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. $\left[Ni\right(CO{)}_4]$
	6. $\left[Ni\right(CN{)}_4{]}^{2-}$

The correct option is:

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

Answer 1

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

[1] $[Fe{F}_6{]}^{4-}$
$F{e}^{2+}=\left[Ar\right]3d^6$

[2] $\left[Ti\right(H_{2}O{)}_{3}C{l}_3]$
$T{i}^{+3}=\left[Ar\right]3d^1$

[3] $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
$C{r}^{+3}=\left[Ar\right]3d^3$

[4] $[FeC{l}_4{]}^{2-}$
$F{e}^{2+}=\left[Ar\right]3d^6$

[5] $\left[Ni\right(CO{)}_4]$
$Ni:\left[Ar\right]3d^{8}4s^2$
Back pairing of electrons due to the presence of strong-field ligands.

[6] $\left[Ni\right(CN{)}_4{]}^{2-}$
$N{i}^{2+}:\left[Ar\right]3d^8$
Electron pairing takes place due to the presence of strong-field ligands.