Knowee
Questions
Features
Study Tools

In metal carbonyls, metal-ligand bonds possess both sigma and pi character. Comment, whetherthe given statement is correct. Give reason for your answer.

Question

In metal carbonyls, metal-ligand bonds possess both sigma and pi character. Comment, whether the given statement is correct. Give reason for your answer.

🧐 Not the exact question you are looking for?Go ask a question

Solution

The given statement is correct. In metal carbonyls, the metal-ligand bonds possess both sigma and pi character.

Reason: Metal carbonyls are coordination compounds where carbon monoxide (CO) acts as a ligand. The bonding between the metal and the ligand occurs through the donation of electron pairs from the CO molecule to the metal center.

The sigma (σ) bond is formed by the overlap of the metal d-orbitals with the antibonding π* orbital of the CO ligand. This results in a strong, directional bond between the metal and the ligand.

Additionally, the pi (π) bond is formed by the overlap of the metal d-orbitals with the π orbital of the CO ligand. This bond is weaker and less directional compared to the sigma bond.

Therefore, in metal carbonyls, the metal-ligand bonds possess both sigma and pi character, contributing to the overall stability and unique properties of these compounds.

This problem has been solved

Similar Questions

five sigma donor pi accepter ligand complexes How they are strong field ligands

What metal orbitals would be involved in sigma bonding according to valence bond theory in[Zn(H2O)6]2+

The number of σσσσσσσσσσ bonds, π bonds and lone pair of electrons in pyridine, respectively are:

A sigma bond is formed by the overlapping of :A s-s orbital alone B s and p orbitals alone C s-s, s-p or p-p orbitals along internuclear axis D None of these

)Five sigma donor ligand complex How to predict wether they are string field ligand or weak

1/1

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.