Explanation

(b)
1.In $Ni(CO)_{4}$, nickel is $sp^{3}$-hybridised because in it oxidation state of NI is zero. So, configuration of
_{28}Ni=$1s^{2}2s^{2}2p^{6},\ 3s^{2}3p^{6}3d^{8},\ 4s^{2}$
3$d^{8}$ 4$s^{2}$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | ⥮ | | | | [/table]
[image:9_image221.emf]
2.In $\left\lbrack Ni(CN)_{4} \right\rbrack^{2 -},\ nickel\ is\ present\$ as $Ni^{2 +}$, so its configuration=$1s^{2},\ 2s^{2}2p^{6},\ 3s^{2}3p^{6}3d^{8}$
3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image222.emf]$CN^{-}$
is strong field ligand, hence it makes $Ni^{2 +}$ electrons to be paired up.
In $\left\lbrack {NiCl}_{4} \right\rbrack^{2 -}\ species,\ nickel\ is\ present\$as $Ni^{2 +}$, so its configuration=$1s^{2},\ 2s^{2}2p^{6},\ 3s^{2}3p^{6}3d^{8}$
3$d$ 4$s$ 4$p$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | | | | | [/table]
[image:9_image223.emf]
$Cl^{-}$
is weak field ligand, hence $Ni^{2 +}$ electrons are not paired.