Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 53

Explain the following facts: (a) Copper and iron have several oxidation states, whereas zinc has only one. (b) Copper and iron form colored ions, whereas zinc does not.

Short Answer

Expert verified
Copper and Iron can have multiple oxidation states due to their electron configurations, with unpaired electrons in different energy levels, while Zinc's electron configuration with fully filled orbitals allows only one oxidation state. Regarding colors, Copper and Iron's ions show color because of their incomplete d-orbitals that facilitate electron transitions upon absorbing certain light wavelengths. In contrast, Zinc's ions are colorless as its fully filled d-orbitals don't allow such transitions.

Step by step solution

01

Understanding oxidation states

Oxidation states represent the degree of oxidation of an atom in a chemical compound. The oxidation state is determined by the number of electrons that an atom loses, gains, or appears to use when forming a chemical compound. Oxidation states help a chemist keep track of electrons in an atom. Iron (Fe) and Copper (Cu) have several oxidation states due to the presence of unpaired electrons in different energy levels. However, In Zinc (Zn), the outermost orbitals are completely filled, which makes it stable and it shows only +2 oxidation state.
02

Understanding color formations in ions

Color in metal ions is due to d-d transitions. When an electron from a lower energy level is excited to a higher energy level, it absorbs energy in the form of light. The light absorbed is subtracted from white light and the color we see is actually the complementary color of the light absorbed. For Iron and Copper, they have incomplete d-orbitals in their ion state and thus they can absorb certain wavelengths of light leading to the color. Zinc, on the other hand, has completely filled d-orbitals even in their ion-state and thus there are no d-d transitions possible and they do not absorb any visible light, hence no color.
03

Conclusion

Different oxidation states are caused by different electron configurations. Copper and iron have several oxidation states due to the presence of unpaired electrons in different energy levels which allow for multiple oxidations states. Zinc only has one oxidation state because all outermost orbitals are completely filled thus adding or removing electrons is not energetically favorable. The color formation in transition metals occurs due to the transitions of electrons between different energy levels - which is possible in Copper and Iron because of their incomplete d-orbitals, while in Zinc, the d-orbitals are completely filled, eliminating the possibility of d-d transitions, resulting in a colorless ion.

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

Explain the following terms: (a) enantiomers, (b) racemic mixtures.

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 distinguishes a transition metal from a representative metal?

What are the systematic names for the following ions and compounds? (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}\) (b) \(\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\) (c) \(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Br}_{2}\right]^{+}\) (d) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\)

Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammin eaquachlorocobalt(III) chloride.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Organic 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