ (Z for Ni = 28)?
(a) paramagnetism, dsp2, square planar, C.N. of Ni = 2
(b) diamagnetism, dsp2, square planar, C.N. of Ni = 4
(c) diamagnetism, sp3, tetrahedral, C.N. of Ni = 4
(d) paramagnetism, sp3, tetrahedral, C.N. of Ni = 4
Show Answer & Explanation

Answer: (b) diamagnetism, dsp2, square planar, C.N. of Ni = 4

Explanation:
1. The complex is treated as a strong field d8 case: two bidentate en give CN = 4, pairing of 3d electrons gives dsp2 hybridisation, a square planar, diamagnetic complex.

Teacher's Note:
1. Strong field d8 gives square planar geometry.
2. Count the donor atoms: 2 × 2 = 4.

Question 20: Arrange in decreasing order of the number of unpaired electrons: I [Fe(H2O)6]2+, II [Fe(CN)6]3−, III [Fe(CN)6]4−, IV [Fe(H2O)6]3+
(a) IV, I, II, III
(b) III, II, I, IV
(c) III, I, II, IV
(d) II, III, I, IV
Show Answer & Explanation

Answer: (a) IV, I, II, III

Explanation:
1. I: Fe2+ d6 high spin, 4 unpaired; II: Fe3+ d5 low spin, 1; III: Fe2+ d6 low spin, 0; IV: Fe3+ d5 high spin, 5
2. Decreasing: IV (5), I (4), II (1), III (0).

Teacher's Note:
1. Identify oxidation state, spin state, then count.
2. Water: weak field; cyanide: strong field.

Question 21: Which one of these ions absorbs energy from the visible spectrum?
(a) [Cu(NH3)4]+
(b) [Cu(NH3)4]2+
(c) [Zn(H2O)6]2+
(d) [Sc(H2O)6]3+
Show Answer & Explanation

Answer: (b) [Cu(NH3)4]2+

Explanation:
1. Absorption needs d–d transitions, so an incomplete d subshell is needed
2. Cu2+ d9 has a partly filled d subshell and is blue; Cu+ d10, Zn2+ d10 and Sc3+ d0 are colourless.

Teacher's Note:
1. d0 and d10 complexes are colourless.
2. d9 complexes are coloured.

Question 22: In the complex [Ni(H2O)2(NH3)4]2+, the magnetic moment (μ) of Ni is
(a) zero
(b) 2.83 BM
(c) 1.73 BM
(d) 3.87 BM
Show Answer & Explanation

Answer: (b) 2.83 BM

Explanation:
1. Ni2+ is d8 and the complex is octahedral: t2g6 eg2, two unpaired electrons
2. μ = √8 = 2.83 BM.

Teacher's Note:
1. Octahedral d8 always has two unpaired electrons.
2. Both NH3 and H2O are not strong enough to pair eg electrons.

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