Is tetraammine dichloro cobalt (III) an optically active compound?


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> Is
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> $$ \text{Co}[{(\text{N}{\text{H}}\_{3})}\_{4}{\text{Cl}}\_{2}]$$
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> optically active? If so, write its stereo-isomers.
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> 
> 


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?  

Thanks!




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


![enter image description here](https://i.stack.imgur.com/dw0s6.jpg)


[image source](http://chemistry.tutorvista.com/inorganic-chemistry/coordination-compound.html)


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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