In which of following option the (X-A-X) adjacent angle difference in cationic part and anionic part is maximum in the solid state? - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
${PCl}_{5}$
B
${Cl}_{2} O_{6}$
C
$N_{2} O_{5}$
D
${BeH}_{2}$

Solution:

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

$N_{2} O_{5} \to \underset{(180^{\circ})}{{NO}_{2}^{+}} + \underset{(120^{\circ})}{{NO}_{3}^{-}}$

Solution:

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