Question

Why does the symmetric C-O stretch of a trans dicarbonyl complex have zero intensity? I was...

Why does the symmetric C-O stretch of a trans dicarbonyl complex have zero intensity?

I was thinking if the stretches are occuring at the same time in opposite directions; the two stretches would sort of cancel one another.

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Answer #1

The symmetric C-O stretch of a trans dicarbonyl complex have zero intensity.

As predicted here, the symmetric stretch is zero due to equal stretch on both sides, which cancels with each other. This can also be explained in terms of the relation,

Intensity (symmetric)/Intensity (antisymmetric) = tan^2(theta/2)

wherein,

theta is the angle between the two carbonyls in the complex,

Now with trans-dicarbonyl complex, the angle between the two carbonyls is 180 degree

That is the term, tan2(theta/2) becomes zero.

Therefore, we do not see any symmetric stretch for the trans-dicarbonyl complex.

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