Question

The hybridization of $P$ is $P{O}_4^{3-}$ is the same as of:

• A. $S$ in $S{O}_3$
  
• B. $N$ in $N{O}_3^{-}$
• C. $I$ in $IC{l}_2^{+}$
• D. $I$ in $IC{l}_4^{-}$

Answer 1

The correct option is C

$I$ in $IC{l}_2^{+}$

Number of hybrid orbitals of $P$ in $P{O}_4^{3-}=\frac{1}{2}[5+0+3]=4\left(s{p}^3\right)$
No. of hybrid orbitals of $N$ in $N{O}_3^{-}=\frac{1}{2}[5+0+1]=3\left(s{p}^2\right)$
No. of hybrid orbitals of $I^{+}$ in $IC{l}_2^{+}(5s^{2}5p^2{x}^{5^p}{y}^{5^P}z:(s{p}^3{)}^2\left(s{p}^3{)}^2\right(s{p}^3{)}^1(s{p}^3{)}^1$
- atom in excited state: $5s^{2}5p^{4}5d_{z^2}^{1}5d_{z^2-y^2}^0$
$I$ - atom is $s{p}^3{d}^2$ hybridised state in $IC{I}_4^{-}$ ∶
$\left(s{p}^3{d}^2{)}^2\right(s{p}^3{d}^2{)}^2\left(s{p}^3{d}^2{)}^1\right(s{p}^3{d}^2{)}^1\left(s{p}^3{d}^2{)}^1\right(s{p}^3{d}^2{)}^1$
-Two lone pairs occupy the axial positions of octahedron and the shape of $IC{l}_4^{-}$ becomes square planar.