Hybridisation of oxygen in O $_{2} F_{2}$ is:

s p

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s p^{2}

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s p^{3}

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s p^{3} d

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Solution

The correct option is C $s p^{3}$

In $O_{2} F_{2}$ molecule, the oxygen atom undergoes $s p^{3}$ hybridization which results in tetrahedral geometry. Two of the hybridized orbitals are occupied by lone pairs in the oxygen and two involved in bonding.

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N O_{3}^{-}, N O_{2}^{+}, N H_{4}^{+}

s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p, a n d s p

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