H2S and H2O Bonding Angles
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
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.
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
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
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
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
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
The larger the central atom is the less the hydrogens have to spread out and
the smaller the resulting angle.
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Update: June 2012