Question

Among the following species, identify the isostructural pairs.
$N{F}_3,N{O}_3^{-},B{F}_3,H_3{O}^{+}$

• A. $[N{F}_3,N{O}_3^{-}]$ and $[B{F}_3,H_3{O}^{+}]$
• B. $[N{F}_3,H_3{O}^{+}]$ and $[N{O}_3^{-},B{F}_3]$
• C. $[N{F}_3,H_3{O}^{+}]$ and $[N{F}_3,B{F}_3]$
• D. None of the above

Answer 1

The correct option is B $[N{F}_3,H_3{O}^{+}]$ and $[N{O}_3^{-},B{F}_3]$

Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus

In $N{F}_3$: $H=\frac{1}{2}(5+3-0+0)=4$
Thus N in $N{F}_3$ is $s{p}^3$ hybridised as 4 orbitals are involved in bonding.

In $N{O}_3^{-}$: $H=\frac{1}{2}(5+0-0+1)=3$
Thus N in $N{O}_3^{-}$ is $s{p}^2$ hybridised as 3 orbitals are involved in bonding.

In $B{F}_3$: $H=\frac{1}{2}(3+3-0+))=3$
Thus B in $B{F}_3$ is $s{p}^2$ hybridised as 3 orbitals are involved in bonding.

In $H_3{O}^{+}$: $H=\frac{1}{2}(6+3-1+0)=4$
Thus O in $H_3{O}^{+}$ is $s{p}^3$ hybridised as 4 orbitals are involved in bonding.

Thus, isostructural pairs are $[N{F}_3,H_3{O}^{+}]$ and $[N{O}_3^{-},B{F}_3]$