Does anyone know the answers to these? Its in reference to the stereochemical is

Question

Does anyone know the answers to these? Its in reference to the stereochemical isomers of [Mo(CO)4L2]] by IR spectroscopy!

If you need more details let me know ... really baffled on these last two.

Conversion between these two isomers is very fast for [Mo(CO)4(PPh3 )2] but slower for [Mo(CO)4(P(OPh)3)2]. Explain this observation, considering the differences in Mo-P bonding and the cone-angles of the ligands. From the CO stretching frequencies of the lowest frequency bands, order the cis-geometry compounds from strongest CO bonds to weakest. Can you explain this trend? How many isomers would you expect if ethylenediamine, rather than piperidine, were reacted with [Mo(CO)6] to give a similar tetracarbonyl product? ferences

Explanation / Answer

From PPh3 and P(OPh)3, electron donating ability of P(OPh)3 is less as compared to PPh3 due to high electonegativity of oxygen.

So M-P in case PPh3 will be strong as compared to that of P(OPh)3. Another reason is P(OPh)3 is bigger ligand than PPh3 and it will create a more steric hindrance in the cis isomer.

Hence, the conversion between the two isomers of Mo(CO)4(PPh3)2 will be fast as compared to other.

In case of P(OPh)3, electron donating tendency is less due to more electronegativity of oxygen. This it will decrease the electron density on metal, thus there will be decrease in back bonding from metal to CO group and hence stretching frequency will be high in this molecule as compared to other i.e., Mo(CO)4(PPh3)2.

If ethylenediamine (en) will react with Mo(CO)6 to form tetracarbonyl compoud, it will form Mo(CO)4(en), only 1 isomer is possible for this.

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