Question

Number/of sigma bonds pi bonds and lone pairs on xe atom in xeof4

Answer 1

$\huge\mathfrak\pink{anSwEr}$

Here, sigma = 5, pi = 1 and n= 1.

Add sigma + n = 5+1 = 6. this is the number of hybrid orbitals that Xe will need to house these electron pairs; so here Xe needs 6 hybrid orbitals.

hybridization state will be sp3d2, or d2sp3.

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Answer 2

The number of sigma bonds, pi bonds and lone pairs on $Xe$ atom in $\[XeO{F_4}\]$ is $\[5,\,1\]$ and $1$ respectively.

Explanation:

• The structure of $\[XeO{F_4}\]$ is given in the image below.
• $Xe$ has 8 valence electrons.
• In the structure, it is clear that four valence electrons form single bonds with four fluorine atoms.
• Two valence electrons on the Xenon atom form a double bond with one oxygen atom.
• Two valence electrons remain on the Xenon atom as a lone pair.
• Therefore, the number of sigma bonds, pi bonds and lone pairs on $Xe$ atom in $\[XeO{F_4}\]$ is given as:

1. Sigma bond - $5$
2. Pi bond - $1$
3. Lone pair - $1$

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