Calculate the sum of bond orders between same bonded atoms in $Q$ and $R$ compounds.

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Solution

$N a_{2} O_{2} + H_{2} S O_{4} c o l d - - \to N a_{2} S O_{4} + H_{2} O_{2}$

$2 N a_{2} O_{2} + 2 H_{2} S O_{4} h o t - - \to 2 N a_{2} S O_{4} + 2 H_{2} O + O_{2}$

The $O - O$ bond order in hydrogen peroxide $(Q)$ is 1 and in oxygen molecule $(R)$ is 2.

Therefore, the sum of bond orders between the same bonded atoms in $Q$ and $R$ compounds is $1 + 2 = 3$ .

Hence, the correct answer is 3.

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