JEE Chemistry Coordination Compounds MCQs Set 14

Answer: (b) 4

Explanation:
1. [Ni(CN)4]2-: Ni2+ d8, square planar, diamagnetic.
2. [Ni(CO)4]: Ni(0) d10, diamagnetic.
3. [NiCl4]2-: tetrahedral, 2 unpaired, paramagnetic.
4. [Fe(CN)6]4-: Fe2+ d6 low spin, diamagnetic.
5. [Cu(NH3)4]2+: Cu2+ d9, paramagnetic.
6. [Fe(CN)6]3-: Fe3+ d5 low spin, 1 unpaired, paramagnetic.
7. [Fe(H2O)6]2+: d6 high spin, 4 unpaired, paramagnetic.
8. Total paramagnetic = 4.

Teacher's Note:
1. Find the oxidation state, d count and field strength of the ligand.
2. Cu2+ is always paramagnetic.

Answer: (a) FeF6 3- > CoF6 3- > [Co(C2O4)3]3-

Explanation:
1. FeF6 3-: Fe3+ is d5, F- is weak, so high spin with 5 unpaired electrons.
2. CoF6 3-: Co3+ is d6, high spin, 4 unpaired electrons.
3. [Co(C2O4)3]3-: Co3+ d6 with oxalate, low spin, 0 unpaired electrons (it is the standard diamagnetic complex).
4. Magnetic moment increases with unpaired electrons, so FeF6 3- > CoF6 3- > [Co(C2O4)3]3-.

Teacher's Note:
1. Spin-only moment = square root of n(n+2).
2. Weak-field F- gives high spin.

Answer: (a) B and D only

Explanation:
1. Cis-platin and related coordination compounds of platinum are used as anticancer drugs.
2. EDTA is used for lead and metal poisoning, and D-penicillamine for copper poisoning, not for tumours.

Teacher's Note:
1. Cis-platin is [Pt(NH3)2Cl2] and is an anticancer drug.
2. The trans isomer is inactive.

Answer: (d) A - I, B - II, C - IV, D - III

Explanation:
1. [Ni(CO)4]: Ni(0) d10 with a strong-field ligand, tetrahedral sp3.
2. [Cu(NH3)4]2+: Cu2+ d9 forms a square planar complex, dsp2.
3. [Fe(CN)6]4-: Fe2+ d6 with strong CN-, inner orbital d2sp3.
4. [Fe(H2O)6]2+: Fe2+ d6 with weak H2O, outer orbital sp3d2.
5. So A - I, B - II, C - IV, D - III.

Teacher's Note:
1. Strong-field ligands lead to inner orbital (d2sp3) complexes.
2. Weak-field ligands lead to outer orbital (sp3d2) complexes.

Answer: (b) 90 degrees

Explanation:
1. In the fac isomer the three Cl ligands occupy one triangular face of the octahedron.
2. Any two ligands on the same face are cis, so every Cl-Co-Cl angle is 90 degrees.
3. A 180 degree angle would need trans Cl atoms, which occurs in the mer isomer.

Teacher's Note:
1. fac means facial, three identical ligands on one face.
2. mer means meridional, with one trans pair.

Answer: (a) X = [Co(H2O)6]2+, Y = CoCl4 2-, Z = tetrahedral

Explanation:
1. Co2+ in water gives the pink octahedral [Co(H2O)6]2+.
2. Concentrated HCl replaces water by Cl-, giving the deep blue tetrahedral [CoCl4]2-.
3. The equilibrium is the well-known pink to blue colour change used to test for water.

Teacher's Note:
1. Weak-field Cl- gives a tetrahedral complex with Co2+.
2. Cobalt chloride paper is pink when wet and blue when dry.

Answer: (b) 480

Explanation:
1. Ti3+ is d1, so CFSE = -0.4 \( \Delta_o \) = -96.0 kJ/mol, giving \( \Delta_o \) = 240 kJ/mol.
2. Energy per ion = 240000/(6 x 10^23) = 4 x 10^-19 J.
3. \( \lambda = hc/E = (6.4 \times 10^{-34})(3 \times 10^{8})/(4 \times 10^{-19}) = 4.8 \times 10^{-7} \) m = 480 nm.

Teacher's Note:
1. For d1 only one electron sits in t2g, so CFSE = -0.4 \( \Delta_o \).
2. Convert kJ per mole to joules per ion before using E = hc/lambda.

Answer: (c) [Fe(CN)6]3-

Explanation:
1. All are Fe3+ complexes with the same metal ion and oxidation state.
2. The splitting depends on the ligand, and CN- is the strongest field ligand among C2O4 2-, F-, CN- and NH3.

Teacher's Note:
1. With the same metal and oxidation state, a stronger ligand gives a larger splitting.
2. CN- sits near the top of the spectrochemical series.

Answer: (b) [Co(NH3)5NO2]2+

Explanation:
1. NO2- is an ambidentate ligand and can bind through N (nitro) or through O (nitrito).
2. So [Co(NH3)5NO2]2+ shows linkage isomerism with two isomers.
3. The other cations have no isomers.

Teacher's Note:
1. Ambidentate ligands such as NO2- and SCN- cause linkage isomerism.
2. Look for such ligands first.

Answer: (d) 6

Explanation:
1. Mn2+ is d5, and H2O is a weak-field ligand, so it is high spin with 5 unpaired electrons.
2. \( \mu = \sqrt{n(n+2)} = \sqrt{35} = 5.92 \) which is 6 B.M. to the nearest integer.

Teacher's Note:
1. Half-filled d5 high spin has the maximum number of unpaired electrons.
2. The nearest integer of 5.92 is 6.