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So, I am having difficulty with the following explanation/question.

(Based on your MO diagram and pi bonding theory) explain why $\ce{CN-}$ is a soft base and prefers to bind to soft acids.

Now, I know that the HOMO is the 5 σ MO and that the electron pair in the 5 σ MO is, to a large extent, more concentrated around the carbon. So, is $\ce{CN-}$ a soft base because it has a low energy HOMO but large magnitude HOMO coefficient? Does polarizability have anything to do with it?

When $\ce{CN-}$ reacts with a soft acid what physical changes in the cyanide molecule would such a reaction lead to?

Not really sure about this one. Would it have to do with the donation of the electron pair in the HOMO and therefore change the HOMO?

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  • $\begingroup$ But since N is more electronegative than C (i.e. it has lower energy AOs than C), wouldn't more electron density be concentrated on N as N contributes more to the bonding MOs of the cyanide ion? $\endgroup$ – Tan Yong Boon Nov 8 '17 at 2:39
  • $\begingroup$ If you look at the electron distribution it surrounds the carbon atom instead of the nitrogen as one would expect. This is caused by the interaction of the atomic orbitals of carbon among each other. They contribute more significantly to this 5σ molecular orbital compared to the influence of the 2p atomic orbital. You should not forget that the energy difference between 2s and 2p in Nitrogen is greater than that in Carbon, so the overlap between them is less significant, and thus the total contribution to the 5σ molecular orbital. $\endgroup$ – MatthewSpire Nov 8 '17 at 3:05
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In one of his great answers, Martin gave us the calculated molecular orbitals for carbon monoxide I am copying below. Note that while cyanide has a nitrogen atom in place of the oxygen and features a single negative charge, we can still qualitatively consider the MO schemes identical; thus I will base my discussion on what you see in the image.

Molecular orbital scheme of carbon monoxide by Martin

In this scheme, the HOMO is referred to as MO number 7; since 5 and 6 are π-type MOs, this corresponds to the 5σ you mentioned.

One thing we immediately note is that the carbon-centred lobe of the HOMO is very diffuse and extends far into space; much farther than any other orbital of the molecule. It is also the orbital that an acid (either proton or Lewis acid) would bond to. In Pearson’s HSAB theory, large and diffuse orbitals — often also called polarisable — are associated with soft acids or bases, as the charge per volume value is considered comparatively low. Therefore, just by looking at the size of this orbital, we can identify it as a ‘soft’ one.

It should also be explained why soft bases tend to create more stable complexes with soft acids. This is a direct consequence of the more diffuse nature of the orbital: the overlap between two orbitals (which will later form a bond) is typically best if the orbitals have similar shapes and sizes. Thus, because the HOMO has a rather diffuse lobe on carbon’s side it will prefer binding to other compounds with rather diffuse orbitals in a Lewis acid-base reaction.

That should be all that is required to explain cyanide’s behaviour under Pearson’s theory.

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  • $\begingroup$ I wonder if the molecular orbital appears "diffuse" because you set the isocontour value to a particular value. $\endgroup$ – Eric Brown Nov 8 '17 at 9:18
  • $\begingroup$ @EricBrown No matter what the isocontour value is set to, there will always be a good deal more density in a larger lobe on the carbon side. $\endgroup$ – Jan Nov 8 '17 at 10:47

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