There is a "rule of thumb" where the max oxidation state of a
3rd period transition element could be estimated by the number of
4s electrons + unpaired 3d electrons
eg.
Max OS of Mn [Ar] 3d5 4s2 =
2 + 5 = +7
Max OS of Fe [Ar] 3d6 4s2 = 2 + 6 = +6
Could someone shed more light on how the paired/unpaired 3d
electrons is a factor in this context?
Heh, quite a hornet's nest being stirred up here.
This is rather beyond the A level syllabus, so do let your
students know that there's no need to worry over it. In fact, for
the Singapore H2 syllabus, H2 Chem students are not even required
to show knowledge of the "rule of thumb" that you
mentioned, which is required only for some of the other A
level syllabuses around the world (even these other A level
syllabuses only required the students to be aware of the rule, but
not require students to explain it).
There is more than one way to look at this, or to understand and
explain this, but let's keep it simple here (ie. other
explanations require more in-depth Uni level concepts, not to
mention the existence of multiple anomalies, eg. Cu, that makes
inorganic chemistry so thoroughly entertaining, if you will).
I'll give two simplified reasons which may help the
student to understand whyfore this "rule of thumb",
but these reasons are by no means exhaustive, and the keen student
will no doubt have many further questions, but it would be
more appropriate to deal with these at the University levels
(rather than A levels).
1st reason : in some ways, pairing of electrons have a
'stabilizing' effect (although in some other ways, the
inter-electron repulsion of paired electrons have an
opposite 'destabilizing' effect; I already warned you that these questions would
beget further questions, quite beyond the ken of the A
levels), in the sense that (similar to the concept of free
radicals being extra unstable and reactive) the unpaired d
electrons are more available and willing to be donated
away for ionization (in the case of lower OSes)
or for occupation of covalent bonding
orbitals with other elements (usually electronegative O
atoms), in the case of higher OSes.
2nd reason : coincidentally (so to speak), it so happens that
simultaneously, the increasing effective nuclear charge,
electronegativity and ionization enthalpies (as we go from
left to right of the d block elements) reaches the 'tipping
point' after Manganese, and as such increasingly prohibits these
elements from losing electrons and exhibiting higher OSes.
As I mentioned previously, these reasons described above
are certainly a simplification (of the wonderous
nuances and exponentially increasing variances of
inorganic chemistry, particularly transition metal chemistry), but
will more than suffice for the A level student (and certainly so
for the Singapore H2 Chemistry student).
