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Question

Consider the molecules CH4, NH3 and H2O. Which of the given statements is false?

A
The H-O-H bond angle in H2O is smaller than the H-N-H bond angle in NH3
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B
The H-C-H bond angle in CH4 is larger than the H-N-H bond angle in NH3
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C
The H-C-H bond angle in CH4, the H-N-H bond angle in NH3 and the H-O-H bond angle in H2O are all greater than 90⁰
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D
The H-O-H bond angle in H2O is larger than the H-C-H bond angle in CH4
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Solution

The correct option is D The H-O-H bond angle in H2O is larger than the H-C-H bond angle in CH4
We can figure out the geometry of all these molecules using VSEPR and hybridisation theory.

In CH4, electron pairs = 4 + 42=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 NH3, electron pairs = 5 + 32=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 H2O, electron pairs = 6 + 22=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 H2O is larger than the H-C-H bond angle in CH4' is false.

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