Question

Consider the molecules $C{H}_4$, $N{H}_3$ and $H_{2}O$. Which of the given statements is false?

• A. The H-O-H bond angle in $H_{2}O$ is smaller than the H-N-H bond angle in $N{H}_3$
• B. The H-C-H bond angle in $C{H}_4$ is larger than the H-N-H bond angle in $N{H}_3$
• C. The H-C-H bond angle in $C{H}_4$, the H-N-H bond angle in $N{H}_3$ and the H-O-H bond angle in $H_{2}O$ are all greater than 90⁰
• D. The H-O-H bond angle in $H_{2}O$ is larger than the H-C-H bond angle in $C{H}_4$

Answer 1

The correct option is D The H-O-H bond angle in $H_{2}O$ is larger than the H-C-H bond angle in $C{H}_4$

We can figure out the geometry of all these molecules using VSEPR and hybridisation theory.

In $C{H}_4$, electron pairs = $\frac{4+4}{2}=4$

Since we have 4 electron pairs and 4 hydrogens on carbon, this will give us a tetrahedral structure with all bond pairs and each H-C-H bond will be at an angle of 109.5⁰

In $N{H}_3$, electron pairs = $\frac{5+3}{2}=4$

Here, we got 4 electron pairs but there are only 3 hydrogens on nitrogen. So, this will give us a tetrahedral geometry with 3 bond pairs and one lone pair. And we know from VSEPR that lone pair-bond pair repulsion is greater than bond pair-bond pair. So, the 3 bond pairs will be skewed away from the lone pair giving a pyramidal shape. The angle of H-N-H bond will be slightly reduced, as a result, to around 107⁰

Finally in $H_{2}O$, electron pairs = $\frac{6+2}{2}=4$

4 electron pairs and 2 bonds with hydrogen. This means we are left with 2 lone pairs. Again, we will have tetrahedral geometry but with 2 lone pairs. And since lone pair-lone pair repulsion is even stronger, the bond pairs will be further skewed away reducing the H-O-H bond angle to about 104.5⁰

So, we can conclude that bond angle of H-C-H > H-N-H > H-O-H > 90⁰. This means that the statement - 'The H-O-H bond angle in $H_{2}O$ is larger than the H-C-H bond angle in $C{H}_4$' is false.