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.

  • $\begingroup$ wikipedia://Polyhedral skeletal electron pair theory ; also hybridization model usually performs poorly outside CHNOF compounds. $\endgroup$
    – permeakra
    Oct 25 '15 at 17:44
  • $\begingroup$ @permeakra the wiki didn't help me too much, other than increasing my knowledge, it didn't had anything related to my question i.e. didn't mentioned anything about the charge on closo-boranes $\endgroup$ Oct 25 '15 at 18:07
  • $\begingroup$ @ChinmayChandak read it again, in particular pay close attention to $B_6H_6^{2-}$ orbital diagram $\endgroup$
    – permeakra
    Oct 25 '15 at 18:17
  • $\begingroup$ @permeakra I read it once again, and as I said, I didn't find anything related to the charge on closo-borane, how is there a 2- charge on closo-borane ? $\endgroup$ Oct 26 '15 at 10:56
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    $\begingroup$ @ChinmayChandak Charge equals number of electron in the particle minus number of electrons in the atoms composing particles. Count electrons on orbital diagram and read corresponding text. In short: when the boran is composed, it is built from n*2p orbitals of BH fragments each coming with one electrons and forming n bonding orbitals (so there is no charge here) and n sp orbitals of BH fragments, forming one large bonding orbitals. Howevere, since BH fragments do NOT give electrons to this orbital, 2 electrons must come from elsewhere, so 2- charge. Seriously, pay attention to electron counts. $\endgroup$
    – permeakra
    Oct 26 '15 at 11:28

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:

BH fragment orbitals

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.):

Bonding MOs in B6H6

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.

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    $\begingroup$ Interesting, but does this apply to all such forms? What happens if we try a tetrahedron? If it's isoelectronic to white phosphorus, then the borane anion would have four negative charges. $\endgroup$ Dec 29 '20 at 13:35
  • $\begingroup$ @OscarLanzi truthfully, it's been a long time since I've studied this, and I've forgotten all of it by now, so I have no idea what would happen. My first instinct is to agree - yes it would need 4 negative charges. One would have four orbitals in a tetrahedral arrangement which (just like in methane) would form one $a_1$ and three $t$ bonding orbitals; the BH units provide 4 electrons so need 4 more to fill them up. Searching for $\ce{B4H4}$ brings up some interesting discussions, eg. sciencedirect.com/science/article/pii/S2210271X1730316X though I merely skimmed. $\endgroup$
    – orthocresol
    Dec 31 '20 at 2:13
  • $\begingroup$ I'm not entirely sure why $\ce{B4H4}$ is unstable, whereas $\ce{B4Cl4}$ isn't. I would perhaps hypothesise that the tetrahedral cage is just too small — having the more electronegative chlorine on the outside helps to reduce the electron density inside the cage. But that's just wild speculation. $\endgroup$
    – orthocresol
    Dec 31 '20 at 2:20

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