Given below are two statements :
Statement I : [Ni(CN)₄]²⁻ is square planar and diamagnetic complex, with dsp² hybridization for Ni but [Ni(CO)₄] is tetrahedral, paramagnetic and with sp³-hybridication for Ni.
Statement II : [NiCl₄]²⁻ and [Ni(CO)₄] both have same d-electron configuration have same geometry and are paramagnetic.
In light the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Both statements fall on one complex: [Ni(CO)₄] is tetrahedral but has no unpaired electron at all.

1. Find each d count. In [Ni(CN)₄]²⁻ and [NiCl₄]²⁻, nickel is Ni²⁺, which is 3d⁸. In [Ni(CO)₄], nickel is Ni(0), 3d⁸ 4s², and the strong-field CO ligands push the two 4s electrons into 3d, giving 3d¹⁰.
2. Apply VBT. Strong-field CN⁻ pairs the d⁸ electrons and frees one 3d orbital, so the complex is dsp². Weak-field Cl⁻ cannot pair them, so the complex uses sp³ and keeps 2 unpaired electrons. A d¹⁰ nickel has every 3d orbital full, so it must use sp³ and has nothing unpaired.

   Complex	Ni and d count	Hybridisation, shape	Magnetism
   [Ni(CN)₄]²⁻	Ni²⁺, d⁸	dsp², square planar	0 unpaired, diamagnetic
   [Ni(CO)₄]	Ni(0), d¹⁰	sp³, tetrahedral	0 unpaired, diamagnetic
   [NiCl₄]²⁻	Ni²⁺, d⁸	sp³, tetrahedral	2 unpaired, paramagnetic

3. Check each statement. Statement I is right about [Ni(CN)₄]²⁻ and about the shape of [Ni(CO)₄], but calling [Ni(CO)₄] paramagnetic makes it false. Statement II is false too: the d counts are d⁸ and d¹⁰, not the same, and [Ni(CO)₄] is not paramagnetic, even though both are tetrahedral.
4. So. Both statements are false, which is option (b).

The idea to keep. Always fix the metal's oxidation state first. Ni(0) in carbonyls becomes d¹⁰, so it is sp³ and diamagnetic. Ni²⁺ is d⁸: strong field gives dsp², weak field gives sp³ with 2 unpaired.

Answer from the NTA final answer key. Solution by PrepWiser.