Question

$N{a}_2{O}_2\underset{Cold}{\overset{Water}{\to }}P+Q$

$N{a}_2{O}_2\underset{Warm}{\overset{Water}{\to }}P+R$

The sum of bond order between same bonded atoms in Q and R compounds is:

Answer 1

$N{a}_2{O}_2+2H_{2}O\stackrel{Cold}{\to }2NaOH+H_2{O}_2$
So the compound Q is $H_2{O}_2$ and the bond order between same bonded atoms is 1 as single bond exixt between $O$-atoms.


$N{a}_2{O}_2+2H_{2}O\stackrel{Warm}{\to }4NaOH+O_2$
So the compound R is $O_2$ and the bond order between same bonded atoms is 2 as double bond exixt between $O$-atoms.


Hence, the sum of bond order between same bonded atoms in compounds Q and R is (1+2) = 3.