0
$\begingroup$

Organic chemistry by jonathan clayden states during his explanation of hybridisation using trigonal boron structure (sp2) and tetrahedral boron structure (sp3):

You want to populate the lowest energy orbitals for greatest stability and sp2 orbitals with their greater s character are lower in energy than sp3 orbitals. Another way to put this is that, if you have to have an empty orbital, it is better to have one with the highest possible energy since it has no electrons in it and so it doesn’t affect the stability of the molecule.

enter image description here

I get the first part of the statement; however, the bolded part confuses me. If a higher energy orbital is empty compared to a lower energy empty orbital, how does it affect the stability of the molecule differently? If they are all empty shouldn't they all not affect the stability of the molecule?

$\endgroup$
2
  • 1
    $\begingroup$ It's just to add the first thing a different perspective. As you can see in the picture if the orbital was sp3 then others would be too - effect is indirect. $\endgroup$
    – Mithoron
    Mar 30 at 1:10
  • $\begingroup$ The stronger a bond the lower the total energy. A sp2 orbital has more s character than a sp3 orbital so the molecule with 3 sp2 orbitals is of lower energy because the electrons in the bond are closer to the nuclei. The empty orbitals have no effect until occupied by excitation or addition of electrons. $\endgroup$
    – jimchmst
    Mar 30 at 23:01

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.