Question

Type of hybridization of complex $\left(E\right)$ is:

• A. $s{p}^3{d}^2$
• B. $d^{2}s{p}^3$
• C. $s{p}^3$
• D. $ds{p}^2$

Answer 1

The correct option is A $s{p}^3{d}^2$

$\underset{\left(A\right)}{2Pb(N{O}_3{)}_2}\stackrel{\Delta }{\to }2PbO+\underset{\left(B\right)}{4N{O}_2\uparrow }+O_2\uparrow \underset{\left(A\right)}{Pb(N{O}_3{)}_2}+H_{2}S\to \underset{\left(F\right)}{PbS\downarrow }\underset{\left(F\right)}{PbS}+dil.HN{O}_3\to Pb(N{O}_3{)}_2+\underset{\left(H\right)}{NO\uparrow }+\underset{\left(G\right)}{H_{2}O}\underset{\left(A\right)}{Pb(N{O}_3{)}_2}+K_{2}Cr{O}_4\to \underset{\left(I\right)}{PbCr{O}_4\downarrow }+2KN{O}_3\underset{\left(B\right)}{N{O}_2}+\underset{\left(H\right)}{NO}\stackrel{\text{very low temp.}}{\to }\underset{\left(C\right)}{N_2{O}_3}\underset{\left(C\right)}{N_2{O}_3}+H_{2}O\to \underset{\left(D\right)}{2HN{O}_2}\underset{\left(D\right)}{2HN{O}_2}+2FeS{O}_4+H_{2}S{O}_4\to F{e}_2(S{O}_4{)}_3+2NO+2H_{2}OFeS{O}_4+NO\to \underset{\left(E\right)}{\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4}$

In $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4,Fe$ is in $+1$ oxidation state.
${}_{26}Fe=\left[Ar\right]3d^{6}4s^2{}^{}{}_{26}F{e}^{1+}=\left[Ar\right]3d^{6}4s^1$
Since $NO$ is a strong field ligand, it will move $4s$ electron to the $3d$ subshell.
$\therefore {}_{26}F{e}^{1+}=\left[Ar\right]3d^{7}4s^{0}4p^{0}4d^0$
Hence, the compound $\left(E\right)$ will form octahedral geometry with $s{p}^3{d}^2$ hybridization (outer orbital complex).