Question
Among the following species, identify the isostructural pairs. $$N{F_3},NO_3^ - ,B{F_3},{H_3}{O^ + },H{N_3}$$
A.
$$\left[ {N{F_3},NO_3^ - } \right]{\text{and}}\left[ {B{F_3},{H_3}{O^ + }} \right]$$
B.
$$\left[ {N{F_3},H{N_3}} \right]{\text{and}}\left[ {NO_3^ - ,B{F_3}} \right]$$
C.
$$\left[ {N{F_3},{H_3}{O^ + }} \right]{\text{and}}\left[ {NO_3^ - ,B{F_3}} \right]$$
D.
$$\left[ {N{F_3},{H_3}{O^ + }} \right]{\text{and}}\left[ {N{H_3},B{F_3}} \right]$$
Answer :
$$\left[ {N{F_3},{H_3}{O^ + }} \right]{\text{and}}\left[ {NO_3^ - ,B{F_3}} \right]$$
Solution :
Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus
In $$N{F_3}:H = \frac{1}{2}\left( {5 + 3 - 0 + 0} \right) = 4$$
Thus $$N$$ in $$N{F_3}$$ is $$s{p^3}$$ hybridized as 4 orbitals are involved in bonding.
In $$NO_3^ - :H = \frac{1}{2}\left( {5 + 0 - 0 + 1} \right) = 3$$
Thus $$N$$ in $$NO_3^ - $$ is $$s{p^2}$$ hybridized as 3 orbitals are involved in bonding
$$B{F_3}:H = \frac{1}{2}\left( {3 + 3 - 0 + 0} \right) = 3$$
Thus $$B$$ in $$B{F_3}$$ is $$s{p^2}$$ hybridized and 3 orbitals are involved in bonding
In $${H_2}{O^ + }:H = \frac{1}{2}\left( {6 + 3 - 1 + 0} \right) = 4$$
Thus $$O$$ in $${H_2}{O^ + }$$ is $$s{p^3}$$ hybridized as 4 orbitals are involved in bonding.
Thus, isostructural pairs are $$\left[ {N{F_3},{H_3}{O^ + }} \right]$$ and $$\left[ {NO_3^ - ,B{F_3}} \right].$$