4HNO3+P4O10−10 oC−−−−→+2N2O5+(HPO3)4 Find out the value of – InterviewSolution

1.	4HNO3+P4O10−10 oC−−−−→+2N2O5+(HPO3)4 Find out the value of expression \|x−y\| for above reaction where, x = maximum number of eaqual N−O bond length in N2O5. y = Difference of oxy linkages (P - O - P) between P4O10 and (HPO3)4.
Answer» $4 H N O_{3} + P_{4} O_{10} - 10^{o} C - --- \to + 2 N_{2} O_{5} + (H P O_{3})_{4}$ 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 $(H P O_{3})_{4}$ .

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