Question

Type of hybridization of complex (E) is :

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

Answer 1

Answer: (A)

$s{p}^3{d}^2$

The compound $A$ is lead nitrate $Pb(N{O}_3{)}_2$. Lead nitrate gives blue colour flame in flame test. When heated, it decomposes and $PbO,N{O}_2$ and $O_2$ gases are released. The colour of $N{O}_2$ (compound $B$) is brown.

Potassium dichromate react with lead(II) nitrate to produce precipitate of lead chromate(II) $PbCr{O}_4$, potassium nitrate.

Compound $A$ with $H_{2}S$ forms black precipitate of lead sulfide $PbS$ (compound $F$).

$N{O}_2$ with $H_2$ gas produce $N{H}_3$, which on reaction with $H_{2}O$ forms $N{H}_{4}OH$ (compound $D$). This on reaction with $H_{2}S{O}_4$ and $FeS{O}_4$ produce Mohr's salt -$\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O$.

In the complex the iron has electronic configuration of $F{e}^{2+}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$ to which two electron pairs are given by two $N{H}_4$ ligand and the two binary ligands $S{O}_4$ give four pair of electrons thus the complex has $s{p}^3{d}^2$ hybridization.