Question

Total number of lone pairs in $XeO{F}_4$ is:

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is A

1

Firstly Central atom in $XeO{F}_4$ is Xe [number of valence electrons in Xe is 8 which is more than in O which is 6]

Number of valance electrons in Xe is 8.

Applying VSEPR,

ep = $\frac{(V+L+A-C)}{2}$ = $\frac{(8+5+0-0)}{2}$ = $\frac{13}{2}$

So, VSEPR theory does not hold here.

Therefore, we will apply AVSEPR theory.

So, ep = $\frac{(V+L+A-C)}{2}$ = $\frac{(8+4)}{2}$ = 6

[We have to ignore oxygen while calculating ep]

Now, because we have 6 ep & 5 ligands [including oxygen]

The number of lone pairs = 1