Nickel (Z=28) combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex$\left\lbrack {NiX}_{4} \right\rbrack^{2 -}$. The number of unpaired electron (s) in the nickel and geometry of this complex ion are respectively

Question

Dr. Abubacker · Accepted Answer

(b)
_{28}Ni=[Ar]$\ 3s^{2}\ 3p^{6}4s^{2}\ 3d^{8}$
$Ni^{2 +} = \lbrack Ar\rbrack\ 3s^{2}\ 3p^{6}\ 3d^{8}$
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Nickel has two unpaired electrons and geometry is tetrahedral due to $sp^{3}$ hybridisation.

Q. Nickel (Z=28) combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex$\left\lbrack {NiX}_{4} \right\rbrack^{2 -}$. The number of unpaired electron (s) in the nickel and geometry of this complex ion are respectively

Explanation

Reviewed by Dr. Abubacker

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