$\mathrm{H}_2 \mathrm{O}$ is dipolar whereas $\mathrm{BeF}_2$ is not. It is because:

$\mathrm{H}_2 \mathrm{O}$ is dipolar whereas $\mathrm{BeF}_2$ is not. It is because:
  1. the electronegativity of $\mathrm{F}$ is greater than that of $\mathrm{O}$
  2. $\mathrm{H}_2 \mathrm{O}$ involves hydrogen bonding whereas $\mathrm{BeF}_2$ is a discrete molecule
  3. $\mathrm{H}_2 \mathrm{O}$ is linear and $\mathrm{BeF}_2$ is angular
  4. $\mathrm{H}_2 \mathrm{O}$ is angular and $\mathrm{BeF}_2$ is linear

Solution

Water $\left(\mathrm{H}_2 \mathrm{O}\right)$ is polar because of the bent shape of the molecule. The shape means most of the negative charge from the oxygen on side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule. This is an example of polar covalent chemical bonding. However, because of linear shape, dipole moments cancel each other in $\mathrm{BeF}_2\left(\mathrm{~F}^{-} \leftarrow \mathrm{Be}^{-} \rightarrow \mathrm{F}\right)$ and thus, it is non-polar. Related Theory Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into $a$ cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals.

Asked in: NEET 2004

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