Chemistry - Bond angles for the hydrides

Solution 1:

The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides.

According to VSEPR, electron domains (i.e. bonds and lone pairs) are arranged around the central atom in geometries that minimize electrostatic repulsion between the electron domains. These geometries also happen to be solutions to the Schrödinger equation for the set of molecular orbitals as linear combinations of atomic orbitals, and in the molecular orbital case, they arise not from electrostatic repulsion, but strictly from the mathematics of the linear combinations producing a minimum energy self-consistent set of orbitals.

VSEPR is qualitatively simpler for small molecules. All of the binary hydrides in Group IV, V, and VI have steric number four around the central atom. This means that the number of lone pairs and the number of bonding domains add up to four. These species can also be described using the $\ce{AXE}$ method, where $\ce{A}$ denotes the central atom, $\ce{X}$ denotes an atom bonded to $\ce{A}$ and $\ce{E}$ denotes a lone pair of electrons.

  • The Group IV hydrides all have four bonds and no lone pairs. We describe this arrangement as $\ce{AX4}$.
  • The Group V hydrides all have three bonds and one lone pairs. We describe this arrangement as $\ce{AX3E}$.
  • The Group VI hydrides all have two bonds and two lone pairs. We describe this arrangement as $\ce{AX2E2}$.

There are several factors that affect the degree of repulsion:

  1. Lone pairs - Lone pair domains are more diffuse and have higher repulsion than bonding domains.
  2. The nature of $\ce{X}$ - Larger and/or more electropositive atoms have higher repulsion.
  3. Bond order - Bonds with higher bond orders have higher repulsion than bonds with lower bond orders.

Since $\ce{X=H}$ in all cases, we only need to worry about the first factor: lone pairs.

The arrangements $\ce{AX4}$, $\ce{AX3E}$, and $\ce{AX2E2}$ all have four electron domains, and thus the base electron domain geometry group is tetrahedral.

The differences in bond angle between $\ce{CH4}$, $\ce{NH3}$, and $\ce{H2O}$ originate from the increasing number of lone pairs:

  • $\ce{CH4}$ has no lone pairs: the bond angles are the ideal tetrahedral bond angles of $109.5^\circ$.
  • $\ce{NH3}$ has one lone pair: the bond angles are contracted a little because of the grater repulsion of the lone pair to $107.8^\circ$.
  • $\ce{H2O}$ has two lone pairs: the bond angle is contracted further due to the repulsion of two lone pairs to $104.5^\circ$.

All of the Group IV hydrides will have perfect tetrahedral geometry due to having four bonds to the same atom and no lone pairs. Thus $\ce{CH4}$, $\ce{SiH4}$, $\ce{GeH4}$, $\ce{SnH4}$, and $\ce{PbH4}$ all have bond angles of $109.5^\circ$.

For Group V, the bond angles are:

$$\begin{array}{c|c} \ce{AH3}&\text{angle}\\ \hline \ce{NH3}&107.8^\circ \\ \ce{PH3}&93.5^\circ \\ \ce{AsH3}&91.8^\circ \\ \ce{SbH3}&91.7^\circ \\ \ce{BiH3}&90.5^\circ \\ \end{array}$$

Notice that there is a large decrease from $\ce{NH3}$ to $\ce{PH3}$ and then very small changes for $\ce{AsH3}$, $\ce{SbH3}$, and $\ce{BiH3}$.

You could make the argument that the larger orbitals of the heavier elements will have high repulsion than the smaller orbitals of nitrogen. This is where VSEPR breaks down. VSEPR works well for some compounds because there is significant $s-p$ mixing when constructing the molecular orbitals. This mixing need not happen for the heavier elements. One molecular orbital solution for $\ce{PH3}$, for example, indicates very little contribution from the $\ce{P}\ 3s$ orbital. We can describe the $\ce{P-H}$ bonds as predominantly $\ce{P}_{3p}\ce{-H}_{1s}$ bonds. Thus the geometry around $\ce{P}$ is determined mostly by the orientation of the $p$ orbitals.

If we look at the Group VI hydrides, we see the same trend: a large decrease from $\ce{H2O}$ to $\ce{H2S}$ and then very little change. The same argument can be made for primarily $\ce{A}_{p}\ce{-H}_{1s}$ bonding in the heavier Group VI hydrides as for the heavier Group V hydrides.

$$\begin{array}{c|c} \ce{AH2}&\text{angle}\\ \hline \ce{H2O}&104.5^\circ \\ \ce{H2S}&92.1^\circ \\ \ce{H2Se}&91.0^\circ \\ \ce{H2Te}&90.0^\circ \\ \end{array}$$

Solution 2:

VSEPR does not do a good job in predicting these geometries; it only works by chance for the second period.

For the 14th group, (the IV group in older nomenclature) things are pretty much as Ben explained them. We have a consistent number of substituents (four) and the same number of lone pairs (zero) around the central atom. Therefore, we can arrange the substituents as far away from each other as possible to generate a tetrahedric structure, and molecular orbital calculations indicate that this will lead to the lowest energy. You can predict what the molecular orbitals will look like from symmetry considerations. Changing the central atom only changes the bond length.

Once we move on to group 15, we suddenly have a lone pair on the central atom in ground state. This lone pair will occupy an s-orbital in the isolated atom, and it would be most stable to keep this lone pair in an s-atom for two reasons:

  • p-orbitals actually extend into a direction of space, so the overlap with a bonding partner will be larger

  • the s-orbital is ‘closer to the nucleus’ (in a non-quantum view) meaning that the negative charge of the electrons is better stabilised

As such, there is a general tendency across the entire periodic table to keep lone pairs in orbitals with an s-character as large as possible. The largest s-character is, of course, a pure s-orbital. Therefore, we should predict a bond angle of $90^\circ$ for all group 15 hydrides — which is well met for every case except ammonia. (This is similarly true for the group 16 ones with the exception of oxygen. One lone pair must reside in a p-orbital since only one s-orbital is available.)

So what’s wrong with ammonia and water? Well, the oxygen and nitrogen atoms are much smaller than their heavier homologues and therefore the hydrogen atoms would be much closer to each other at an ideal $90^\circ$ angle. This, however, would induce new steric stress between the hydrogen atoms, destabilising the entire structure. Therefore, the hydrogens shift out just far enough to where they are adequately far away from each other while still allowing the central atom to have as much s-character in its lone pair as possible. The sweet spot happens to be $104.5^\circ$ for water and $107.8^\circ$ for ammonia.

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