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

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

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

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

Correct Answer: B

Solution :

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