# Bond orders in the 3c-4e bond and 3c-2e bond

According to D. P. M. Mingos, in his book Essential Trends in Inorganic Chemistry, the bond order of the three-centre four-electron bond is approximately 0.717, which is one divided by the square root of 2. However, in the case of the three-centre two-electron bond, the bond order is 0.5. This would imply that the former is stronger than the latter. However, both of them have similar-looking molecular orbital diagrams. In the case of the former, the nonbonding MO is filled while in the case of the latter, only the bonding MO is filled. Thus, from the MO perspective, why would these two bonds have different bond orders?

• Not sure about this one, but I suspect it depends on how one defines the bond order. See eg IUPAC Gold Book for a complicated definition. Does Mingos not give any kind of justification? – orthocresol Nov 24 '17 at 0:44
• Mingos has indeed justified the bond order for the 3c-4e bond mathematically (by using the definition of bond order to be the product of the coefficients of the orbitals contributing to the bond times the number of electrons occupying the orbital). However, he did not justify the bond order of o.5 for the 3c-2e bond. – Tan Yong Boon Nov 24 '17 at 0:51
• Oh, ok. I have a hunch I might be able to answer this one, but again, not right now... Good questions, though. – orthocresol Nov 24 '17 at 0:53
• I would guess that at least in the simplest cases of three-center bonding, the bond order is the same with two or four electrons should be the same; the second electron pair is in an essentially nonbonding orbital. – Oscar Lanzi Nov 24 '17 at 1:40
• Revisiting this I don't see how the maths works out. I have seen this definition before, but if I assume the usual form of the MOs for three-centre bonding, the population of the middle MO should not affect the bond order between adjacent atoms because the coefficient of the central atom is zero. Probably need more details from the book. – orthocresol Nov 24 '17 at 22:37