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Name: Duarte
Status: student
Age: 16
Location: N/A
Country: N/A
Date: 12/2/2004

Hi! I've just come across a little problem while studying molecular geometry. I know that in H2S the H-S-H angle is smaller than the H-O-H angle in H2O and that in H2Se it would be even smaller. What I do not understand is why these angles are smaller, given that we are talking about elements of the same group. Does electronegativity play any important role here? At least it is greater in O than in S and Se...Or is it all about atomic radius?


A very astute observation. To explain this, it would be important for you to understand hybridization theory. The bond angle of H-O-H is 105, and that of H- S-H is 92. This would suggest that S is hardly hybridized at all, and in fact using p-orbitals (with little s-characteristics) to form the S-H bonds, whereas O is sp3 hybridized.

To understand why S is not hybridized, we need to understand that the large size of S allows the electron pairs to be far from each other so that the energy incurred from repulsion is not very large and need not be minimized. Since O is a lot smaller, a more stable, less energetic configuration, would have to be made - hence the hybridization.

Greg (Roberto Gregorius)

Very perceptive question. The short answer is that each of the molecules H2O, H2Se, H2Te has two lone, non-bonding pairs of electrons. They originate from the atoms O, Se, Te which are in successive rows of the periodic table. As a result the electron pairs are attracted by the nuclei less and less strongly. As a consequence of this lower attraction the lone pairs occupy more space. This in turn causes the bond angle of the H's to become more acute. This reasoning is just one example of a semi-empirical bonding model called: Valence Shell Electron Pair Repulsion (VSEPR). This model can be understood by any student who is taking high school chemistry. It is discussed in many texts, and on many web sites. One for example is:

A web search on the term "VSEPR" or the acronym spelled out will reveal a lot of "hits" and you can choose the one that fits the depth you wish to investigate the topic. VSEPR is really quite good considering the simplicity of the rules governing it.

Vince Calder

Hi Duarte!

This situation usually is explained by the bonding theories.In fact there are a decrease in the bonding angles that go like that:


The VSEPR (valence-shell-electron-pair-repulsion theory), proposes that the stereochemistry of an atom in a molecule is determined primarily by the repulsive interactions among all the electron pairs in its valence shell.

In that case molecules adopt geometries in which their valence electron pairs position themselves as far from each other as possible and lone pair influence on geometry.

Also the size of the central atom has some effect on bond angles, like you noticed. And again electronegativity is also another factor affecting bond angles.

Consider that this is somehow a not precise explanation maybe appropriate for your age and grade. Nowadays there are very advanced experimental methods that measure bond lengths and angles. There are also up-to-date theories that try to explain in a satisfactory way this measures and predict others with great accuracy.

Thanks for asking NEWTON!

(Dr. Mabel Rodrigues)

The "tetrahedral" bonds for O, S, and Se atoms results from the sp3 hybridization of the electronic orbitals. This results in four equivalent orbitals, equally spaced about the sphere of the central atom. However, when you make two bonds with these orbitals there are two bonded sp3 orbitals and two unbonded electron pairs. The electron pairs take up less space than a hydrogen atom and to accommodate them the bonds move slightly apart, increasing the H-X-H angle to something greater than the ideal tetrahedral angle.

The larger the central atom is the less the hydrogens have to spread out and the smaller the resulting angle.

Greg Bradburn

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