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In the MO theory, is it possible to have a molecule with bond order less than 0? The least bond order I have come across as of now is 0 (for di-atomic noble gases), indicating that the molecule cannot exist and the atoms get pulled apart.

Is it possible to create a scenario (at least theoretically) where we deal with a negative bond order by manipulating the system's conditions or any of that sort?

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No, it's not possible, theoretically either.

If you see the Molecular orbital theory, maximum number of anti bonding electrons = number of bonding electrons.

And bond order is (Number of bonding electrons - number of anti bonding electron)/2

So, the minimum value that you can get is 0 and not negative, as anti bonding electrons can't exceed the number of bonding electrons.

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  • $\begingroup$ There is more to it. $\endgroup$ – Ivan Neretin Apr 3 '18 at 6:34
  • $\begingroup$ What more to it? $\endgroup$ – Aryan Stark Apr 3 '18 at 6:35
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    $\begingroup$ A molecular orbital is not necessarily associated with particular bond. It may spread over many atoms. In this case the number of bonding electrons of one bond ceases to be well-defined. $\endgroup$ – Ivan Neretin Apr 3 '18 at 7:08
  • $\begingroup$ @IvanNeretin could you elaborate? $\endgroup$ – J_B892 Apr 4 '18 at 5:11
  • $\begingroup$ You'll need a different definition of bond order, like that of Huckel method or something, that's what I mean. With that definition, I see no fundamental reason to believe that negative bond orders are forbidden. $\endgroup$ – Ivan Neretin Apr 4 '18 at 5:48
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As a ground state, I don't think so. For the example of diatomic molecules, every combination of atomic orbitals forms equal numbers of bonding and anti-bonding orbitals, and the bonding orbitals are by definition lower in energy. To get to a zero bond order configuration, you have to fill all the anti-bonding and bonding orbitals, so the next orbital up would have to be dominated by higher energy atomic orbitals, and so would be primarily a bonding orbital, and then the bond order is positive again. Adding more atoms might change things, but I'd be surprised. Of course, excited states could have a negative bond order.

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I don't think that bond order can be zero. Think of bond order as the number of bonds an atom is forming in the molecule. So, least, it can form no bond . There would be no physical significance of an atom forming negative number of bonds. That would be absurd.

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    $\begingroup$ Di atomic noble gases have a bond order of 0 $\endgroup$ – J_B892 Apr 3 '18 at 5:46
  • $\begingroup$ Yes I completely agree. But zero here signifies that they do not form bond with each other. If they did it would be unstable. This shows that diatomic noble gases donot exist that is why their bond order is zero. $\endgroup$ – Ashutosh Anand Apr 3 '18 at 5:51
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    $\begingroup$ MO theory is full of seemingly absurd things. $\endgroup$ – Ivan Neretin Apr 3 '18 at 5:51

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