Chapter 28: Problem 9

Give reasons for (a) CO stretching frequencies are higher in \(\left[\mathrm{F}_{3} \mathrm{PNi}(\mathrm{CO})_{3}\right]\) than in \(\left[\mathrm{Me}_{3} \mathrm{PNi}(\mathrm{CO})_{3}\right]\). (b) Extent of \(\pi\) -bonding in \(\left[\mathrm{Ag}(\mathrm{CNR})_{4}\right.\) is higher than in \(\left[\mathrm{Fe}(\mathrm{CNR})_{6}\right]^{2+}\).

Short Answer

Expert verified

The CO stretching frequencies are higher in \([\mathrm{F}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\) than in \([\mathrm{Me}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\) due to the higher electronegativity of F compared to CH3, which strengthens the CO bond. The extent of \(\pi\)-bonding is higher in \(\left[\mathrm{Fe}(\mathrm{CNR})_{6}\right]^{2+}\) than \(\left[\mathrm{Ag}(\mathrm{CNR})_{4}\right.\) due to greater availability of d orbitals for \(\pi\)-bonding in Fe2+.

Step by step solution

Analyzing the CO stretching frequencies

The stretching frequency of a bond is directly related to the strength of the bond. A stronger bond vibrates or 'stretches' at a higher frequency. Here, it has been observed that the CO stretching frequencies in \([\mathrm{F}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\) are higher than in \([\mathrm{Me}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\). This suggests that CO bonds in the first compound are stronger than the latter.

Comparing the molecular structures

The two compounds differ only in the ligands attached to the P atom, F (fluorine) in one case and Me (methyl group) in the other. Fluorine is more electronegative than carbon (central atom in a methyl group). Therefore, \([\mathrm{F}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\) is able to pull electron density away from the Ni atom more effectively than \([\mathrm{Me}_{3}\mathrm{PNi}(\mathrm{CO})_{3}]\). This increases the positive charge on Ni in the first compound which then strengthens its bonds with CO, resulting in higher stretching frequencies.

Analyzing the \(\pi\)-bonding extent

The second part of the problem deals with the extent of \(\pi\)-bonding in \(\left[\mathrm{Ag}(\mathrm{CNR})_{4}\right.\) versus \(\left[\mathrm{Fe}(\mathrm{CNR})_{6}\right]^{2+}\). The extent of \(\pi\)-bonding depends on how many orbitals are available for overlapping and forming \(\pi\)-bonds.

Comparing the d orbitals

The Fe in \(\left[\mathrm{Fe}(\mathrm{CNR})_{6}\right]^{2+}\)has five d orbitals while the Ag in \(\left[\mathrm{Ag}(\mathrm{CNR})_{4}\right.\) has only one free d orbital due to its electronic configuration ([Kr] 4d10 5s1). Therefore, Fe2+ can participate in \(\pi\)-bonding with six CNR groups while Ag can only do so with one CNR group. This leads to higher extent of \(\pi\)-bonding in \(\left[\mathrm{Fe}(\mathrm{CNR})_{6}\right]^{2+}\) than in \(\left[\mathrm{Ag}(\mathrm{CNR})_{4}\right.\).

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Analyzing the CO stretching frequencies

Comparing the molecular structures

Analyzing the \(\pi\)-bonding extent

Comparing the d orbitals

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Making Measurements

Nuclear Chemistry

Chemical Reactions

Physical Chemistry

Chemistry Branches

Study anywhere. Anytime. Across all devices.