Using MO theory, give an explanation for the C-C bond length in cyanogen
See my answer to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.

In this case, your molecule is analogous to the all-carbon system, butadiene.  In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals.  Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.

Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals.  We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons.  Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons.  For the whole molecule we have 8+4+6=18 electrons as expected.

As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.

There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals.  Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).



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Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)



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We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top).  Don't worry about the "S"'s and "A"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.  

Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons.  This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate.  Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.