question_answer

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

A)  both square planar

B)  tetrahedral and square planar respectively

C)  both tetrahedral

D)  square planar and tetrahedral respectively

Correct Answer: C

Solution :

                 \[Ni{{(CO)}_{4}}-\]Oxidation state of\[Ni=0\]. Oxidation state of\[Ni\]in\[Ni{{(PP{{h}_{3}})}_{2}}C{{l}_{2}}\]is \[x+0\times 2+(-2)=0\] \[\therefore \]  \[x=+2\]