Question

Which one of the following are isoelectronic and isostructural?

• A. $N{O}_3^{-},C{O}_3^{2-}$
• B. $S{O}_3,N{O}_3^{-}$
• C. $Cl{O}_3^{-},C{O}_3^{2-}$
• D. $C{O}_3^{2-},S{O}_3$

Answer 1

The correct option is A $N{O}_3^{-},C{O}_3^{2-}$

Isoelectronic species have same number of electrons and Isostructural species have same hybridisation at the central atom.
Steric Number $\left(x\right)$ = $\frac{1}{2}\left(M+V-C+A\right)$.
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion
$N{O}_3^{-}$
$\text{number of electron}=7+8\times 3+1=32$
$\text{Steric Number (x)}=\frac{1}{2}(0+5-0+1)$.
$\text{Steric Number (x)}=3$.
Hybridisation =$s{p}^2$

$C{O}_3^{2-}$
$\text{number of electron}=6+8\times 3+2=32$
$\text{Steric Number (x)}=\frac{1}{2}(0+4-0+2)$.
$\text{Steric Number (x)}=3$.
Hybridisation =$s{p}^2$

$S{O}_3$
$\text{number of electron}=16+8\times 3=40$
$\text{Steric Number (x)}=\frac{1}{2}(0+6-0+0)$.
$\text{Steric Number (x)}=3$.
Hybridisation =$s{p}^2$

$Cl{O}_3^{-}$
$\text{number of electron}=17+8\times 3+1=42$
$\text{Steric Number (x)}=\frac{1}{2}(0+7-0+1)$.
$\text{Steric Number (x)}=4$.
Hybridisation =$s{p}^3$

$N{O}_3^{-},C{O}_3^{2-}$ are isoelectronic and isostructural