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# Bond angle of HCN

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1. ok so i went with the approach of counting bonded pairs and lone pairs, but hat failed... 4bp's = 120deg, 1lp= 120 - 2 = 118 deg. Answers sa 180deg, is it because its H --C-- N: ... and the lp on Nitrogen doesn't come into play?, and even though C and N share 3 pairs of bonding pairs is it because of the fact that there are only 3 atoms, they make a linear shape, hence 180deg? And what is the actual justification behind this?

Thanks
2. Carbon has a valency of 4
Hydrogen - 1
Nitrogen - 3

Carbon is in the middle.

1 of Carbons valence bonds will be used to make a covalent bond with Hydrogen

3 of carbon's valence electrons will be used to make 3 covalent bonds with nitrogen (since nitrogen has a valency of 3) therefore a triple bond exists.

The carbon has no lone pairs of electrons, so you would assume it would be 120deg

But remember that the electron density cloud is much larger on the nitrogen side becasue of the triple bond as opposed to the single bond on the other side, this makes the bonding pairs repel eachother as much as possible making the molecule linear = 180deg

i think i may have made an error somewhere, so feel free to correct me anyone
3. hmm kinda makes sense... cheers
4. correct answer but wrongly explained.

the CN bond is triple
the HC bond is single

therefore there are only two regions of electron density. These arrange at 180º to one another.

If you want to invoke hybridisation then it is sp hybridisation making two sp orbitals at 180º to one another forming two sigma bonds with H and N respectively. The remaining two 'p' orbitals laterally overlap with two orbitals from the nitrogen makng pi bonds.
5. (Original post by charco)

the CN bond is triple
the HC bond is single

therefore there are only two regions of electron density. These arrange at 180º to one another.

If you want to invoke hybridisation then it is sp hybridisation making two sp orbitals at 180º to one another forming two sigma bonds with H and N respectively. The remaining two 'p' orbitals laterally overlap with two orbitals from the nitrogen makng pi bonds.
ok charco, explain to me, the angle of CCO in propanone ... with justifications not beyond the realms of A2 if you wouldn't mind
6. sp2 hybridisation on the central carbon therefore 120º
7. (Original post by charco)
sp2 hybridisation on the central carbon therefore 120º
Darn, this is for AS retakes, and we havent studied hybridisation in A2 nor AS, so is there any other way of explaining?~

thanks
8. (Original post by Mos Def)
Darn, this is for AS retakes, and we havent studied hybridisation in A2 nor AS, so is there any other way of explaining?~

thanks
Not really.

Why they've taken hybridisation out of the syllabus I'll never know.
9. :s
10. OK simplified....

There is a single bond (sigma) between the carbon-carbon atoms
there is a double bond (sigma + pi) between the carbon and the oxygen.
Therefore at the central carbon there are three regions of electron density.
These orientate at 120º to one another.

It is the three sigma bonds that determine the geometry. Three sigma = 120º planar.

The pi system develops by lateral overlap between the remaining p orbital of the carbon and a suitable orbital on the oxygen.
11. (Original post by charco)
There is a single bond (sigma) between the carbon-carbon atoms
there is a double bond (sigma + pi) between the carbon and the oxygen.
Therefore at the central carbon there are three regions of electron density.
These orientate at 120º to one another.

It is the three sigma bonds that determine the geometry. Three sigma = 120º planar.

The pi system develops by lateral overlap between the remaining p orbital of the carbon and a suitable orbital on the oxygen.
Which is a hybridisation-lite LCAO model, interesting.

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