Question

Match List - I with List - II and select the correct answer using the codes given below the lists:

List-I (Compound)	List-II (of Central atom)
(A) $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$	(1) $s{p}^3$
(B) $[PtC{l}_4{]}^{2-}$	(2) $s{p}^3{d}^2$
(C) $\left[Ni\right(CO{)}_4]$	(3) $ds{p}^2$
(D) $\left[Co\right(ox{)}_3{]}^{3-}$	(4) $d^{2}s{p}^3$

• A. $\left(A\right)-2,\left(B\right)-1,\left(C\right)-3,\left(D\right)-4$
• B. $\left(A\right)-2,\left(B\right)-3,\left(C\right)-1,\left(D\right)-4$
• C. $\left(A\right)-4,\left(B\right)-1,\left(C\right)-3,\left(D\right)-2$
• D. $\left(A\right)-4,\left(B\right)-3,\left(C\right)-1,\left(D\right)-2$

Answer 1

The correct option is B $\left(A\right)-2,\left(B\right)-3,\left(C\right)-1,\left(D\right)-4$

$\left[Ni\right(N{H}_3{)}_6{]}^{2+}$ involves $s{p}^3{d}^2$ hybridization and has octahedral geometry. It is an outer orbital complex.
$[PtC{l}_4{]}^{2-}$ involves $ds{p}^2$ hybridization and has square planar geometry.
$\left[Ni\right(CO{)}_4]$ involves $s{p}^3$ hybridization and has tetrahedral geometry.
$\left[Co\right(ox{)}_3{]}^{3-}$ involves $d^{2}s{p}^3$ hybridization and has octahdral geometry.It is an inner orbital complex.