Which of the following molecules possess a "See-saw" shape due to the presence of 4 bond pairs and 1 — Chemical Bonding Chemistry Question

💡 Solution & Explanation

$SF_4$ has 6 valence $e^-$, uses 4 for bonds leaving 1 lp. $XeO_2F_2$ has 8 valence $e^-$, uses 4 for $2 \times O$ (double bonds) and 2 for $2 \times F$, leaving 2 $e^-$ (1 lp). Both have Steric Number 5 with 1 lp, taking equatorial positions and resulting in a see-saw shape. $SiF_4$ and $PCl_4^+$ have 0 lone pairs and are perfectly tetrahedral.