# Stretching Vibrational Frequency in Coordinate Compounds

My Intellectual Friends,

I came across this today, actually now in fact.

What is this? How do I approach this question?

Note by Vishwak Srinivasan
3 years, 4 months ago

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Carbonyl's antibonding molecular orbital(pi) can form a back bond with a filled $$t_2g$$ metal orbital. All this means is that when the back bond is weakening, the frequency will increase.

Hence one must compare the compounds and see if the different ligands can affect the back bond. The P-Phenyl ABMO is of comparable energy with the filled $$t_2g$$ metal orbital. Chlorine takes away most of Phosphorus's electron density, hence the metal isn't very rich in ions. This isn't the case with phenyl, they allow phosphorus to give electron density to the metal and hence increase chances of backbonding. Ligands that increase the chances of back bonding between the metal and the carbonyl decrease frequency. This happens as back bonding lowers the bond order which lowers frequency.

So I think the order will be (iv) > (iii) > (ii) > (i)

This is an interesting question. Do you know the answer?

- 2 years, 7 months ago

Perhaps you may google out for Carbonyl Complexes and their properties

- 3 years, 4 months ago

I searched so. I even searched for Stretching Vibrational Frequency and also found a Wikipedia page. Unfortunately, couldn't understand much. So I posted it here.

- 3 years, 4 months ago

from where did you found this ? As this is not in JEE Syllabus

- 3 years, 4 months ago

I found this in one of my mock tests. By Allen

- 3 years, 4 months ago

- 3 years, 4 months ago

Share with others too dude.

- 3 years, 4 months ago