Question

Match each set of hybrid orbitals from LIST - I with complex(es) given in LIST - II.

LIST - I	LIST - II
P. $ds{p}^2$	1. $[Fe{F}_6{]}^{4-}$
Q. $s{p}^3$	2. $\left[Ti\right(H_{2}O{)}_{3}C{l}_3]$
R. $s{p}^3{d}^2$	3. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
S. $d^{2}s{p}^3$	4. $[FeC{l}_4{]}^{2-}$
	5. $Ni(CO{)}_4$
	6. $\left[Ni\right(CN{)}_4{]}^{2-}$

• A. P - 5; Q - 4, 6; R - 2, 3; S - 1
• B. P - 5, 6; Q - 4; R - 3; S - 1, 2
• C. P - 6; Q - 4, 5; R - 1, 3; S - 2, 3
• D. P - 4, 6; Q - 5, 6; R - 1, 2; S - 3

Answer 1

The correct option is C P - 6; Q - 4, 5; R - 1, 3; S - 2, 3


Similary in $FeC{l}_4{]}^{2-}$, the hybridisation is $s{p}^3$ due to the coordination number 4.

In $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$, the configuration of $C{r}^{3+}$ is $d^3$. Here, $N{H}_3$ can act as both strong field and weak field ligand. So it exist both $d^{2}s{p}^3$ and $s{p}^3{d}^2$ configuration.

Back pairing of electrons due to presence of strong field ligand


Electron pairing take place due to presence of S.F.L.