The ground state electronic configuration of $S$ is $3 s^{2} 3 p^{4}$. How does it form the compound

Question

The ground state electronic configuration of $S$ is $3 s^{2} 3 p^{4}$. How does it form the compound $\mathrm{SF}_{6}$ ?\

Professor PC Thomas Classes · Answer

(b) : $S$ can go into excited state with 6 unpaired electrons due to presence of vacant $3 d$-orbitals, which are overlapped by six F electrons.\