The correct order of bond enthalpy of given molecules is
The correct order of bond enthalpy of given molecules is
$\mathrm{O}_2 < \mathrm{N}_2 < \mathrm{H}_2$
$\mathrm{N}_2 < \mathrm{O}_2 < \mathrm{H}_2$
$\mathrm{H}_2 < \mathrm{N}_2 < \mathrm{O}_2$
$\mathrm{H}_2 < \mathrm{O}_2 < \mathrm{N}_2$
Solution
The correct order of bond enthalpy $=\mathrm{H}_2 < \mathrm{O}_2 < \mathrm{N}_2$ This is due to increasing bond strength due to single, double and triple bonds respectively $(\mathrm{H}-\mathrm{H}, \mathrm{O}=\mathrm{O}, \mathrm{N} \equiv$ N).