Question

$4HN{O}_3+P_4{O}_{10}\stackrel{-10{}^{o}C}{\to }+2N_2{O}_5+(HP{O}_3{)}_4$
Find out the value of expression $|x-y|$ for above reaction
where, x = maximum number of eaqual $N-O$ bond length in $N_2{O}_5$.
y = Difference of oxy linkages (P - O - P) between $P_4{O}_{10}$ and $(HP{O}_3{)}_4$.

Answer 1

x = Maximum number of equal $N-O$ bond length in $N_2{O}_5$,
$\therefore x=4$
Terminal $N-O$ bonds due to delocalisation


$P-O-P$ linkage in $P_4{O}_{10}$ is 6 and $(HP{O}_3{)}_4$ is 4. $\therefore y=6-4=2$
So, | x - y | = 4 - 2 = 2