The complex showing a spin‑only magnetic moment of 2.82 B.M.is

Engineering

Chemistry

Crystal Field Theory and Valence Bond Theory

Question

Ni(PPh₃)₄

Ni(CO)₄

[NiCl₄]^2–

[Ni(CN)₄]^2–

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[NiCl₄]^–2

Hyb = sp³

No. of unpaired electrons = 2

$μ = \sqrt{2 (4)} = \sqrt{8} = 2 .82 BM$

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