$\mathrm{H}_2 \mathrm{O}$ is polar, whereas $\mathrm{BeF}_2$ is not because

1. electronegativity of $\mathrm{F}$ is greater than that of $\mathrm{O}$
2. $\mathrm{H}_2 \mathrm{O}$ involves $\mathrm{H}$-bonding, whereas $\mathrm{BeF}_2$ is a discrete molecule
3. $\mathrm{H}_2 \mathrm{O}$ is angular and $\mathrm{BeF}_2$ is linear
4. $\mathrm{H}_2 \mathrm{O}$ is linear and $\mathrm{BeF}_2$ is angular.

Solution

Because of linear shape, dipole moments cancel each other in $\mathrm{BeF}_2(\mathrm{~F}=\mathrm{Be} \sigma \mathrm{F})$ and thus, it is non-polar, whereas $\mathrm{H}_2 \mathrm{O}$ is $\mathrm{V}$-shaped and hence, it is polar.

Asked in: NEET 2020 (Phase 1)

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