question_answer

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)  \[s{{p}^{3}},\text{ }ds{{p}^{2}},ds{{p}^{2}}\]   

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)  \[ds{{p}^{2}},\text{ }s{{p}^{3}},\text{ }sp\]

Correct Answer: B

Solution :

 (I) In\[Ni{{(CO)}_{4}},\]nickel is\[s{{p}^{3}}-\]hybrid because in it oxidation state of\[Ni\] is zero. So configuration of \[_{28}Ni=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{8}},4{{s}^{2}}\] (II) In\[{{[ni{{(CN)}_{4}}]}^{2-}},\]nickel is present as\[N{{i}^{2+}},\] so its configuration \[=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{8}}\] \[C{{N}^{-}}\]is strong field ligand, hence it makes\[N{{I}^{2+}}\] electrons to be paired up. (Ill) In\[{{[NiC{{l}_{4}}]}^{2-}}\]species, nickel is present as \[N{{i}^{2+}},\]so its configuration \[=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},3{{d}^{8}}.\] \[C{{l}^{-}}\]is weak field ligand, hence\[N{{i}^{2+}}\]electrons are not paired.