$$ \text{Co}[{(\text{N}{\text{H}}_{3})}_{4}{\text{Cl}}_{2}]$$

optically active? If so, write its stereo-isomers.

I know that for a compound to be optically active, it should rotate the plane of Plane Polarized Light, but is there a way to do this without experiment?

  • $\begingroup$ Please do not use formatting in the titles; because of url slug issue. (I know the Q seems a bit scary now, but it's better than the version with all those "mark-ups" in the face. :D) $\endgroup$
    – M.A.R.
    Commented Jan 24, 2015 at 14:52
  • $\begingroup$ Co(II) or Co(III)? Edited title is now in disagreement with formula in question! $\endgroup$ Commented Jan 26, 2015 at 13:19

2 Answers 2


$\ce{Co[(NH3)_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.

enter image description here

image source

If a molecule contains a plane of symmetry it cannot be optically active or chiral.

In the cis isomer there is a plane of symmetry containing the $\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active. With the trans isomer there are several planes of symmetry (one contains the $\ce{Co}$ atom and the four $\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.


The 6 ligands are at the 6 vertices of an octahedron.

There will be cis and trans isomers, but neither isomer is optically active.

For the trans isomer, there will be a central axis containing Cl-Co-Cl, and the 4 N atom lie in a plane of symmetry, bisecting the Cl-Co-Co line segment.

For the cis isomer, there is a 90 degree Cl-M-Cl angle, with the line segment Cl-Cl being an edge of the octahedron. The plane perpendicularly bisecting the Cl-Cl segment is a plane of symmetry. The plane passes through the Co and 2 of the N atoms as well.


Your Answer

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

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