Question

Although geometries of ${NH}_3$ and $H_{2}O$ molecules are distorted tetrahedral, bond angle in water is less than that of ammonia.

• A. True
• B. False

Answer 1

The correct option is A True

• In $N{H}_3$, the $N$ atom is surrounded by three bond pairs ($N-H$) and one lone pair of electrons, which is present on the $N$ atom. The repulsion shown by one lone pair in $N{H}_3$ towards the three bond pair is very less because the number of bonding electrons is more as compared to a lone pair of an electron, as a result, there will be less repulsion between the lone pair and bonding electrons. So, the bond angle is 107°,nearer to normal tetrahedral bond angle of 109°.5’.
• But in the case of $H_{2}O$, the $O$ atom is surrounded by two bond pairs and two lone pairs of electrons. Consequently, the repulsion will be stronger and greater between the lone pairs & bond pairs. The bond pairs will be repelled more by the lone pairs, resulting in a decrease of bond angle to 104°, as compared to normal tetrahedral bond angle 109°.5’.
• Therefore instead $N{H}_3$ and $H_{2}O$ have tetrahedral geometry, still, they have different bond angles