Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

The electronic configuration of $Xe$ is $\left[Kr\right]4d^{10}5s^{2}5p^6$

$Xe$ contributes its 6 valence electrons in bonding.

$O$ contributes 2 electrons and fluorine contributes 4 electrons.

Hence, 1 lone pair of Xe is left.