1) $N a_{2} O_{2} W a t e r - --- \to 0^{o} C P + Q$

2) $N a_{2} O_{2} W a t e r - --- \to 25^{o} C P + R ↑ + S$
Calculate the sum of bond orders between same bonded atoms in Q and R compounds.

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Solution

1) $N a_{2} O_{2} W a t e r - --- \to 0^{o} C N a O H (P) + H_{2} O_{2} (Q)$

2) $N a_{2} O_{2} W a t e r - --- \to 25^{o} C N a O H (P) + O_{2} ↑ (R) + H_{2} O (S)$

Bond order of $[O - O]$ in $H_{2} O_{2} = 1$
Bond order of $[O - O]$ in $O_{2} = 2$
So, $s u m = 3$

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N a_{2} O_{2} W a t e r - --- \to 25^{o} C P + R ↑ + S

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\frac{a (q - r)}{p} + \frac{b (r - p)}{q} + \frac{c (p - q)}{r} = 0

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