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Question

The percentage of p-character in each orbital of the central atom used for bonding in NH3 is:
  1. 25%
  2. 75%
  3. >75%
  4. 33.3%


Solution

The correct option is C >75%
As we know that the hybridization of N in NH3 is sp3, the percent p character must be 3/4×100=75%.

This happens when the structure doesn't have any lone pairs and it is a simple tetrahedral structure. But, here it has a pyramidal shape.

And due to lone pair-bond pair repulsion, the bond angle decreases from an ideal 109.5 degree to 107 degree.

As bond angle decreases, % s character also decreases. But % p character increases. As a result of which here % p will be more than 75% and % s will be less than 25%.

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