Consider the following four complexes of nickel: (I) [ Ni ( CO )₄] (II) [ Ni ( CN )₄]²⁻ (III) [ NiCl ₄]²⁻ (IV) [ Ni ( H ₂ O )₆]²⁺ The sum of the number of unpaired electrons present in all four complexes is:

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Correct answer: B. 4

For [ Ni ( CO )₄] : Nickel is in 0 oxidation state with electronic configuration 3 d ^8 4 s ^2 . CO is a strong field ligand, which forces the 4 s electrons to pair up in the 3 d orbitals, resulting in a 3 d ¹⁰ configuration. Number of unpaired electrons = 0 . For [ Ni ( CN )₄]²⁻ : Nickel is in +2 oxidation state ( 3 d ^8 ). CN ^- is a strong field ligand, causing pairing of electrons against Hund's rule. This leaves one empty 3 d orbital ( dsp ^2 hybridization). Number of unpaired electrons = 0 . For [ NiCl ₄]²⁻ :

Consider the following four complexes of nickel: (I) [ Ni ( CO )₄] (II) [ Ni ( CN )₄]²⁻ (III) [ NiCl ₄]²⁻ (IV) [ Ni ( H ₂ O )₆]²⁺ The sum of the number of unpaired electrons present in all four complexes is:

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