Question

Among $[Ni(CO{)}_4{]}^{2-},[Ni(CN{)}_4{]}^{2-},[NiC{l}_4{]}^{2-}$ species, the hybridisation states of the $Ni$ atom are, respectively (At. no. of $Ni=28$ )

• A. $s{p}^3,ds{p}^2,ds{p}^2$
• B. $s{p}^3,ds{p}^2,s{p}^3$
• C. $s{p}^3,s{p}^3,ds{p}^2$
• D. $ds{p}^2,s{p}^3,sp$

Answer 1

Answer: (B)

(I) In $Ni(CO{)}_4$, nickel is $s{p}^3$-hybrid because in it oxidation state of $Ni$ is zero. So configuration of
${}_{28}N{i}_{10}=1s^{2}2s^{2}2p^6,3s^{2}3p^{6}3d^8,4s^2$

(II) In ${\left[Ni(CN{)}_4\right]}^{2-}$, nickel is present as $N{i}^{2+}$, so its configuration
$=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^8$

$C{N}^{-}$ is strong field ligand, hence it makes $N{i}^{2+}$ electrons to be paired up.
(III) In ${\left[NiC{l}_4\right]}^{2-}$ species, nickel is present as $N{i}^{2+}$, so its configuration
$=1s^2,2s^{2}2p^6,3s^{2}3p^6,3d^8$

$C{l}^{-}$is weak field ligand, hence $N{i}^{2+}$ electrons are not paired.