3
$\begingroup$

The cyanide ion, when acting as a nucleophile, typically attacks via carbon as its HOMO has a greater contribution from carbon. But how can we deduce this from the MO diagram?

To me it appears that the nitrogen $\mathrm{2p}$ orbitals are energetically closest to the $3\sigma$ HOMO, so its coefficient should be larger in the HOMO. What is wrong with this line of reasoning?

MO diagram.

Improve this question
$\endgroup$
7
  • $\begingroup$ related chemistry.stackexchange.com/q/18449/16683 $\endgroup$ – orthocresol♦ Oct 17 '16 at 16:58
  • $\begingroup$ can u please @martin-γƒžγƒΌγƒγƒ³ answer the question ..as in the linked question it is mentioned that We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. i dont know how to tag , can someone pls tag him. $\endgroup$ – aks0854 Oct 17 '16 at 18:01
  • $\begingroup$ @akso854 There are 4 MOs with the same $\mathrm{\Sigma}$ symmetry. So the exact energies are going to matter here. I suspect that you're getting a much larger contribution from the carbon $\mathrm{2s}$ orbital, and that seems consistent with the shape of the boundary surface in the diagrams. $\endgroup$ – Zhe Oct 17 '16 at 18:30
  • $\begingroup$ thanks @Zhe ..i also think the same. 2s orbital is behind greater lobe on carbon in HOMO. $\endgroup$ – aks0854 Oct 17 '16 at 18:57
  • 1
    $\begingroup$ Whilst I don't think this post in unclear in it's meaning (and so should not be closed), it could definitely do with some tidying up to use better grammar and formatting. You are much more likely to get an answer if you present your posts in a nice way. $\endgroup$ – bon Oct 18 '16 at 10:40
1
$\begingroup$

The 2s and 2pz atomic orbitals are of the same symmetry in the molecule. That means that the 2 sigma* and 3 sigma are actually hybrids. The former is a hybrid were most of the electron density points to the bond (making it partially bonding), while in the latter most of the electron density points outside, i.e. a lone pair in the carbon.

Improve this answer
$\endgroup$

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.