Question

Match the following:

List-I	List-II
(A) ${sp}^3$	(I) ${\left[{Co\left({NH}_3\right)}_6\right]}^{3+}$
(B) ${dsp}^2$	(II) $\left[{Ni\left(Co\right)}_4\right]$
(C) ${sp}^3{d}^2$	(III) $\left[{Pt\left({NH}_3\right)}_2{Cl}_2\right]$
(D) $d^2{sp}^3$	(IV) ${\left[Co{F}_6\right]}^{3-}$
	(V) $\left[{Fe\left(Co\right)}_5\right]$

• A. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
• B. (A)-(V), (B)-(II), (C)-(IV), (D)-(III)
• C. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
• D. (A)-(II), (B)-(III), (C)-(I), (D)-(V)

Answer 1

Answer: (C)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

$Pt$ atom in $\left[Pt\right(N{H}_3{)}_{2}C{l}_2]$ is $ds{p}^2$ hybridized. It has 4 metal-ligand bonds.

$Ni$ atom in $\left[Ni\right(CO{)}_4]$ is $s{p}^3$ hybridized It has 4 metal-ligand bonds.

$Co$ atom in $[Co{F}_6{]}^{3-}$ is $s{p}^3{d}^2$ hybridized. It is outer orbital or high spin or spin-free complex.

$Co$ atom in $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ is $d^{2}s{p}^3$ hybridized. It is inner orbital or low spin or spin paired complex.

Hence, option $C$ is correct.