$H_2O$ is dipolar, whereas $BeF_2$ is not. It is because:-
A
$H_2O$ involves hydrogen bonding whereas $BeF_2$ is a discrete molecule
B
$H_2O$ is linear and $BeF_2$ is angular
C
$H_2O$ is angular and $BeF_2$ is linear
D
The electronegativity of F is greater than that of O
The answer is C.
Solution
Both O–H bonds and Be–F bonds are polar. Whether the molecule as a whole has a dipole moment depends on its shape, because bond dipoles are vectors. $BeF_2$: Be has 2 bond pairs and no lone pair ($sp$ hybridisation), so the molecule is linear (F–Be–F at $180^\circ$). The two equal bond dipoles point in opposite directions and cancel, so $\mu = 0$. $H_2O$: O has 2 bond pairs and 2 lone pairs ($sp^3$), so the molecule is angular (bent, bond angle about $104.5^\circ$). The two bond dipoles do not cancel, and $\mu = 1.85$ D. So $H_2O$ is dipolar because it is angular, while $BeF_2$ is non-polar because it is linear.
ConceptShape decides net dipole moment
Dipole Moment and Molecular Shape in past papers
2 questions from this chapter have appeared across 2 exam years.
