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Bond Order and Enthalpy

The bond ener...

Question

The bond energies in NO, $N O^{+}$ and $N O^{-}$ follow the order:

N O^{+} > N O > N O^{-}

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N O > N O^{+} > N O^{-}

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N O^{-} > N O > N O^{+}

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N O^{+} > N O^{+} > N O

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Solution

The correct option is A $N O^{+} > N O > N O^{-}$
Bond energy is directly proportional to Bond Order. Bond order of $N O^{+}, N O a n d N O^{-}$ are $3, 2.5 a n d 2$ .
So option (A) is right.

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Similar questions

Bond energies in $N O, N O^{+}$ and $N O^{-}$ are such as

N O

N O^{+}

N O^{-}

Bond energies in $N O, N O^{+}$ and $N O^{-}$ are such as

N - O

N O_{2}^{+}

N O_{2}^{-}

N O_{3}^{-}

N O_{2}^{+}, N O_{2}^{-}

N O_{3}^{-}

N - O

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