Half sigma bonds

Question

$\ce{N2+}$ has $2$ $\pi$-bonds and a half sigma bond.

First of all, how can a "half" sigma bond exist? Usually when we encounter non integral bond orders, it is mostly because of resonance which creates partial $\ce{\pi-\pi}$ bonds and doesnt involve orbitals participating in sigma bond.

To understand this, I tried to draw the lewis structure of $\ce{N2+}$. While a triple bond and two lone pairs exist in $\ce{N2}$, I couldnt figure out how $\ce{N2+}$ looks like.

Then I wrote down the molecular configuration using MOT. I noticed that the last electron in $\ce{N2+}$ goes into $\ce{\sigma_{2p}}$ orbital.

I suspect that this might have got something to do with partial sigma bonds, but I'm not really able to see through it clearly. Can someone help me out?

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In case someone is wondering about the source, it is the 54th question asked in Jee mains 2020 10th January, Shift -I

Answer 1

First of all, how can a "half" sigma bond exist?

Usually, you expect double bonds to be shorter and stronger than the corresponding single bonds, and triple bonds even shorter and stronger. The OP already mentioned bond-orders of 1.5 that occur for conjugate double bond systems, and those have properties in between single and double bonds.

Perhaps the most famous case for a "half" sigma bond is the hydrogen molecule ion, $\ce{H2+}$. It has a bond distance of 100 pm (compare 74 pm for dihydrogen) and a bond energy of 2.77 eV (compare 4.52 eV for dihydrogen). Source: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

So in terms of comparing the bonding in the hydrogen molecule ion with the hydrogen molecule, it makes sense to talk of a partial bond in terms of bond length and strength. It might make sense to talk of "half" a sigma bonds because the single bond has 2 electrons, but in this case there is only one.

The reason this case is famous, at least in courses introducing quantum chemistry, is that it is a single-electron system, so easier to work with theoretically and computationally than the dinitrogen cation.

Then I wrote down the molecular configuration using MOT. I noticed that the last electron in N2+ goes into σ2p orbital.

Here is the molecular orbital diagram, and a depiction of the highest occupied molecular orbital.

Source: https://www.chemtube3d.com/orbitalsnitrogen/

As you can see, this orbital has two nodes. In comparison, the orbitals corresponding to the pi bonds have only one node. Using arguments similar to the ordering of orbitals in Hueckel theory, you would expect the sigma orbital to be higher in energy.

However, comparing the energies of states in all diatomics, you see that it is more complicated (the difference between the energy of 2s and 2p changes across the 2nd period elements, and with it the amount of mixing between 2s and 2p of same symmetry):

Source: https://chem.libretexts.org/Courses/Heartland_Community_College/HCC%3A_Chem_161/9%3A_Molecular_Geometry_and_Bond_Theory/9.8%3A_M.O._Theory_and_the_Period_2_Diatomic_Molecules

[OP in comments] However why would the loss of electron be from a stable bond pair? Why cant it be one of the lone electrons? Also why does it have to be a sigma bond and not a pi bond, if at all the loss is not from one of the lone pairs?

You are using the valence bond model (or the Lewis structure) to "name" electrons. There is no 1:1 mapping of "lone pairs" to the molecular orbital diagram. It does say that the two combination of 2s orbitals "cancel each other out", i.e. are non-bonding; however, the shape of these orbitals does not match the valence bond model, where sp-orbitals pointing away from the center of the molecule are counted as lone pairs.

Answer 2

"Half" a bond is new to me, but it makes sense if a (full) bond consists of two electrons: one from each atom bonded together. Then half of that would be one electron in the bond, for 50% of a full bond.

It would get interesting in the boranes where you have three-center bonds, or a pi system where you have an electron deficiency. And then, if you have an anion where an electron is added to antibonding orbitals, especially where you have an extended pi system... But then you can use some other numerical system to indicate which bonds are weakened or strengthened.

Those odd-electron molecules do require some extra imagination.