# Why sp3 hybridisation in NH3 but bonding with pure p orbitals in case of other hydrides of group 15 [duplicate]

According to my book, the following line was given relating to the bond angle in Group 15 elements

The bond angle in PH3, AsH3 and SbH3 is close to 92° which suggests that the orbitals used for bonding are close to pure p-orbitals.

I wanted to ask if there is any reason for having sp3 hybridised orbitals for N, but not so for other elements of that group in their hydrides. Or is it completely based on experimental observations?

## 2 Answers

The answer to this question is actually much more complicated than most textbooks make it seem. There's a great explanation in Abright, Burdett and Whango "Orbital Interactions in Chemistry" Ch. 9, which I'll try to summarize here.

As explained in this answer: Why is the inversion barrier larger in PH3 than it is in NH3?, the lone pair orbital in a planar XH3 molecule would be pure $$p$$. Thus the bonding orbitals average to $$sp^2$$. As the bond angle decreases from 180, the loss of symmetry makes possible mixing between that lone pair orbital (originally a pure central atom $$p$$ orbital) and an anti-bonding orbital involving the central atom $$s$$ orbital. This "s-p mixing" stabilizes the lone pair orbital at the expense of the (empty) antibonding orbital. This mixing necessarily also involves the bonding orbital that involves the central atom $$s$$ atomic orbital.

In nitrogen, the low energy of the atomic $$s$$ orbital and the strong overlap interaction of that $$s$$ orbital with the hydrogen $$s$$ orbitals means that the bonding orbital made from the atomic $$s$$ orbital is quite low energy and the antibonding orbital is quite high energy. As a result, mixing with the $$p$$ lone pair orbital is less favorable, so less mixing occurs and the angle does not deviate as far from 180.

With the larger central atoms, the interaction of the $$s$$ orbital with the hydrogen $$s$$ orbitals is less, so the antibonding orbital is not as high energy. Likewise, the $$s$$ bonding orbital is not as low energy. So mixing with the lone pair $$p$$ orbital is more favorable, and the bond angle deviates farther from 180. The figure below (from Albright, figure 9.9) plots the energy of the occupied orbitals. The lowest line represents the $$s$$ bonding orbital and the top line the lone pair.

The key point here is that more $$s$$ character is mixed into the lone pair orbital as you move down from NH3.

I would recommend you read Linus Pauling’s Nature of the Chemical Bond, an excellent book for VBT.

Basically, the reason for hybridisation is to increase bond strength, which is maximised at 2 when one has 25% s and 75% p character.

However, having a lone pair in an s orbital is more stable than having it in a hybridised orbital (due to promotion energy), so in cases with lone pairs we end up with the lone pairs having more s character and the bonding pairs having more p character. This is the same for any compound with lone pairs.

Case in point- the lone pairs in water have 44% s character, while the bonding pairs have 6% each.

To answer your question, nitrogen is much more electronegative than hydrogen, so the interaction is somewhat ionic. Hence, the bond angle increases just like in water- the predicted bond angle for water would be about 90 degrees, but that is obviously not the case.

At the end of the day, VBT is much more complex than one might think if one learns it in organic chemistry.

• I would like to add that a pure s orbital has bond strength of 1, while a pure p orbital has bond strength of sqrt 3. – ANZGC FlyingFalcon Feb 16 '19 at 15:49
• Also, hybridisation is a mathematical concept, so the notion of it costing energy is simply Pauling’s way of explaining why lone pairs always have more s character. – ANZGC FlyingFalcon Feb 16 '19 at 15:55
• While Pauling's ideas are very important in historical context and are useful for introductory qualitative understanding of bonding, the quantitative aspects are largely out of date. For example, your number of 44% s character in H2O lone pairs is MUCH higher than is generally accepted now based on MO theory and computational approaches. – Andrew Feb 16 '19 at 18:31
• I guess Pauling still advocated for bunny ears on water, at least that is what this number reflects, that has been disproved though on various computational and experimental levels. – Martin - マーチン Feb 16 '19 at 18:49
• That’s definitely true, but considering the OP asked for a valence bond treatment that’s the best hybridisation” answer I could give. – ANZGC FlyingFalcon Feb 17 '19 at 15:59