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In a question I was solving, it was mentioned that:
[...] closo-boranes are compounds having the general formula $\ce{B_nH_n^2-}$ [...] an example is $\ce{B6H6^2-}$.
Why do these boranes possess a (2−) charge? I am aware that in electron-deficient compounds such as these, there are different types of bonds between boron atoms (such as three-centre two-electron bonds), but I don't know why this should lead to a negative charge.
The application of MO theory to (electron-deficient) cluster compounds such as boranes is generally summarised in a set of rules, known as Wade's rules. We can look at the closo structure $\ce{[B6H6]^2-}$ in more detail to see where the two negative charges come from.
In closo boranes, each BH fragment points outwards from a vertex of a polyhedron. For $\ce{[B6H6]^2-}$, it turns out that this polyhedron is the familiar octahedron, which has six vertices.
What we must do now is to count how many electrons and how many orbitals each BH fragment contributes to the entire cluster. This can be done fairly easily:
Here we have neglected the boron 2s orbitals. One 2p orbital on boron has the correct symmetry to overlap with the hydrogen 1s orbital, forming bonding (blue) and antibonding (red) orbitals; the other two 2p orbitals remain non-bonding (gold).
So, each BH fragment contributes two valence electrons to the cluster. Since there are six BH fragments, the total valence electron count is (so far) 12. Saying this implies that the boron 2s electrons aren't valence electrons, which is, in a sense, true; they do not participate much in the delocalised bonding in the entire cluster.
If we use the three bonding and nonbonding orbitals to form properly symmetrised MOs across the whole cluster, it turns out we get seven bonding MOs across the entire cluster. I'm not keen to ChemDraw these MOs, so I've taken a picture from Housecroft's Inorganic Chemistry (4th ed.):
The lowest energy MO ($\mathrm{a_{1g}}$) comes from an in-phase combination of all six bonding fragment MOs. The three next lowest ($\mathrm{t_{1u}}$) come from overlap of two bonding fragment MOs with four nonbonding 2p orbitals on the other boron atoms. Three more ($\mathrm{t_{2g}}$) come from "sideways", or "tangential", overlap between four boron 2p orbitals.
In general, it turns out that for a polyhedron with $n$ vertices (at least, where $n \geq 6$ — see comments), there will be $n+1$ bonding MOs which can be made using the fragment MOs at each vertex.
Now, in order to get a stable cluster, we need to be able to fill all these $n+1$ bonding MOs with electrons. In the case of $n = 6$, we have 7 MOs to fill, and so we need 14 electrons. However, as established previously, we only have 12 valence electrons!
The solution is therefore to tack on two extra electrons, and that's why the closo borane requires a (2−) negative charge. If it didn't have this negative charge, it wouldn't be able to fill all of the bonding MOs in an octahedron, and it's quite likely that it would choose to adopt a different geometry that is no longer closo.
