Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: formaldehyde23 on March 03, 2014, 08:50:26 PM
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Hi everyone,
I'm confused about why H2S has a smaller angle (90 degrees) than H2O (104.5 degrees) in terms of hybridization.
I know the non-hybridization explanation: H2S has a smaller angle because the S atom is larger than the O atom. So, the electrons do not need to spread as far apart in order to reach stability.
In terms of hybridization, I know that S uses its 3p orbitals whereas O uses sp^3 orbitals. There is more s character through the hybridization of the oxygen atom. If this is correct reasoning , how would greater s-character lead to a greater bond angle?
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I've seen this explained using both valence bond and molecular orbital theories. I tend to favor MO for just about everything but in this case they both essentially boil down to the relative energy gaps between 2s-2p (oxygen) and 3s-3p (sulfur), the latter of which is larger. Admittedly the VB approach is far easier to conceptualize, and explain here: because the 3s-3p energy gap is large, hybridization of the orbitals costs too much energy, so bonding occurs primarily between the hydrogens and "pure" sulfur p-orbitals, which have 90 degree geometric separation. The MO approach is functionally similar, but you have to do a symmetry treatment on the sulfur atomic orbitals and linear combination of hydrogen atomic orbitals, which requires some group theory knowledge.