Question

Match List I with List II

List I (Complexes)		List II (Hybridisation)
A	$[Ni(CO{)}_4]$	I	$s{p}^3$
B	$[Cu(N{H}_3{)}_4{]}^{2+}$	II	$ds{p}^2$
C	$[Fe(N{H}_3{)}_6{]}^{2+}$	III	$s{p}^3{d}^2$
D	$[Fe(H_{2}O{)}_6{]}^{2+}$	IV	$d^{2}s{p}^3$

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

Answer 1

Answer: (C)

For ${\left[Fe{\left(N{H}_3\right)}_6\right]}^{+2},{\Delta }_0<P$, hence the pairing of electrons does not occur in $t_{2g}$. Therefore complex is outer orbital and its hybridisation is $s{p}^3{d}^2$.
Match List I with List II

List I (Complexes)	List II (Hybridisation)
$[Ni(CO{)}_4]$	$s{p}^3$
$[Cu(N{H}_3{)}_4{]}^{2+}$	$ds{p}^2$
$[Fe(N{H}_3{)}_6{]}^{2+}$	$s{p}^3{d}^2$
$[Fe(H_{2}O{)}_6{]}^{2+}$	$s{p}^3{d}^2$