STD 11 - 4. Chemical bonding and molecular structure — JEE STD 11 Science — Question

Ground state of electronic configuration $X e=[K r] 4 d^{10} 5 s^2 5 p^6$

Excited state of electronic configuration $Xe =[ Kr ] 4 d ^{10} 5 s ^2 5 p ^3 5 d ^3$

The electronic configuration of oxygen $=1 s^2 2 s^2 2 p^4$

In $XeO _3$, the hybridisation of $Xe$ is $sp ^3$, It form $3$ $\sigma$ bond and three $\pi$ bonds with three oxygen atoms and contain one lone pair of electron.

The oxygen atom form bond with $Xe$ like $s p^3+p_x, s p^3+p_y, s p^3+p_z$, but never form $s p^3+s$