Question

Which of the following compounds has the highest bond angle?

• A. $N{H}_3$
• B. $H_{2}O$
• C. $C{H}_4$
• D. All are same

Answer 1

The correct option is C $C{H}_4$

$N{H}_3$ and $H_{2}O$ and $C{H}_4$ have 1, 2 and zero lone pairs respectively with the hybridization of $s{p}^3$ and tetrahedral geometry. But the shapes of the three molecules respectively are pyramidal, bent and tetrahedral.

So, in $N{H}_3$ and $H_{2}O$, the bond angles are less than ${109}^{\circ }{28}^{\prime }$ due to LP-BP and LP-LP repulsions respectively.

The bond angle in $C{H}_4$ is ${109}^{\circ }{28}^{\prime }$ as it is a regular tetrahedron.

$N{H}_3$
$H_{2}O$
$C{H}_4$