Match List I with List II
| List I | List II |
| A. \([Ni(CO)_4]\) | I. \(sp^3\) |
| B. \([Ni(CN)_4]^{2-}\) | II. \(dsp^2\) |
| C. \([Fe(CN)_6]^{4-}\) | III. \(sp^3d^2\) |
| D. \([MnF_6]^{4-}\) | IV. \(d^2sp^3\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-I, B-II, C-IV, D-III
By valence bond theory the hybridisation depends on the number of ligands and on whether the ligands are strong field (CO, \( \ce{CN^{-}} \): they pair the d electrons and free inner d orbitals) or weak field (\( \ce{F^{-}} \): no pairing, outer 4d orbitals used).
A. \( \ce{[Ni(CO)4]} \) — This matches I. sp3. Ni (Z = 28) is [Ar]3d84s2, oxidation state 0; CO pairs all ten electrons into 3d10, and the empty 4s and three 4p orbitals form four sp3 hybrids (tetrahedral, diamagnetic).
B. \( \ce{[Ni(CN)4]^{2-}} \) — This matches II. dsp2. x + 4(−1) = −2 gives \( \ce{Ni^{2+}} \): 3d8. \( \ce{CN^{-}} \) pairs the eight electrons in four 3d orbitals, leaving one 3d orbital for dsp2 (square planar, diamagnetic).
C. \( \ce{[Fe(CN)6]^{4-}} \) — This matches IV. d2sp3. x + 6(−1) = −4 gives \( \ce{Fe^{2+}} \); Fe (Z = 26) [Ar]3d64s2 loses the two 4s electrons: 3d6. \( \ce{CN^{-}} \) pairs them in three 3d orbitals, so two inner 3d orbitals are free: d2sp3 (inner orbital, octahedral, diamagnetic).
D. \( \ce{[MnF6]^{4-}} \) — This matches III. sp3d2. x + 6(−1) = −4 gives \( \ce{Mn^{2+}} \); Mn (Z = 25) [Ar]3d54s2 loses 4s2: 3d5. \( \ce{F^{-}} \) does not cause pairing, all five 3d orbitals stay occupied, and 4s, 4p and 4d orbitals give sp3d2 (outer orbital, five unpaired electrons).
Shortcut: A-I and B-II leave Options (2) and (3); C-IV then leaves only Option (2).