Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 62

Copper is also known to exist in +3 oxidation state, which is believed to be involved in some biological electron transfer reactions. (a) Would you expect this oxidation state of copper to be stable? Explain. (b) Name the compound \(\mathrm{K}_{3} \mathrm{CuF}_{6}\) and predict the geometry of the complex ion and its magnetic properties. (c) Most of the known Cu(III) compounds have square planar geometry. Are these compounds diamagnetic or paramagnetic?

Short Answer

Expert verified
Copper in a +3 state is not very stable due to the removal of d electrons. The compound \(K3CuF6\) is named as Potassium Hexafluorocuprate (III) and is likely to adopt a square planar geometry, making it a diamagnetic complex. Similarly, most Cu (III) compounds with square planar geometries are diamagnetic, due to their completely filled d-orbitals.

Step by step solution

01

- Analyzing Copper's Oxidation State

Chromium and copper are known exceptions to the Aufbau principle. Copper generally prefers to be in a +2 state (\(Cu^{2+}\)), which is more stable due to it having a completed d orbital. However in a +3 state (\(Cu^{3+}\)), copper would have to remove one of these d electrons, making it less stable. Hence, without biological intervention to maintain the +3 state, one would not expect it to be very stable.
02

- Naming and Predicting geometry

The compound \(\mathrm{K}_{3} \mathrm{CuF}_{6}\) would be named as Potassium Hexafluorocuprate(III). The copper ion in this compound exists in a +3 oxidation state, as indicated by the compound's overall charge and the charge of the other components. As for its geometry, copper (III) complexes usually adopt a four-coordinate, square planar geometry, similar to other d^8 complexes such as Ni(II) and Rh(I). The geometry reflects the influence of the Jahn-Teller distortion. Predicting the magnetic properties, due to the electronic configuration of \(\mathrm{Cu}^{3+}\) all of the electrons are paired so it is diamagnetic.
03

- Determining Magnetic Properties

Most Cu(III) compounds have square planar geometries. These compounds are likely to be diamagnetic because the \(Cu^{3+}\) ion has a \(d^8\) configuration. In a square planar complex, all of the d-orbitals are filled completely, so there are no unpaired electrons, making the complex diamagnetic.

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

Define the following terms: coordination compound, ligand, donor atom, coordination number, chelating agent.

Write the formulas for each of these ions and compounds: (a) tetrahydroxozincate(II), (b) pentaaquochlorochromium(III) chloride, (c) tetrabromocuprate(II), (d) ethylenediaminetetraacetatoferrate(II).

In a dilute nitric acid solution, \(\mathrm{Fe}^{3+}\) reacts with thiocyanate ion \(\left(\mathrm{SCN}^{-}\right)\) to form a dark-red complex: $$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+\mathrm{SCN}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}$$ 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 \(1.0 \times 10^{-3} M \mathrm{KSCN}\) 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 \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NCS}\right]^{2+}\)

A student in 1895 prepared three chromium coordination compounds having the same formulas of \(\mathrm{CrCl}_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\) with these properties: Write modern formulas for these compounds and suggest a method for confirming the number of \(\mathrm{Cl}^{-}\) ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates, which are compounds that have a specific number of water molecules attached to them. The \(\mathrm{Cr}\) has a coordination number of 6 in all three compounds.)

Why is zinc not considered a transition metal?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

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