$4 H N O_{3} + P_{4} O_{10} - 10^{o} C - --- \to + 2 N_{2} O_{5} + (H P O_{3})_{4}$
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 $(H P O_{3})_{4}$ .

Open in App

Solution

x = Maximum number of equal $N - O$ bond length in $N_{2} O_{5}$ ,
$∴ x = 4$
Terminal $N - O$ bonds due to delocalisation

$P - O - P$ linkage in $P_{4} O_{10}$ is 6 and $(H P O_{3})_{4}$ is 4. $∴ y = 6 - 4 = 2$
So, | x - y | = 4 - 2 = 2

Suggest Corrections

Similar questions

4 H N O_{3} + P_{4} O_{10} - 10^{o} C - --- \to + 2 N_{2} O_{5} + (H P O_{3})_{4}

| x - y |

N - O

N_{2} O_{5}

P_{4} O_{10}

(H P O_{3})_{4}

P - O

P_{4} O_{10}

P_{4} O_{10}

4 H N O_{3} + P_{4} O_{10} ⟶ 4 H P O_{3} + X

X

P_{4} O_{10}

P - O

Join BYJU'S Learning Program