In my textbook it is written that the order of basic strength of pnictogen hydrides is

$$\ce{NH3 > PH3 > AsH3 > SbH3 > BiH3}$$

I tried but could not find any explanation as to why this happens. What is the explanation?


3 Answers 3


Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons. It is the lone pair of electrons that makes these molecules nucleophilic or basic. As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (the nitrogen lone pair is contained in the n=2 shell, while the bismuth lone pair is in the n=6 shell). As the electron density of the lone pair is spread over a greater volume and is consequently more diffuse, the lone pair of electrons becomes less nucleophilic, less basic.

Commment: But greater the EN of the central atom , lesser would be its tendency to donate electrons right?

Response to Comment: The ENs of the central atoms are N (3.04), P (2.19), As (2.18), Sb (2.05), Bi (2.02). All of these atoms, except for nitrogen have similar ENs and I think the electron density argument is valid for them. However, in the case of nitrogen and phosphorous there is a significant EN difference that would tend to argue in the direction opposite to my answer. However there is another important difference between N and P. The lone pair in ammonia is in an sp3 orbital. In all of the other cases the central atom is essentially unhybridized (~90° H-X-H angles) and the lone pair exists in an s orbital. Therefore the lone pair electron density in ammonia (being in an sp3 orbital) is effectively increased compared to phosphine where the lone pair is in an s orbital. So even though the EN difference between N and P is significant, when hybridization differences of the central atom are taken into account the electron density argument still explains the trend in basicity.

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    $\begingroup$ Usually, my teacher says that the electronegativity arguments are valid for elements in the same period. Trends involving traversal along a group are better explained by the size of the central atom, not sure if that is logically valid. $\endgroup$
    – Anmoldeep
    Sep 22, 2021 at 7:55

The hydrides of nitrogen family have one lone pair of electrons on their central atom. Therefore,they act as Lewis bases.As we go done the group, the basic character of these hydrides decreases. Nitrogen atom has the smallest size among the hydrides.Therefore the lone pair is concentrated on a small region and electron density on it is the maximum.Consequently, its electron releasing tendency is maximum.As the size of the central atom increases down the family, the electron density also decreases. As a result, the electron donating capacity or the basic strength decreases down the group.


Realise that there is a notable drop in basicity from nitrogen to phosphorus and then a slow and continuous further diminishing. The notable drop is due to the difference of the molecular structures of ammonia and phosphane.

As I answered elsewhere on this network, the ‘ideal’ bonding situation from an orbital point of view would be to just use p-orbitals to bond and leave the lone pair in an s-type orbital which generally has a lower energy. Doing this allows for a greater σ bonding stabilisation. And that is what all pnictogens below phosphorus (and phosphorus to a certain extent, although the bond angle is still $3^\circ$ off) do: their lone pair can be considered a pure s-type orbital at first approximation.

Ammonia, however, has space constraints since nitrogen is so much smaller than phosphorus and all the other pnictogens. If it were to assume $90^\circ$ bond angles, the hydrogen atoms would end up too close together resulting in steric strain. Thus, nitrogen moves the hydrogens apart by not bonding with pure p orbitals but with $\mathrm{sp^3}$ hybrid orbitals (in nitrogen’s case, the hybridisation is actually very close to $25~\%$ s-character and thus to $\mathrm{sp^3}$!). This means that the lone pair must also be localised in an $\mathrm{sp^3}$ type orbital which in turn means that it is not spherical and diffuse but ‘points in a direction’. This pointing allows for acidic hydrogens to easily attach to it and thus explain nitrogen’s and ammonia’s high basicity. I would also like to highlight an answer of Martin on the topic of $\ce{NH3}$ versus $\ce{PH3}$ hybridisation.

Further down the group, i.e. behind phosphorus, a simple diffusity argument is sufficient to explain the slightly reducing basicity. The higher the period, the more populated orbitals lie beneath the s-type lone pair thus the further out it is (also the greater the atom becomes). And the more diffuse a lone pair is the less likely it becomes to participate in bonding or to be basic in any way.

  • $\begingroup$ I realize that this is an older post so I might not get an answer here, but could you also explain the lower basicity of orbitals with more s character using an energetic argument? When I learned about the higher acidity of alkynes, one explanation was that the lone pair in the conjugate base was more stabilized because it was in a lower-energy sp orbital (as opposed to the sp3 orbital of an alkane. Could you apply the same argument here, or is it not applicable since PH3's lone pair is in a 3s orbital, not an orbital made of 2s and 2p orbitals like in ammonia? $\endgroup$
    – Akash
    Sep 2, 2022 at 13:42
  • $\begingroup$ @Akash Yes, that is true; the entire line of thinking is ultimately energetic and orbitals with a higher s component are lower in energy than those with a lower s component. And then as you stated. $\endgroup$
    – Jan
    Oct 1, 2022 at 15:12

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