Why does bond angle decrease in the order H2O, H2S, H2Se?

I know that bond angle decreases in the order $\ce{H2O}$, $\ce{H2S}$ and $\ce{H2Se}$. I wish to know the reason for this. I think this is because of the lone pair repulsion but how?

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths: \begin{array}{lcc} \text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\ \hline \ce{H2O} & 104.5 & 96 \\ \ce{H2S} & 92.3 & 134 \\ \ce{H2Se}& 91.0 & 146 \\ \hline \end{array}

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being $\ce{sp^3}$ hybridized, but due to lone pair - lone pair electron repulsions, the lone pair-X-lone pair angle opens up slightly in order to reduce these repulsions, thereby forcing the $\ce{H-X-H}$ angle to contract slightly. So instead of the $\ce{H-O-H}$ angle being the perfect tetrahedral angle ($109.5^\circ$) it is slightly reduced to $104.5^\circ$. On the other hand, both $\ce{H2S}$ and $\ce{H2Se}$ have no orbital hybridization. That is, The $\ce{S-H}$ and $\ce{Se-H}$ bonds use pure $\ce{p}$-orbitals from sulfur and selenium respectively. Two $\ce{p}$-orbitals are used, one for each of the two $\ce{X-H}$ bonds; this leaves another $\ce{p}$-orbital and an $\ce{s}$-orbital to hold the two lone pairs of electrons. If the $\ce{S-H}$ and $\ce{Se-H}$ bonds used pure $\ce{p}$-orbitals we would expect an $\ce{H-X-H}$ interorbital angle of $90^\circ$. We see from the above table that we are very close to the measured values. We could fine tune our answer by saying that in order to reduce repulsion between the bonding electrons in the two $\ce{X-H}$ bonds the angle opens up a bit wider. This explanation would be consistent with the $\ce{H-S-H}$ angle being slightly larger than the corresponding $\ce{H-Se-H}$ angle. Since the $\ce{H-Se}$ bond is longer then the $\ce{H-S}$ bond, the interorbital electron repulsions will be less in the $\ce{H2Se}$ case alleviating the need for the bond angle to open up as much as it did in the $\ce{H2S}$ case.

The only new twist on all of this that some universities are now teaching is that water is not really $\ce{sp^3}$ hybridized, the $\ce{sp^3}$ explanation does not fit with all of the experimentally observed data, most notably the photoelectron spectrum. The basic concept introduced is that "orbitals only hybridize in response to bonding." So in water, the orbitals in the two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized, but one lone pair resides in a nearly pure p-orbital and the other lone pair is in a roughly $\ce{sp}$ hybridized orbital.

• Neat answer Ron. The H-S-H bond can open up a bit because the other side of the p-orbital is more empty as a result of the S-H bond, but not too much of course because there is still electron density there. Is that the force stopping it from going all the way to a 180 degree bond? Or are there other forces involved? (hope this was a bit clear, just curious) – Jori Aug 7 '14 at 21:34
• Thanks Jori. Each S-H bond uses one p orbital and each p orbital is oriented roughly 90 degrees from the other. Px and Py, or PX and Pz, or Py and Pz - pick which two you'd like to use to make the two S-H bonds, but they are all 90 degrees from one another. There's no way to get bonds 180 degrees apart using pure p orbitals. – ron Aug 7 '14 at 21:39
• Yes the bonds can bend a little, but two p orbitals (or more accurately, their wavefunctions) cannot interact since they are orthogonal to each other. Also, yes, there will always be electron density throughout the entire p orbital. Bonding will shift the density somewhat, but it will still exist throughout the entire orbital. A picture would probably help a lot. – ron Aug 7 '14 at 22:05
• Hybridisations are a self-convinient model to explain certain facts. Why have we ignored the hybridisations in H2S H2Se? Was it just to support our experimental observations or does it have a concrete reason? – the_random_guy42 Mar 23 '15 at 8:12
• In addition to "the sp3 explanation does not fit with all of the experimentally observed data," it is also inconsistent with our theory of molecular orbitals, from which the diagrams in porphyrin's answer come. To quote Albright's "Orbital interactions in chemistry", the idea that S uses pure p orbitals for bonding in SH2 "is far removed from reality." – Andrew Feb 5 '19 at 13:22

The question asks why water has a larger angle than other hydrides of the form $\ce{XH2}$ in particular $\ce{H2S}$ and $\ce{H2Se}$. There have been other similar questions, so an attempt at a general answer is given below.
There are, of course, many other triatomic hydrides, $\ce{LiH2}$, $\ce{BeH2}$, $\ce{BeH2}$, $\ce{NH2}$, etc.. It turns out that some are linear and some are V shaped, but with different bond angles, and that the same general explanation can be used for each of these cases.

It is clear that as the bond angle for water is neither $109.4^\circ$, $120^\circ$, nor $180^\circ$ that $\ce{sp^3}$, $\ce{sp^2}$ or $\ce{sp}$ hybridisation will not explain the bond angles. Furthermore, the UV photoelectron spectrum of water, which measures orbital energies, has to be explained as does the UV absorption spectra.

The way out of this problem is to appeal to molecular orbital theory and to construct orbitals based upon $\ce{s}$ and $\ce{p}$ orbitals and their overlap as bond angle changes. The orbital diagram was worked out a long time ago is now called a Walsh diagram (A. D. Walsh J. Chem. Soc. 1953, 2262; DOI: 10.1039/JR9530002260). The figure below sketches such a diagram, and the next few paragraphs explain the figure.

The shading indicates the sign (phase) of the orbital, 'like to like' being bonding otherwise not bonding. The energies are relative as are the shape of the curves. On the left are the orbitals arranged in order of increasing energy for a linear molecule; on the right those for a bent molecule. The orbitals labelled $\Pi_\mathrm{u}$ are degenerate in the linear molecule but not so in the bent ones. The labels $\sigma_\mathrm{u}$, $\sigma_\mathrm{g}$ refer to sigma bonds, the $\mathrm{g}$ and $\mathrm{u}$ subscripts refer to whether the combined MO has a centre of inversion $\mathrm{g}$ (gerade) or not $\mathrm{u}$ (ungerade) and derive from the irreducible representations in the $D_\mathrm{\infty h}$ point group. The labels on the right-hand side refer to representations in the $C_\mathrm{2v}$ point group.

Of the three $\Pi_\mathrm{u}$ orbitals one forms the $\sigma_\mathrm{u}$, the other two are degenerate and non-bonding.
One of the $\ce{p}$ orbitals lies in the plane of the diagram, the other out of the plane, towards the reader.

When the molecule is bent this orbital remains non-bonding, the other becomes the $\ce{3a_1}$ orbital (red line) whose energy is significantly lowered as overlap with the H atom's s orbital increases.
To work out whether a molecule is linear or bent all that is necessary is to put electrons into the orbitals. Thus, the next thing is to make a list of the number of possible electrons and see what diagram predicts. \begin{array}{rcll} \text{Nr.} & \text{Shape} & \text{molecule(s)} & \text{(angle, configuration)} \\ \hline 2 & \text{bent} & \ce{LiH2+} & (72,~\text{calculated})\\ 3 & \text{linear} & \ce{LiH2}, \ce{BeH2+} &\\ 4 & \text{linear} & \ce{BeH2}, \ce{BH2+} &\\ 5 & \text{bent} & \ce{BH2} & (131, \ce{[2a_1^2 1b_2^2 3a_1^1]})\\ 6 & \text{bent} & \ce{^1CH2} & (110, \ce{[1b_2^2 3a_1^2]})\\ & & \ce{^3CH2} & (136, \ce{[1b_2^2 3a_1 1b_1^1]})\\ & & \ce{BH2^-} & (102)\\ & & \ce{NH2+} & (115, \ce{[3a_1^2])}\\ 7 & \text{bent} & \ce{NH2} & (103.4, \ce{[3a_1^2 1b_1^1]})\\ 8 & \text{bent} & \ce{OH2} & (104.31, \ce{[3a_2^2 1b_1^2]})\\ & & \ce{NH2^-} & (104)\\ & & \ce{FH2^+} &\\ \hline \end{array}

Other hydrides show similar effects depending on the number of electrons in $\ce{b2}$, $\ce{a1}$ and $\ce{b1}$ orbitals; for example: \begin{array}{ll} \ce{AlH2} & (119, \ce{[b_2^2 a1^1]}) \\ \ce{PH2} & (91.5, \ce{[b_2^2 a_1^2 b_1^1]}) \\ \ce{SH2} & (92)\\ \ce{SeH2} & (91)\\ \ce{TeH2} & (90.2)\\ \ce{SiH2} & (93)\\ \end{array}

The agreement with experiment is qualitatively good, but, of course the bond angles cannot be accurately determined with such a basic model only general trends.

The photoelectron spectrum (PES) of water shows signals from $\ce{2a1}$, $\ce{1b2}$, $\ce{3a1}$, $\ce{1b1}$ orbitals, ($21.2$, $18.7$, $14.23$, and $\pu{12.6 eV}$ respectively) the last being non-bonding as shown by the lack of structure. The signals from $\ce{3b2}$ and $\ce{3a1}$ orbitals show vibrational structure indicating that these are bonding orbitals.

The range of UV and visible absorption by $\ce{BH2}$, $\ce{NH2}$, $\ce{OH2}$ are $600 - 900$, $450 - 740$, and $150 - \pu{200 nm}$ respectively. $\ce{BH2}$ has a small HOMO-LUMO energy gap between $\ce{3a1}$ and $\ce{1b1}$ as the ground state is slightly bent. The first excited state, is predicted to be linear as its configuration is $\ce{1b_2^2 1b_1^1}$ and this is observed experimentally.

$\ce{NH2}$ has a HOMO-LUMO energy gap from $\ce{3a_1^2 1b_1^1}$ to $\ce{3a_1^1 1b_1^2 }$, so both ground and excited states should be bent, the excited state angle is approx $144^\circ$. Compared to $\ce{BH2}$, $\ce{NH2}$ is more bent so the HOMO-LUMO energy gap should be larger as observed.

$\ce{OH2}$ has a HOMO-LUMO energy gap from $\ce{3a_1^2 1b_1^2}$ to $\ce{3a_1^2 1b_1^1 4a_1^1 }$, i.e. an electron promoted from the non-bonding orbital to the first anti-bonding orbital. The excited molecule remains bent largely due to the strong effect of two electrons in $\ce{3a1}$ counteracting the single electron in $\ce{4a1}$. The bond angle is almost unchanged at $107^\circ$, but the energy gap will be larger than in $\ce{BH2}$ or $\ce{NH2}$, again as observed.

The bond angles of $\ce{NH2}$, $\ce{NH2-}$ and $\ce{NH2+}$ are all very similar, $103^\circ$, $104^\circ$, and $115^\circ$ respectively. $\ce{NH2}$ has the configuration $\ce{3a_1^2 1b_1^1}$ where the $\ce{b1}$ is a non bonding orbital, thus adding one electron makes little difference, removing one means that the $\ce{3a_1}$ orbital is not stabilised as much and so the bond angle is opened a little.

The singlet and triplet state $\ce{CH2}$ molecules show that the singlet has two electrons in the $\ce{3a1}$ orbital and has a smaller angle than the triplet state with just one electron here and one in the non-bonding $\ce{b1}$, thus the triplet ground state bond angle is expected to be larger than the singlet.

As the size of the central atom increases, its nucleus becomes more shielded by core electrons and it becomes less electronegative. Thus going down the periodic table the $\ce{X-H}$ bond becomes less ionic, more electron density is around the $\ce{H}$ atom thus the $\ce{H}$ nucleus is better shielded, and thus the $\ce{X-H}$ bond is longer and weaker. Thus, as usual with trends within the same family in the periodic table, the effect is, basically, one of atomic size.

Molecules with heavier central atom, $\ce{SH2}$, $\ce{PH2}$, etc. all have bond angles around $90^\circ$. The decrease in electronegativity destabilises the $\Pi_\mathrm{u}$ orbital raising its energy. The $\ce{s}$ orbitals of the heavier central atoms are larger and lower in energy than those of oxygen, hence these orbitals overlap with the $\ce{H}$ atom's $\ce{s}$ orbital more weakly. Both these factors help to stabilise the linear $3\sigma_\mathrm{g}$ orbital and hence the $\ce{4a1}$ in the bent configuration. This orbital belongs to the same symmetry species as $\ce{3a1}$ and thus they can interact by a second order Jahn-Teller interaction. This is proportional to $1/\Delta E$ where $\Delta E$ is the energy gap between the two orbitals mentioned. The effect of this interaction is to raise the $\ce{4a1}$ and decrease the $\ce{3a1}$ in energy. Thus in going down the series $\ce{OH2}$, $\ce{SH2}$, $\ce{SeH2}$, etc. the bond angle should decrease which is what is observed.

Example have been given for $\ce{XH2}$ molecules, but this method has also been used to understand triatomic and tetra-atomic molecules in general, such as $\ce{NO2}$, $\ce{SO2}$, $\ce{NH3}$, etc..

Adding a bit to the answers above, one factor that isn't shown in the Walsh diagram is that as the angle decreases, there is increased mixing between the central atom valence s and p orbitals, such that the 2a$$_1$$ orbital has increased p contribution and the 3a$$_1$$ has increased s. This is where one gets the result that Ron mentioned at the end of his answer that the lone pairs on water reside in a pure p (1b$$_1$$) and an sp (3a$$_1$$) orbital. That means the bonding orbitals shift from one pure s (2a$$_1$$) and one pure p (1b$$_2$$) to one sp (2a$$_1$$) and one p (1b$$_2$$) (ignoring the extreme case where 3a$$_1$$ actually gets lower in energy than 1b$$_2$$, which isn't really relevant). Mixing occurs to a greater extent in $$\ce{SH2}$$ relative to $$\ce{OH2}$$ because the 3s and 3p orbitals of S are closer in energy to each other than 2s and 2p on O.

If we hybridize the two bonding orbitals so that they are equivalent and do the same for the two nonbonding orbitals, we find that they start as bonding = 50% s/50% p (ie $$sp$$ hybrid) and nonbonding = 100% p and shift towards an endpoint of bonding and nonbonding both being 25% s/75% p (ie $$sp^3$$ hybrid).

Thus, the common introductory chemistry explanation that "bonding in $$\ce{SH2}$$ is pure p" is not supported by the MO analysis. Instead, $$\ce{SH2}$$ is closer to $$sp^3$$ than $$\ce{H2O}$$ is. The bonding orbitals in $$\ce{H2O}$$ are somewhere between $$sp^2$$ and $$sp^3$$. So it is correct to say that "the bonds in $$\ce{SH2}$$ have less s character than those in $$\ce{OH2}$$", but not to say that they are "pure p".

The fact that the $$\ce{SH2}$$ bond angle is around 90 degrees is not because its bonds are made from p orbitals only. That coincidence is a red herring. Instead, the fact that the bond angle is smaller than the canonical $$sp^3$$ is because the bonding and nonbonding orbitals are not equivalent. That means that the particular p orbitals involved in each $$sp^3$$ group do not have to have the same symmetry as in, for example, a tetrahedral molecule like CH4.

I will try to give u a most appropriate and short answer that u can understand easily See h20 has 104.5 degrees bond angle , h2s has 92degrees , h2se has 91degrees and h2te has 90degrees bond angles Draw diagrams of these u will find all of them have tetrahedral shape with 2 lone pairs , assume that no hybridization occurs and all these central atoms are using pure p orbitals for bonding then because of repulsions by lone pairs the bond angle should be 90degrees between 2 surrounding atoms , now according to dragos rules when central atom belongs to 3rd period or higher and electro negativity of surrounding atoms is 2.5 or less then central atom uses almost pure p orbitals . So the final answer is the extenend of hybridization decreases In this case which leads to decrease in bond angle . note that only in h2te no hybridization is observed.

We know that as the electronegativity of central atom increases, the bond angles also increase. The relevant electronegative order is $$\ce{O > S > Se}\,,$$ hence the bond angle order of $$\ce{H2O>H2S>H2Se}\,.$$

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer. – Martin - マーチン Feb 3 '16 at 9:07