Chapter: Chemical Bonding· Chemistry
In \(\mathrm{BeF_2}\), Be is sp hybridised with no lone pairs, so the molecule is linear (180°); in \(\mathrm{H_2O}\), O is sp³ hybridised with two lone pairs, so the molecule is angular (104°35′).
\(\mathrm{BeF_2}\): Be is \(1s^{2}2s^{2}\). In the excited state (\(2s^{1}2p^{1}\)) the 2s and one 2p orbital mix to give two sp hybrid orbitals pointing in opposite directions (180°). Each overlaps with a half-filled 2p orbital of F. There are only two bond pairs and no lone pairs on Be, so F–Be–F is linear.
\(\mathrm{H_2O}\): O is \(1s^{2}2s^{2}2p^{4}\). Its 2s and three 2p orbitals form four sp³ hybrid orbitals: two contain lone pairs and two (half-filled) form O–H bonds with H atoms. The four electron pairs are arranged roughly tetrahedrally, but only two positions have atoms. The two lone pairs repel the bond pairs strongly and reduce the angle from 109°28′ to 104°35′, so H–O–H is bent (angular, V-shaped).