molecular orbital theory - 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?

Answer

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.