Question

how many angles greater than 90° are present in XeOF4

Answer 1

0 angles greater than 90° are present in XeOF4 found 
all angles are less than 90° are present in XeOF4 

Answer 2

Answer:

Two angles greater than 90° are present in $XeOF_{4}$.

Explanation:

The number of bond pairs in $XeOF_{4}$ is 5 and the number of lone pairs of electrons is 1.

Therefore, the hybridization = 5 + 1 = 6

That is, hybridization is $sp^{3}d^2$.

So, the geometry of $XeOF_{4}$ is octahedral and its shape is square pyramidal with O atom at the apex.

Because of the presence of the lone pair, all the F - Xe = O bond angles become less than 90° because of the lone pair-bond pair repulsion.

This makes two F - Xe - F bond angles 120° and the remaining two F - Xe -F bond angles 60°.

Therefore, there are two angles greater than 90° present in $XeOF_{4}$.

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