Correct order of dissociation energy of $N_{2}$ and $N_{2}^{+}$ is:

N_{2} > N_{2}^{+}

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N_{2} = N_{2}^{+}

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N_{2} < N_{2}^{+}

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none of these

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Solution

The correct option is B $N_{2} > N_{2}^{+}$
$N_{2} > N_{2}^{+}$
According to molecular orbital theory, $N_{2}$ have $10$ electrons in bonding electron orbitals and $4$ electrons in anti bonding electron orbitals.
Bond order of $N_{2}$ is $\frac{10 - 4}{2} = 3$
when it comes to $N_{2}^{+}$ , it has $9$ electrons in bonding electron orbitals and $4$ electrons in ant bonding electron orbitals. Hence bond order of $N_{2}^{+}$ is $\frac{9 - 4}{2} = 2.5$
Bond order is directly proportional to bond dissociation energy.
Hence, Bond dissociation energy of $N_{2}$ is greater $N_{2}^{+}$ .

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Similar questions

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

The correct order of bond strength is $N_{2}^{+}, N_{2}$ and $N_{2}^{-}$

N_{2}

N_{2}

N_{2}^{+}

\lim_{n \to \infty} \{\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + . . . + \frac{n}{1 - n^{2}}\}

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