$N_{2}$ has a greater dissociation energy than $N_{2}^{+}$ , whereas $O_{2}$ has a lower dissociation energy than $O_{2}^{+}$ .
If true enter 1, else enter 0.
1

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Solution

The correct option is A 1
$N_{2}$ loses bonding electron hence, destabilised; $O_{2}$ loses anti-bonding electron hence, stabilized.

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Similar questions

N_{2}

{N_{2}}^{+}

O_{2}

{O_{2}}^{+}

N_{2}

N_{2}^{+}

O_{2}

O_{2}^{+}

N_{2}

{N_{2}}^{+}

O_{2}

{O_{2}}^{+}

The relationship between the dissociation energy of $N_{2}$ and is : ${N_{2}}^{+}$

F_{2} < {C l}_{2} < O_{2} < N_{2}

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