IIT-JEE 2010 · Paper 2, Chemistry question 1

2.82 B.M. is $\sqrt{n(n+2)}$ with $n=2$ unpaired electrons.

• $\ce{Ni(CO)4}$ and $\ce{Ni(PPh3)4}$: Ni(0), $3d^{10}$ after pairing, tetrahedral: diamagnetic.
• $\ce{[Ni(CN)4]^{2-}}$: Ni(II), $d^8$, strong-field cyanide: square planar, diamagnetic.
• $\ce{[NiCl4]^{2-}}$: Ni(II), $d^8$, weak-field chloride: tetrahedral, two unpaired electrons, $\sqrt8=2.83$ B.M.

(B)