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

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. Jan 24 '15 at 14:52
  • $\begingroup$ Co(II) or Co(III)? Edited title is now in disagreement with formula in question! $\endgroup$ – Klaus-Dieter Warzecha Jan 26 '15 at 13:19

$\ce{Co[(NH3)_4Cl2]}$ 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.

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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.

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