Question

The correct order of bond angles (smallest first) in $H_{2}O,N{H}_3,B{F}_3$ and $Si{H}_4$ is:

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

Answer 1

The correct option is C $H_{2}O<N{H}_3<Si{H}_4<B{F}_3$

$H_{2}O$
$H_{2}O$ has a bond angle of around ${105}^o$. Two lone pairs of electrons are present on the $O$ atom. Since lone pair - lone pair repulsions are greater than bond pair - bond pair repulsions, the bond pairs try to move away from the lone pairs thereby decreasing the bond angle.
$N{H}_3$
The central atom $N$ has one lone pair of electrons. Steric number will be the sum of number of sigma bonds and number of lone pair i.e. 3 +1 = 4.
So, its shape is pyramidal. The bond angle is around ${107}^o$ due to lone pair -bond pair repulsions.
In $Si{H}_4$, no lone pairs of electrons are present on the central atom. It has four bond pairs, giving it a tetrahedral geometry and bond angle of 109°5'.
In $B{F}_3$, the central atom B has zero lone pairs of electrons and three bond pairs. So, the shape is trigonal planar with a bond angle of 120°.
$\therefore$ (c) option is correct.
$H_{2}O<N{H}_3<Si{H}_4<B{F}_3$
$\left({104.5}^{\circ }\right)$ $\left({107}^{\circ }\right)$ $\left({109.5}^{\circ }\right)$ $\left({120}^{\circ }\right)$