Question

Among $[Ni(CO)_4]^{2-}, [Ni(CN)_4]^{2-}, [NiCl_4]^{2-}$ species, the hybridisation states of the $Ni$ atom are, respectively (At. no. of $ Ni=28 $ )

• A. $ s{{p}^{3}},\text{ }ds{{p}^{2}},ds{{p}^{2}} $
• B. $ s{{p}^{3}},\text{ }ds{{p}^{2}},\text{ }s{{p}^{3}} $
• C. $ s{{p}^{3}},\text{ }s{{p}^{3}},\text{ }ds{{p}^{2}} $
• D. $ ds{{p}^{2}},\text{ }s{{p}^{3}},\text{ }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
$_{28}Ni_{10} =1 s^{2} 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{8}, 4 s^{2}$

(II) In $\left[ Ni ( CN )_{4}\right]^{2-}$, nickel is present as $Ni ^{2+}$, so its configuration
$=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{8}$

$CN ^{-}$ is strong field ligand, hence it makes $Ni ^{2+}$ electrons to be paired up.
(III) In $\left[ NiCl _{4}\right]^{2-}$ species, nickel is present as $Ni ^{2+}$, so its configuration
$=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{8}$

$Cl ^{-}$is weak field ligand, hence $Ni ^{2+}$ electrons are not paired.