What is the bond order of #"H"_2^+#?

1 Answer
Truong-Son N.
Aug 7, 2017

#1//2#. What does this mean? Is the bond stronger or weaker than in #"H"_2#? Is #"H"_2^(+)# more stable or less stable than #"H"_2#?

Well, we begin from the #"H"_2# molecule, with the simplest molecular orbital (MO) diagram you would ever find (I am not exaggerating):

http://www.meta-synthesis.com/

The #1s# atomic orbitals had overlapped (and being spherical, they can only overlap head-on) to form:

  • one #sigma_(1s)# bonding MO.
  • one #sigma_(1s)^"*"# antibonding MO.

Changes to bond order (corresponding to bond strength) can be summarized as follows:

  • Putting electrons into a bonding MO increases the bond order by #1/2# per electron (strengthening the bond).
  • Putting electrons into an antibonding MO decreases the bond order by #1/2# per electron (weakening the bond).

And vice versa for taking out electrons.

Therefore, the #"H"_2^(+)# cation, with one less electron (examine the charge!), would lose one from the #bb(sigma_(1s))# bonding MO.

Thus, the bond order is #bbul(1//2)#. What does this mean? Is the bond stronger or weaker than in #"H"_2#?

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