The hybridisation of the orbitals of oxygen in $H_{2} O_{2}$ is:

{s p}^{3} d

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s p

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{s p}^{2}

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{s p}^{3}

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Solution

The correct option is C ${s p}^{3}$
The oxygen atom in hydrogen peroxide contains $2$ lone pairs and $2$ bond pairs. Due to this, the oxygen atom undergoes $s p^{3}$ hybridization which results in open book structure.

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Similar questions

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

C (1) H_{2} = C (2) H - C (3) \equiv C H

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

A l H_{3}

A l H_{4}^{-}

Question 03

The types of hybrid orbitals of nitrogen in $N O_{2}^{+}, N O_{3}^{-} a n d N H_{4}^{+}$ respectively are expected to be

(a) $s p, s p^{3} a n d s p^{2}$

(b) $s p, s p^{2} a n d s p^{3}$

(d) $s p^{2}, s p^{3} a n d s p$

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