Question

Among the following species, identify the iso-structural pairs.

$N{F}_3,N{O}_3^{-},B{F}_3,H_3{O}^{+},N_{3}H$

• A. [$N{F}_3,N{O}_3^{-}$] and [$B{F}_3,H_3{O}^{+}$]
• B. [$N{F}_3,B{F}_3$] and [$N{O}_3^{-},H_3{O}^{+}$]
• C. [$N{F}_3,H_3{O}^{+}$] and [$N{O}_3^{-},B{F}_3$]
• D. [$N{F}_3,N{O}_3^{-}$] and [$B{F}_3,H_3{O}^{+}$]

Answer 1

The correct option is C

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

Lets first calculate the electron pairs of different compounds.

$\to N{F}_3\to$ central atom $\to$ N (5 valence electrons)

3 Ligands

ep = $\frac{(V+L+A-C)}{2}=\frac{(5+3)}{2}$ = 4

The geometry will be tetrahedral [with 3bp and 1 lp]

$\to N{O}_3^{-}\to$ Central atom $\to$ N (5 - valence)

3 Ligands

ep = $\frac{(V+L+A-C)}{2}=\frac{(5+3+1)}{2}$ [VSEPR is not applied]

Try AVSEPR Theory

ep = $\frac{(V+L+A-C)}{2}=\frac{(5+1)}{2}$ = 3[ignore oxygen to compute ep]

Geometry is Trigonal Planar.

$B{F}_3\to$ Central atom $\to$ B (3 - valence)

3 Ligands

ep = $\frac{(V+L+A-C)}{2}=\frac{(3+3)}{2}$ = 3

Geometry is Trigonal planar

$H_3{O}^{+}\to$ Central atom $\to$ 0 (6 - valence)

3 Ligands

Ep = $\frac{(V+L+A-C)}{2}=\frac{(6+3+0-1)}{2}$ = 4

Geometry is Tetrahedral $\to N_{3}H\Rightarrow$ Here H cannot be central atom Its structure comes out to be

$H-N\stackrel{....}{-}N\stackrel{....}{=}N$ [Linear]

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