Bond Angles in Hydrogen Sulphide - A Level or higher Watch

Dynasty
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At my Cambridge interview I was asked to explain the bond angles in water, which is pretty straightforward. After that, I was asked to tell the interviewer whether the angles between the two hydrogen's was bigger in Hydrogen Sulphide than the respective angle in water.

Now, Sulphur and Oxygen have the same number of outer electrons, so the same number of lone pairs. I said that is may be bigger because of the larger atomic radius, so the bonding electrons would be further away from the two lone pairs, and thus feel less repulsion, so the angle would be larger. The interviewer declined to tell me which was correct, and that I should research it that night. A quick look on wikipedia gave the bond angle to be 92 degrees (water is 104.5 degrees).


This confused me, as I can't think why this would be - surely as there are effectively 4 pairs of electrons, with less repulsion, it should be nearer tetrahedral than water, but this doesn't seem to be the case.

Can anyone enlighten me? Cheers
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Peridot_1234
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hydrogen Suphide - H2S

H ----- S ------ H

Linear

therefore 180 degrees

(oops!! typo!)
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Dynasty
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This begs to differ
http://upload.wikimedia.org/wikipedi...dimensions.svg
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EierVonSatan
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(Original post by Peridot_1234)
hydrogen Suphide - H2S

H ----- S ------ S

Linear

therefore 180 degrees
No, its 92 degrees - this means the p orbitals in S are still pretty much normal to one another (px and py lie on the x and y axis orthogonal). This means the hybridisation that takes place in H2O is not occurring- the reason for this is down simply to the size of sulfer, the bonding pairs to the hydrogen's are far apart enough in space that there's not much repulsion and therefore no need for the system to spend energy and hybridise
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Dynasty
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Ok, cheers
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Akkuz
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(Original post by Peridot_1234)
hydrogen Suphide - H2S

H ----- S ------ S

Linear

therefore 180 degrees
Thats incorrect.

It's 2 hydrogens, 1 sulphur atom.

And you need to take into account the lone pairs on the Sulphur atom.
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DeanK2
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(Original post by Dynasty)
At my Cambridge interview I was asked to explain the bond angles in water, which is pretty straightforward. After that, I was asked to tell the interviewer whether the angles between the two hydrogen's was bigger in Hydrogen Sulphide than the respective angle in water.

Now, Sulphur and Oxygen have the same number of outer electrons, so the same number of lone pairs. I said that is may be bigger because of the larger atomic radius, so the bonding electrons would be further away from the two lone pairs, and thus feel less repulsion, so the angle would be larger. The interviewer declined to tell me which was correct, and that I should research it that night. A quick look on wikipedia gave the bond angle to be 92 degrees (water is 104.5 degrees).


This confused me, as I can't think why this would be - surely as there are effectively 4 pairs of electrons, with less repulsion, it should be nearer tetrahedral than water, but this doesn't seem to be the case.

Can anyone enlighten me? Cheers
Good question.
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DeanK2
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Pretty tough if you did not know hybridisation.
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DeanK2
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(Original post by EierVonSatan)
No, its 92 degrees - this means the p orbitals in S are still pretty much normal to one another (px and py lie on the x and y axis orthogonal). This means the hybridisation that takes place in H2O is not occurring- the reason for this is down simply to the size of sulfer, the bonding pairs to the hydrogen's are far apart enough in space that there's not much repulsion and therefore no need for the system to spend energy and hybridise
:confused:

Surely sulfer must hybridise? How else could the bonds be identical, and at such an angle?
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EierVonSatan
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(Original post by DeanK2)
:confused:

Surely sulfer must hybridise? How else could the bonds be identical, and at such an angle?
Why must it? The bond angles are nearly 90 degrees to one another as the p orbitals are.
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DeanK2
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(Original post by EierVonSatan)
Why must it? The bond angles are nearly 90 degrees to one another as the p orbitals are.
I don't know how you could form the sigma bond without hybridisation?

EDIT: (I also needed to involve d orbitals to obtain the 90 degrees. I really cannot see what you mean - do you have a example?).
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EierVonSatan
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(Original post by DeanK2)
I don't know how you could form the sigma bond without hybridisation?
Okay, its hybridised (sp) with the hydrogen...using only sulfer's p orbitals and hydrogen's s orbital
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DeanK2
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(Original post by EierVonSatan)
Okay, its hybridised (sp) with the hydrogen...using only sulfer's p orbitals and hydrogen's s orbital
:p: I was doing a different compound!! :confused: .

I had Oxygen instead of hydrogen in some places? Arghh!

Sorry.

I really don't get why I wasn't using hydrogen.

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nigga123456
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ur ed
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nigga123456
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(Original post by Peridot_1234)
hydrogen Suphide - H2S

H ----- S ------ H

Linear

therefore 180 degrees

(oops!! typo!)

ur just simply ed
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