Match List-I with List-II.
| List-I | List-II |
| COMPOUND | MAGNETIC MOMENT, HYBRIDIZATION |
| (A) [MnCl4]2- | (I) 0.0 BM, d2sp3 |
| (B) [Cr(CO)6] | (II) 5.92 BM, sp3 |
| (C) [FeCl6]4- | (III) 0.0 BM, dsp2 |
| (D) [Ni(CN)4]2- | (IV) 4.90 BM, sp3d2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
By valence bond theory, the hybridisation and magnetic moment of a complex follow from the d electron count of the metal and the ligand strength: weak field ligands (\( \ce{Cl^{-}} \)) leave the electrons unpaired, strong field ligands (CO, \( \ce{CN^{-}} \)) pair them and free inner d orbitals. The spin-only magnetic moment is:
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
(A) \( \ce{[MnCl4]^{2-}} \) — This matches (II) 5.92 BM, sp3. x + 4(−1) = −2, so Mn is +2. Mn (Z = 25) is [Ar]3d54s2; losing the two 4s electrons gives \( \ce{Mn^{2+}} \): 3d5. Weak field \( \ce{Cl^{-}} \) uses 4s and 4p (tetrahedral, sp3), n = 5, μ = √35 = 5.92 BM.
(B) \( \ce{[Cr(CO)6]} \) — This matches (I) 0.0 BM, d2sp3. CO is neutral, so Cr is 0. Cr (Z = 24) is [Ar]3d54s1; the strong field CO pairs all six electrons into three 3d orbitals, leaving two 3d orbitals free: d2sp3, n = 0, μ = 0 BM.
(C) \( \ce{[FeCl6]^{4-}} \) — This matches (IV) 4.90 BM, sp3d2. x + 6(−1) = −4, so Fe is +2. Fe (Z = 26) is [Ar]3d64s2; losing the two 4s electrons gives \( \ce{Fe^{2+}} \): 3d6. Weak field \( \ce{Cl^{-}} \) leaves 4 unpaired electrons and the outer 4d orbitals are used: sp3d2, μ = √24 = 4.90 BM.
(D) \( \ce{[Ni(CN)4]^{2-}} \) — This matches (III) 0.0 BM, dsp2. x + 4(−1) = −2, so Ni is +2. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8. Strong field \( \ce{CN^{-}} \) pairs the eight electrons into four 3d orbitals, leaving one free: dsp2 (square planar), μ = 0 BM.