Explanation

(d)
$\left\lbrack Ni(Cl)_{4} \right\rbrack^{2 -}$ oxidation state of Ni is +2
So, configuration of $Ni^{2 +} = 1s^{2},\ 2s^{2}2p^{6},\ 3s^{2}3p^{6}3d^{8}$
In $Ni^{2 +}$ 3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image247.emf]
Thus, due to $sp^{3}$-hybridisation of $Ni^{2 +}$ in $\left\lbrack NiCl_{4} \right\rbrack^{2 -}$, the shape of $\left\lbrack NiCl_{4} \right\rbrack^{2 -}$is tetrahedral.