6
$\begingroup$

Why do square planar coordination compounds of type $\ce{[Mabcd]}$ not show optical activity, although they contain 4 different ligands (i.e. chiral central metal atom)?

$\endgroup$
0

1 Answer 1

10
$\begingroup$

Even though the central atom has 4 different ligands this does not necessarily mean that the compound is optically active. The condition for that rather the following (from Wikipedia):

A molecule is achiral (not chiral) when an improper rotation, that is a combination of a rotation and a reflection in a plane, perpendicular to the axis of rotation, results in the same molecule[...] For tetrahedral molecules, the molecule is chiral if all four substituents are different.

So for 4-coordinated carbon atoms which are tetrahedral having four "ligands" actually establishes optical activity. But in the case of a square planar metal complex you always have a mirror plane, namely the one made up by the central atom and the ligands:

enter image description here

And as a result you don't get optical activity. But it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited.

$\endgroup$
1
  • $\begingroup$ "it might be possible that some vibrational modes destroy the mirror plane symmetry element of the metal complex and this might lead to some optical activity depending on how well this mode excited." Hello, thanks for your answer! Could you please give an example of this "vibrational optical activity"? Also, is it possible to predict it theoretically? $\endgroup$ Commented Apr 23, 2018 at 1:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.