List-II (Hybridisation of central metal ion): (I) d2sp3 (II) sp3 (III) sp3d2 (IV) dsp2
Choose the correct answer from the options given below :
Answer: A → III, B → II, C → I, D → IV
Solution
Crystal field splitting fixes the hybridisation of each complex.
[CoF6]3−: Co3+ is 3d6; the weak-field F− causes no pairing, so the outer 4d orbitals are used — sp3d2.
[NiCl4]2−: Ni2+ is 3d8; weak-field Cl− gives a tetrahedral geometry with sp3 hybridisation.
[Co(NH3)6]3+: Co3+ is 3d6; the strong-field NH3 forces all six d-electrons to pair into three orbitals, freeing two inner 3d orbitals for bonding — d2sp3.
[Ni(CN)4]2−: Ni2+ is 3d8; strong-field CN− pairs the electrons and leaves one 3d orbital free for a square-planar dsp2 arrangement.
So A pairs with sp3d2, B with sp3, C with d2sp3, and D with dsp2.
JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Match the columns