JEE Main 2025 · 22 January · morning · Single correct · +4 / −1
From the magnetic behaviour of [NiCl₄]²⁻ (paramagnetic) and [Ni(CO)₄] (diamagnetic), choose the correct geometry and oxidation state.
- (a) [NiCl₄]²⁻: Ni(0), tetrahedral [Ni(CO)₄]: Ni(0), square planar
- (b) [NiCl₄]²⁻: Ni^(II), tetrahedral [Ni(CO)₄]: Ni(0), tetrahedral
- (c) [NiCl₄]²⁻: Ni^(II), tetrahedral [Ni(CO)₄]: Ni^(I I), square planar
- (d) [NiCl₄]²⁻: Ni^(II), square planar [Ni(CO)₄]: Ni(0), square planar
Answer and solution
Answer (b) [NiCl₄]²⁻: Ni^(II), tetrahedral [Ni(CO)₄]: Ni(0), tetrahedral
Solution
Chemistry ma’am · 4 steps
Find nickel's oxidation state first, then let the unpaired electrons tell you the hybridisation.
- Oxidation states. In [NiCl₄]²⁻, four Cl⁻ give -4, so x - 4 = -2 and nickel is +2. In [Ni(CO)₄], CO is a neutral ligand and the complex has no charge, so nickel is 0.
- Chloride complex. Ni²⁺ is 3d⁸. Cl⁻ is a weak-field ligand, so it does not pair the electrons and 2 stay unpaired, which explains the paramagnetism. No 3d orbital is free, so nickel uses sp³ hybrids and the shape is tetrahedral.
- Carbonyl complex. Ni(0) is 3d⁸ 4s². CO is a strong-field ligand: it pushes the two 4s electrons into 3d, giving 3d¹⁰ with every electron paired, which explains the diamagnetism. The empty 4s and 4p orbitals form sp³ hybrids, so this complex is also tetrahedral.
Complex Oxidation state of Ni d-electrons and unpaired Hybridisation Geometry [NiCl₄]²⁻ +2 3d⁸, 2 unpaired sp³ tetrahedral [Ni(CO)₄] 0 3d¹⁰, 0 unpaired sp³ tetrahedral - So. Both complexes are tetrahedral, with Ni^(II) in [NiCl₄]²⁻ and Ni(0) in [Ni(CO)₄]. That matches option (b).
The idea to keep. For four-coordinate nickel: a paramagnetic complex must be sp³ tetrahedral. A d⁸ square planar complex is always diamagnetic. d¹⁰ Ni(0) has no empty 3d orbital, so it can only be tetrahedral.
Answer from the NTA final answer key. Solution by PrepWiser.