Subject: isolobal vs. isoelectronic

From: Krzysztof Radacki <[email-masked]>
Date: Wed, 17 Oct 2001 18:18:49 +0100

Dear CCL'ers,

just at this weekend I had nice discussion about isolobal particles. 
The starting point were small isoelectronic boron-clusters with
different apical fragments.
CH_2 is isoelectronic to BH_3. I've never thought about it before, but
I've always presumed that it's mean also that they are isolobal.
The argument from my opponent was the definition of isolobality: "the
fragments are isolobal if they have equal number of orbitals with
approximately the same symmetry, shape and energy".
So he had argued that BH_3 has only one free orbital (not involved in
B-H bond) and CH_2 two of them. So BH_3 is isolobal to CH_3(+) but only
isoelectronic to CH_2.
I've been really confused (as I've said never thought about it before),
and asked him how to explain that already R. Hoffmann had shown in his
Nobel-lecture that orbitals of d^6-ML_5 and CH_3(+) are isolobal and
just on example of chemistry CH_2 <--o--> Cr(CO)_5 what leads to
isolobal analogy between CH_2 <--o--> CH_3(+). (there was also
comparison of CH_3(+), CH_2 and CH(-) as donor but I haven't used this
argument)
He hasn't explained but we've stopped our discussion in this point.
Everything good but I can't get rid of it.

Coming back to BH_3 and CH_2.
Does really BH_3 give only one orbital to cluster. Or does CH_2 gives
only two.
My point of view is:
CH_2 group has two hydrogen atoms - one in exo- and one in
endo-position. The exo-one, as every exo-ligand, does not participate in
cluster-Aufbau. Endo are parts of cluster so CH_2 gives 2 empty orbitals
and one occupied with H-atom; BH_3 gives one empty, 2nd occupied with
endo-H and 3rd occupied with mi-H. It means that both fragments are
3-orbitals donors.

I know - this is first-year question - but better late than never   :)

Any comments are wellcomed.
Krzys