Explanation

(c)
$\left\lbrack NiCl_{4} \right\rbrack^{2 -}\$; oxidation number of Ni,
x$-$4=$-$2
$\therefore$ x=+2
$Ni_{(28)} = \lbrack Ar\rbrack 3d^{8},4s^{2}$
3$d^{8}$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ [/table]
$\$
[image:9_image227.emf]
$Ni^{2 +}\lbrack Ar\rbrack$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ [/table]
$\left\lbrack NiCl_{4} \right\rbrack^{2 -}$
[image:9_image228.emf]
$\mathbf{s}\mathbf{p}^{\mathbf{3}}\mathbf{\ }$-hydrid orbitals, tetrahedral
$Cl^{-}$ is a weak ligand and thus unpaired electrons are not paired. Lone pairs from $4Cl^{-}$are accommodated in four $sp^{3}$ hybrid orbitals.
N=unpaired electron=2, paramagnetic
Magnetic moment (spin only)
$= \sqrt{N(N + 2)}\ BM = \sqrt{8} = 2.828\ BM$