Among the following compounds the compound with zero bond order is:

Ne₂

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F₂

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O₂

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N₂

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Solution

The correct option is A
Ne₂

In neon molecule, the number of electrons present in antibonding orbitals is 8 and the number of electrons present in bonding orbitals is also 8.

So bond order = 1/2 [N_b- N_a] = 1/2 [8 - 8] = 0

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Similar questions

F_{2}, N_{2}, C l_{2}

O_{2}

N_{2}, O_{2}, O_{2}^{-}

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

N_{2}^{+}

N_{2}

O_{2}^{+}

O_{2}

What is the effect of the following processes on the bond order in $N_{2}$ and $O_{2}$ ?

(i). $N_{2} \to N_{2}^{+} + e^{-}$

(ii) $O_{2} \to O_{2}^{+} + e^{-}$

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