Sequence of ond energy in $N O, N O^{+}$ and $N O^{-}$ is:

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $N O^{+} > N O > N O^{-}$
Bond order $\propto$ Bond energy
$Bond order = \frac{(No. of bonding electrons - No. of anti bonding electrons)}{2}$
For $N O^{+}, B . O = \frac{8 - 2}{2} = 3 N O, B . O = \frac{8 - 3}{2} = 2.5 N O^{-}, B . O = \frac{8 - 4}{2} = 2$
Hence, Bond energy is, $N O^{+} > N O > N O^{-}$

Suggest Corrections

Similar questions

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

N O

N O^{+}

N O^{-}

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

Join BYJU'S Learning Program