Question

hybridization of oxygen in OF2 is

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Answer 1

your answers is Sp3

it has bent structure.

oxygen has 2 bond pairs and 2 lone pairs result into the Sp3

Answer 2

Answer:

The hybridization of oxygen in $OF_{2}$  is $sp^{3}$.

Explanation:

• $OF_{2}$ or oxygen difluoride is a chemical compound formed by the reaction shown below:

• $2F_{2}+2NaOH$→ $OF_{2}$ +$2NaF+ H_{2}O$

• $OF_{2}$ is a colourless gaseous compound with a strong odour and often acts as an oxidizer.

• If we look at the electronic configuration of halogen Fluorine and oxygen in their ground state,

$\tensor^{}_{9}{\mathrm{F}}{} =\end{document}$ $1s^{2}2s^{2}2p^{5}$

$\tensor^{}_{8}{\mathrm{O}}{} =\end{document}$ $1s^{2}2s^{2}2p^{4}$

• After forming bonds with $F$ and fulfiling the octet rule,

the outermost shell of the central Oxygen atom has 4 hybridised orbitals i.e. $2s,2p_{x} ,2p_{y} ,2p_{z}$.

• [As the steric number is 4 i.e. it has 2 bond pairs and 2 lone pairs]

• The hybridization is $sp^{3}$.

Hence the hybridization of oxygen in $OF_{2}$ is $sp^{3}$.