In which of following option the (X-A-X) adjacent angle difference in cationic part and anionic part is maximum in the solid state?

# In which of following option the (X-A-X) adjacent angle difference in cationic part and anionic part is maximum in the solid state?

1. A

${\mathrm{PCl}}_{5}$

2. B

${\mathrm{Cl}}_{2}{\mathrm{O}}_{6}$

3. C

${\mathrm{N}}_{2}{\mathrm{O}}_{5}$

4. D

${\mathrm{BeH}}_{2}$

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### Solution:

The N is having sp hybridized with the two bound pairs and the cation of the ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ is ${\mathrm{NO}}_{2}^{+}$. Similar to this, the ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$anion is $N{O}_{3}$, with the N having hybridized with the three bound pairs.
Thus, the ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ cation is ${\mathrm{NO}}_{2}^{+}$ and the ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ anion is $N{O}_{3}$, and the N atom in the cation has hybridised due to the two bound pairs in the sp state.

${\mathrm{N}}_{2}{\mathrm{O}}_{5}\to \underset{\left({180}^{\circ }\right)}{{\mathrm{NO}}_{2}^{+}}+\underset{\left({120}^{\circ }\right)}{{\mathrm{NO}}_{3}^{-}}$

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