Question

Which of the following hydrides has the largest bond angle?

H2S

H2Te

H2O

H2Se

​

Answer 1

Answer:

H2Se

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Answer 2

Here are the H−X−H bond angles and the H−X bond lengths:

molecule bond angle/∘ bond length/pm H

2

O 104.5 96 H

2

S 92.3 134 H

2

Se 91.0 146

The traditional textbook explanation would argue that the orbitals in the water molecule is close to being 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 H−X−H angle to contract slightly. So instead of the H−O−H angle being the perfect tetrahedral angle (109.5∘) it is slightly reduced to 104.5∘. On the other hand, both H

2

S and H

2

Se have no orbital hybridization. That is, The S−H and Se−H bonds use pure p-orbitals from sulfur and selenium respectively. Two p-orbitals are used, one for each of the two X−H bonds; this leaves another p-orbital and an s-orbital to hold the two lone pairs of electrons. If the S−H and Se−H bonds used pure p-orbitals we would expect an H−X−H interorbital angle of 90∘. 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 X−H bonds the angle opens up a bit wider. This explanation would be consistent with the H−S−H angle being slightly larger than the corresponding H−Se−H angle. Since the H−Se bond is longer then the H−S bond, the interorbital electron repulsions will be less in the H

2

Se case alleviating the need for the bond angle to open up as much as it did in the H

2

S case.