Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 14

No compounds are known in which scandium is in the +2 oxidation state. Suggest an explanation.

Short Answer

Expert verified
Scandium generally forms a stable +3 oxidation state by losing three electrons, leading to an electronic configuration similar to the noble gas argon. The +2 oxidation state of scandium would require losing only two electrons, which leaves a less stable electronic configuration with one electron remaining in the 3d subshell. This results in no known compounds with scandium in the +2 oxidation state due to its less favorable electron configuration and reduced stability.

Step by step solution

01

Electronic configuration of scandium

Scandium has an atomic number of 21. Its electronic configuration can be written as \(1s^{2}\, 2s^{2}2p^{6}\, 3s^{2}3p^{6}\, 4s^{2}3d^{1}\). This configuration shows that scandium has two electrons in the 4s subshell and one electron in the 3d subshell.
02

Oxidation states

Oxidation states indicate the number of electrons that an atom can gain or lose to form a compound with other elements. The oxidation state of an element can often be predicted by looking at its position in the periodic table and its electronic configuration.
03

Examining +3 oxidation state of scandium

When scandium forms ions, it generally loses three electrons to achieve a stable configuration. These three electrons are the two in the 4s subshell and the one in the 3d subshell. Thus, scandium forms a +3 ion with an electronic configuration \(1s^{2}\, 2s^{2}2p^{6}\, 3s^{2}3p^{6}\) like the noble gas argon, leading to a stable +3 oxidation state.
04

Exploring the +2 oxidation state

If scandium were to form a +2 oxidation state, it would have to lose only two electrons, leaving one electron in the 3d subshell. The resulting ion would have an electronic configuration of \(1s^{2}\, 2s^{2}2p^{6}\, 3s^{2}3p^{6}\, 3d^{1}\). This configuration is less stable than the one with a +3 oxidation state, as the remaining electron in the 3d subshell is not energetically favorable.
05

Conclusion

Scandium tends to achieve a more stable electronic configuration by losing three electrons, resulting in a +3 oxidation state. The +2 oxidation state, which would require only losing two electrons, leaves an unfavorable electron configuration with less stability. Therefore, no known compounds with scandium in the +2 oxidation state exist.

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

By writing formulas or drawing structures related to any one of these three complexes, \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Cl}\) \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{ONO})_{2}\right]\) cis-[ \(\left.\mathrm{V}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) illustrate (a) geometric isomerism, (b) linkage isomerism, (c) optical isomerism, (d) coordination-sphere isomerism.

Give brief statements about the relevance of the following complexes in living systems: (a) hemoglobin, (b) chlorophylls, (c) siderophores.

(a) A compound with formula \(\mathrm{RuCl}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) is dissolved in water, forming a solution that is approximately the same color as the solid. Immediately after forming the solution, the addition of excess \(\mathrm{AgNO}_{3}(a q)\) forms \(2 \mathrm{~mol}\) of solid \(\mathrm{AgCl}\) per mole of complex. Write the formula for the compound, showing which ligands are likely to be present in the coordination sphere. (b) After a solution of \(\mathrm{RuCl}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) has stood for about a year, addition of \(\mathrm{AgNO}_{3}(a q)\) precipitates \(3 \mathrm{~mol}\) of \(\mathrm{AgCl}\) per mole of complex. What has happened in the ensuing time?

A certain complex of metal \(\mathrm{M}\) is formulated as \(\mathrm{MCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\). The coordination number of the complex is not known but is expected to be 4 or 6. (a) Would conductivity measurements provide information about the coordination number? (b) In using conductivity measurements to test which ligands are bound to the metal ion, what assumption is made about the rate at which ligands enter or leave the coordination sphere of the metal? (c) Suppose you experimentally determine that this complex exists in aqueous solution as a single species. Suggest a likely coordination number and the number and type of each ligand.

Oxyhemoglobin, with an \(\mathrm{O}_{2}\) bound to iron, is a low-spin \(\mathrm{Fe}(\mathrm{II})\) complex; deoxyhemoglobin, without the \(\mathrm{O}_{2}\) molecule, is a high-spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of \(\mathrm{O}_{2}\) in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15 -minute exposure to air containing 400 ppm of CO causes about \(10 \%\) of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and \(\mathrm{O}_{2}\) to hemoglobin? (e) \(\mathrm{CO}\) is a strong-field ligand. What color might you expect carboxyhemoglobin to be?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear 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