question_answer

The geometry of \[Ni\,{{(CO)}_{4}}\] and \[[Ni\,{{(PP{{h}_{3}})}_{2}}]\,C{{l}_{2}}\] are

A) both square planar

B) tetrahedral and square planar respectively

C) both tetrahedral

D) square planar and tetrahedral respectively

Correct Answer: C

Solution :

[c]

\[Ni\,{{(CO)}_{4}};\]Oxidation state of \[Ni=0\]

\[Ni=I{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}},4{{s}^{2}},3{{d}^{8}}\]

As \[CO\]is a strong ligand and hence pairing of electrons occurs.

Oxidation state of \[Ni\] in \[[Ni\,{{(PP{{h}_{3}})}_{2}}]\,C{{l}_{2}}\] is \[x+0\times 2+(-2)=0\]

\[x=+\,2\]

As both the complexes has \[s{{p}^{3}}\text{-}\]hybridisation, thus their geometry is tetrahedral.