# Why is the bond angle H-P-H smaller than H-N-H?

$\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.

But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?

Starting point: 2s orbitals are lower in energy than 2p orbitals.

The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron.

In the analogous case for phosphorus (phosphine, $\ce{PH_3}$), the $\ce{H-P-H}$ bond angle is 93.5 degrees. This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.

So the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.

There are two choices for the central atom:

remain unhybridized: [$\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another. As a result,

• substituents will be arranged closer together and destabilizing steric interactions will be increased
• due to the absence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be higher in energy
• the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital

or

hybridize: [$\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another. As a result,

• steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart
• due to the presence of s-character in the $\ce{X-H}$ bonds emanating from $\ce{X}$, the electrons in these orbitals will be lower in energy
• the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character

An example:

Let's now consider the example of $\ce{NH_3}$ and $\ce{NF_3}$. Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine. Because of this electron redistribution, the $\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density. Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital. The orbital will wind up with higher p-character and the s-character that has been "saved" can be used to stabilize other electrons (the lone pair!). Our prediction would be that $\ce{NF_3}$ should have more p-character in its $\ce{N-F}$ bonds than $\ce{NH_3}$ has in its $\ce{N-H}$ bonds. As a result we would expect the $\ce{F-N-F}$ bond angle in $\ce{NF_3}$ to be smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH_3}$. Indeed the bond angle in $\ce{NF_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!

Back to our problem:

Nitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2). In our $\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\ce{X~=~N}$ than when $\ce{X~=~P}$. Using the same reasoning used in our example, we would then expect the $\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed. The fact that phosphorus, being a second row element, has longer $\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.

In the case of ammonia, the shorter $\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\ce{N-H}$ bonds makes the hybridized case the lowest energy. Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.

• It is not true that hybridisation costs energy, as it is only a mathematical concept, which has no real physical meaning, it is impossible to assign an energy value to hybridisation. It is also not possible, that an element chooses to hybridise. Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule. – Martin - マーチン Jul 4 '14 at 1:48
• @Martin Thanks for the comment. Martin wrote 1) "It is not true that hybridisation costs energy" - I agree, see my edited answer. 2) "Hybridisation is a result of a certain - the thermodynamically most stable - geometric arrangement of atoms in a molecule" - I agree, in fact I said that in my initial version, "Whichever of these two pathways is most exothermic..." – ron Jul 4 '14 at 3:08
• Old post, I know, so there's a good chance I don't get an answer here, but does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance here? Yes, the cone angle is smaller, but an $sp^3$-orbital is shorter than a p-orbital, and with small atoms like hydrogen, I would think a small change in bond length would have a pretty significant effect on sterics around them phosphorus atom. – SendersReagent Mar 24 '16 at 22:03
• @DGS asked, "does the decrease in H-P-H angle due to lack of hybridization necessarily mean more steric hindrance"? If the bond angle was the only thing changing, then increased steric interaction would result. But, as you note the bond length changes too (1.42 angstroms in phosphine) making it hard to answer your question without calculations. – ron Mar 24 '16 at 23:22
• @Martin-マーチン "It is not true that hybridisation costs energy, as it is only a mathematical concept..." I agree. Yet this might lead to some confusion. Hybridisation is conceptually taught (at secondary school level) via two mechanisms. The first option distinguishes initial promotion ($\neq$ hybridisation) from the mixing itself ($=$ hybridisation). The second way is to combine promotion and mixing by calling both together as hybridisation. – Linear Christmas Jul 29 '17 at 21:10

First note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better.

Second note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves unnatural or anomalous.

If you compare nitrogen with phosphorus, you will note, that the former is much smaller than the latter, i.e. van der Waals radii $r(\ce{N})=155~\mathrm{pm};\ r(\ce{P})=180~\mathrm{pm}$ (ref. wikipedia), covalent radii $r(\ce{N})=71~\mathrm{pm};\ r(\ce{P})=107~\mathrm{pm}$ (ref. wikipedia). Therefore also the orbitals in nitrogen are smaller, and $\ce{s}$ and $\ce{p}$ orbitals will occupy more of the same space than in phosphorus. As a result the $\ce{N-H}$ bond distance will naturally also be shorter.

A lone pair is usually most stable in an orbital that has high $\ce{s}$ character. Bonds will most likely be formed with the higher lying $\ce{p}$ orbitals. The orientation of these towards each other is exactly $90^\circ$.

In ammonia this would lead to very close $\ce{H\cdots{}H}$ contacts, which are repulsive and therefore the hydrogen atoms are pushed away from each other. This is possible since in the second period the $\ce{s-p}$ splitting is still very small and the nitrogen $\ce{s}$ orbital is accessible for the hydrogen atoms. This will ultimately result in mixing $\ce{s}$ and $\ce{p}$ orbitals for nitrogen in the respective molecular orbitals. This phenomenon can be referred to as hybridisation - the linear combination of orbitals from the same atom. This term is therefore somewhat independent from its most common usage.

It is also very important to know, that the molecular wavefunction of a molecule has to reflect its overall symmetry. In this case it is $C_{3v}$, which means there is a threefold rotational axis and three vertical mirror planes (the axis is element of these planes). This gives also rise to degenerate orbitals. A canonical orbital picture has to reflect this property (BP86/cc-pVDZ; valence orbitals are ordered with increasing energy from left to right).

Note that the lowest lying valence molecular orbital is formed only from $\ce{s}$ orbitals (There is one additional $\ce{1s^2-N}$ core orbital.)
Now Natural Bond Orbital (NBO) Theory can be used to transform these delocalised molecular orbitals to a more common and familiar bonding picture, making use of atomic hybrid orbitals. This method is called localising orbitals, but it has the expense of losing the energy eigenvalue that may be assigned to canonical orbitals (NBO@BP86/cc-pVDZ; valence NBO cannot be ordered by energy levels).

In this theory you will find three equivalent $\ce{N-H}$ bonds, that are composed of $32\%~\ce{1s-H}$ and $68\%~\ce{s^{$0.87$}p^3-N}\approx\ce{sp^3-N}$ orbitals. Note that the lone pair orbital at nitrogen has a slightly higher $\ce{s}$ orbital contribution, i.e. $\ce{s^{1.42}p^3-N}\approx\ce{sp^3-N}$.

So the thermodynamically most favoured angle is found to be $107^\circ$ due to a compromise between optimal orbital overlap and least internuclear repulsion.

The canonical bonding picture in phosphine is very similar to ammonia, only the orbitals are larger. Even in this case it would be wrong to assume, that there is no hybridisation present at all. However, the biggest contribution to the molecular orbitals stems from the $\ce{p}$ orbitals at phosphorus.

Applying the localisation scheme, one end up with a different bonding picture. Here are three equal $\ce{P-H}$ bonds that are composed of $48\%~\ce{1s-H}$ and $52\%~\ce{s^{$0.5$}p^3-P}$ orbitals. The lone pair at phosphorus is composed of $57\%\ce{s} + 43\%\ce{p}$ orbitals.

One can see the difference of the molecules also in their inversion barrier, while for ammonia the inversion is readily available at room temperature, $\Delta E \approx 6~\mathrm{kcal/mol}$, it is very slow for phosphine, $\Delta E \approx 40~\mathrm{kcal/mol}$.

This is mostly due to the fact, that the nitrogen hydrogen bonds have already a significant $\ce{s}$ orbital contribution, which can be easily increase, to form the planar molecule with formally $\ce{sp^2}$ hybrids.

• @ martin : so hybridization is totally hypothetical and not practical,right? – user5764 Jul 5 '14 at 6:36
• @user36790 In the most common usage, this is true, but it can give you somewhat more insight into the shape of molecules. In a more genaral meaning, hybridisation just refers to the mixing of orbitals at the same atom in a molecule, and that might be observable. – Martin - マーチン Jul 5 '14 at 7:27

Since Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.

• d-orbitals are too far removed to do anything in phosphorus and sulphur. Check out Martin's answer. – Jan Nov 11 '15 at 15:00