Question

Which of the following has the least bond angle?

• A. $N{H}_3$
• B. $Be{F}_2$
• C. $H_{2}O$
• D. $C{H}_4$

Answer 1

The correct option is C $H_{2}O$

In beryllium flouride there are two bond pairs and no lone pairs on the central atom beryllium. So, it exists in linear structure with bond angle of ${180}^o$.

All $N{H}_3$, $H_{2}O$ and $C{H}_4$ have four electron pairs. So, the expected structure is tetrahedral for all the three molecules.

In $C{H}_4$ there are four bond pairs and no lone pairs on the central atom.

In $N{H}_3$ there are three bond pairs and one lone pair on the central atom.

In $H_{2}O$ there are two bond pairs and two lone pairs on the central atom.

The repulsions in a molecule follow the order
lone pair - lone pair > lone pair - bond pair > bond pair - bond pair.

Due to the strong repulsions, the presence of lone pairs on central atom decreases the bond angle in the molecule.



Due to the presence of two lone pairs, the bond angle in the water molecule decreases more than ammonia (one lone pair) from expected ${109}^o{28}^{\prime }$.

Therefore, water has the least bond angle.