Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 85

Given the colors observed for \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\) (permanganate ion (see Exercise \(23.84 ),\) what can you say about how the energy separation between the ligand orbitals and the empty \(d\) orbitals changes as a function of the oxidation state of the transition metal at the center of the tetrahedral anion?

Short Answer

Expert verified
As the oxidation state of the transition metal at the center of the tetrahedral anion increases (from +5 in VO₄³⁻, to +6 in CrO₄²⁻, and +7 in MnO₄⁻), the energy separation between the ligand orbitals and the empty d orbitals increases. This is inferred from the observation that the colors of the complexes change from yellow (orthovanadate ion) to orange/yellow (chromate ion) and finally to violet (permanganate ion), which indicates that the energy of the absorbed light increases with the oxidation state of the transition metal.

Step by step solution

01

Identify the transition metal ions and their oxidation states

The complexes given are orthovanadate ion (VO₄³⁻), chromate ion (CrO₄²⁻), and permanganate ion (MnO₄⁻). The transition metals in these complexes are V, Cr, and Mn, and their oxidation states are +5, +6, and +7, respectively.
02

Determine the colors of the complexes

The colors observed for the complexes are: - Orthovanadate ion (VO₄³⁻): Yellow - Chromate ion (CrO₄²⁻): Yellow/orange - Permanganate ion (MnO₄⁻): Violet
03

Relate the colors to energy differences

The color of a complex is related to the energy difference between the ligand orbitals and the d orbitals. This energy difference corresponds to the energy of the absorbed light, causing the complementary color to be observed. The energy of the absorbed light is given by the equation: \[ E = h \cdot f\] Where \(E\) is the energy, \(h\) is Planck's constant, and \(f\) is frequency of the absorbed light. As we go from yellow to violet, we note that the frequency (and hence energy) of the absorbed light increases. On the color wheel, yellow is opposite of violet. Therefore, a yellow compound absorbs light in the violet region, and vice versa.
04

Derive a relationship between energy separation and oxidation state

Since the energy of the absorbed light increases as we move from orthovanadate ion (yellow) to permanganate ion (violet), we can infer that the energy separation between the ligand orbitals and the empty d orbitals also increases. This implies that, as the oxidation state of the transition metal at the center of the tetrahedral anion increases (from +5 in VO₄³⁻, to +6 in CrO₄²⁻, and +7 in MnO₄⁻), the energy separation between the ligand orbitals and the empty d orbitals increases.

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 total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a 0.100-L sample of the water with a solution of EDTA \(^{4-} .\) The EDTA \(^{4-}\) chelatesthe two cations: $$\begin{array}{c}{\mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-}} \\\ {\mathrm{Ca}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-}}\end{array}$$ It requires 31.5 \(\mathrm{mL}\) of 0.0104 \(\mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second 0.100-L sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with 18.7 \(\mathrm{mL}\) of 0.0104 \(M[\mathrm{EDTA}]^{4-} .\) Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in \(\mathrm{mg} / \mathrm{L} .\)

Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the \(+3\) rather than in the \(+2\) oxidation state (for metals that form stable \(+3\) ions in the first place). Suggest an explanation, keeping in mind the Lewis acid-base nature of the metal-ligand bond.

Which type of magnetic material cannot be used to make permanent magnets, a ferromagnetic substance, an anti-ferromagnetic substance, or a ferrimagnetic substance?

The complex \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) contains five unpaired electrons. Sketch the energy-level diagram for the \(d\) orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex?

Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) hexamminechromium(III) nitrate (b) tetraamminecarbonatocobalt(III) sulfate (c) dichlorobis(ethylenediamine)platinum(IV) bromide (d) potassium diaquatetrabromovanadate(III) (e) bis(ethylenediamine) zinc(II) tetraiodomercurate(II)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Inorganic 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