Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 13

For each of the following compounds, determine the electron configuration of the transition-metal ion. \((\mathbf{a})\) TiO, \((\mathbf{b}) \mathrm{TiO}_{2},(\mathbf{c}) \mathrm{NiO},(\mathbf{d}) \mathrm{ZnO}\) .

Short Answer

Expert verified
The electron configurations of the transition-metal ions in the given compounds are: a) \(\mathrm{Ti^2+ : [Ar] \ 3d^2}\) b) \(\mathrm{Ti^4+ : [Ar]}\) c) \(\mathrm{Ni^2+ : [Ar] \ 3d^8}\) d) \(\mathrm{Zn^2+ : [Ar] \ 3d^{10}}\)

Step by step solution

01

a) TiO

Before finding the electron configuration of the transition-metal ion, let's first determine the oxidation state of Ti in TiO: Oxygen has an oxidation state of -2, so Ti must have an oxidation state of +2 to balance the charges in the compound. Now, we can write the electron configuration of the neutral Ti atom and then remove the appropriate number of electrons to account for the oxidation state: The electron configuration of the neutral Ti atom is: \[ \mathrm{[Ar] 3d^2 4s^2}\] The Ti²⁺ ion would have an electron configuration where we remove two electrons: \[ \mathrm{[Ar] 3d^2}\]
02

b) TiO₂

To find the oxidation state of Ti in TiO₂, we need to consider that there are two oxygen atoms present in the compound: Each oxygen atom has an oxidation state of -2, making a total oxidation state of -4. Therefore, Ti must have an oxidation state of +4 to balance the charges in the compound. Using the electron configuration of the neutral Ti atom from the previous step, we can remove four electrons to account for the oxidation state: The Ti⁴⁺ ion would have an electron configuration: \[ \mathrm{[Ar] }\]
03

c) NiO

First, we determine the oxidation state of Ni in NiO: As it's been mentioned before, oxygen has an oxidation state of -2. Therefore, Ni must have an oxidation state of +2 to balance the charges in the compound. The electron configuration of a neutral Ni atom is: \[ \mathrm{[Ar] 3d^8 4s^2}\] The Ni²⁺ ion would have an electron configuration where we remove two electrons: \[ \mathrm{[Ar] 3d^8}\]
04

d) ZnO

Lastly, let's determine the oxidation state of Zn in ZnO: Oxygen exhibits an oxidation state of -2. Thus, Zn must have an oxidation state of +2 to balance the charges in the compound. The electron configuration of a neutral Zn atom is: \[ \mathrm{[Ar] 3d^{10} 4s^2}\] The Zn²⁺ ion would have an electron configuration where we remove two electrons: \[ \mathrm{[Ar] 3d^{10}}\] In summary, the electron configurations of the transition-metal ions in the given compounds are: a) \(\mathrm{Ti^2+ : [Ar] \ 3d^2}\) b) \(\mathrm{Ti^4+ : [Ar]}\) c) \(\mathrm{Ni^2+ : [Ar] \ 3d^8}\) d) \(\mathrm{Zn^2+ : [Ar] \ 3d^{10}}\)

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

(a) Using Werner's definition of valence, which property is the same as oxidation number, primary valence or secondary valence? (b) What term do we normally use for the other type of valence? (c) Why can \(\mathrm{NH}_{3}\) serve as a ligand but BH \(_{3}\) cannot?

Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group.

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{Na}_{2}\left[\mathrm{CdCl}_{4}\right]\) (b) \(\mathrm{K}_{2}\left[\mathrm{MoOCl}_{4}\right]\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}\) (d) \(\left[\mathrm{Ni}(\mathrm{CN})_{5}\right]^{3-}\) (e) \(\mathrm{K}_{3}\left[\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) (f) \(\left[\mathrm{Zn}(\mathrm{en})_{2}\right] \mathrm{Br}_{2}\)

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge-transfer color, the violet color of MnO \(_{4}^{-}\) is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\) occurs at a wavelength of 565 \(\mathrm{nm} .\) The \(\mathrm{CrO}_{4}^{2-}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for MnO \(_{4}^{-}?\) Explain. (e) The VO \(_{4}^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

(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 AgNO \(_{3}(a q)\) forms 2 mol of solid 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 mol of AgCl per mole of complex. What has happened in the ensuing time?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

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