Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 35

Predict the number of unpaired electrons in the following complex ions: (a) \(\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-},(\mathrm{b})\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\)

Short Answer

Expert verified
The number of unpaired electrons in the complex ion \([\mathrm{Cr}(\mathrm{CN})_{6}]^{4-}\) is 0, and in the complex ion \[\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}\] is 3.

Step by step solution

01

Determining the Electronic Configuration of Chromium (Cr)

The atomic number of chromium is 24. According to the Aufbau principle, the electronic configuration of a Cr atom in ground state is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^4\]. However, chromium is one of the exceptions to this rule and promotes one electron from 4s to 3d in order to achieve a half-filled (more stable) 3d subshell and a fully filled 4s subshell. This gives us the more accurate electronic configuration \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\]
02

Evaluating the Effect of Ligands and Charge on the Complex Ion \([\mathrm{Cr}(\mathrm{CN})_{6}]^{4-}\)

In the complex ion \[\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]^{4-}\], Cr must lose three electrons to form a Cr3+ ion (because the net charge of the complex ion (-4) must be offset by the Cr and ligands, and since six CN- ligands each add -1 charge, the Cr must have charge +3). Starting from the ground state configuration of Cr (\[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\]), this results in a configuration of \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\] (after reduction), and thus 3 unpaired electrons. However, since CN- is a strong field ligand, it will induce electron pairing in the 3d orbital. Therefore, all electrons in the Cr3+ ion become paired and there are no unpaired electrons left.
03

Evaluating the Effect of Ligands and Charge on the Complex Ion \[\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}\]

In the complex ion \[\left[\mathrm{Cr}\left(\mathrm{H}_{2}\mathrm{O}\right)_{6}\right]^{2+}\], Cr must also lose three electrons to form a Cr3+ ion because six water molecules don't add any charge to the ion, therefore, the +2 charge of the complex ion is due to Cr3+. This reduction also results in the configuration \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\] with 3 unpaired electrons. However, H2O is a weak field ligand and does not induce electron-pairing in the 3d orbital. Hence, the number of unpaired electrons remains 3.

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is more labile than the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) complex. Suggest an experiment that would prove that \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a labile complex.

Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

What are the differences between geometric isomers and optical isomers?

In a dilute nitric acid solution, \(\mathrm{Fe}^{3+}\) reacts with thiocyanate ion (SCN \(^{-}\) ) to form a dark-red complex: $$ \mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+} $$ The equilibrium concentration of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, \(1.0 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) was mixed with \(1.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{KSCN}\) and \(8.0 \mathrm{~mL}\) of dilute \(\mathrm{HNO}_{3}\). The color of the solution quantitatively indicated that the \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\) concentration was \(7.3 \times 10^{-5} M .\) Calculate the formation constant for \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)

Define the following terms: stereoisomers, geometric isomers, optical isomers, and plane-polarized light.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free