Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 65

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?

Short Answer

Expert verified
The \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion has an octahedral coordination geometry due to the 6 ammonia ligands. The weak field ligands lead to minimal splitting of the \(d\) orbitals. The 5 \(d\) electrons will singly occupy the orbitals in the following manner: \(d_{xy}\), \(d_{yz}\), \(d_{xz}\), \(d_{z^2}\), and \(d_{x^2-y^2}\). This results in a high-spin complex with 5 unpaired electrons.

Step by step solution

01

Analyzing the given complex

The given complex is \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) which contains Mn (Manganese) as the central metal ion and 6 NH\(_3\) (Ammonia) ligands surrounding it.
02

Determine the coordination geometry

Since there are six ligands surrounding the metal ion, the coordination geometry is octahedral.
03

Apply the crystal field theory and determine the energy-level splitting

In an octahedral field, the \(d\) orbitals split into two sets: - \(t_{2g}\) orbitals (lower energy): \(d_{xy}\), \(d_{yz}\), and \(d_{xz}\) - \(e_{g}\) orbitals (higher energy): \(d_{z^2}\) and \(d_{x^2-y^2}\) Ammonia (NH\(_3\)) is a weak field ligand, which means it will not split the \(d\) orbitals significantly.
04

Add electrons to the energy levels accordingly

Mn\(^{2+}\) has 5 \(d\) electrons. Since ammonia is a weak field ligand and does not cause significant splitting, the electrons will first occupy the lower energy \(t_{2g}\) orbitals, and then singly occupy the higher energy \(e_{g}\) orbitals, following Hund's rule. The energy level placement will look like this: - \(d_{xy}\): 1 electron - \(d_{yz}\): 1 electron - \(d_{xz}\): 1 electron - \(d_{z^2}\): 1 electron - \(d_{x^2-y^2}\): 1 electron
05

Determine if the complex is a high-spin or low-spin complex

As the electrons are maximumly singly occupied in the higher energy orbitals, the complex is a high-spin complex. The energy-level diagram for the \(\left[\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\) complex ion has 5 unpaired electrons in the \(d\) orbitals, making it a high-spin complex.

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) If a complex absorbs light at \(610 \mathrm{nm},\) what color would you expect the complex to be? (b) What is the energy in joules of a photon with a wavelength of 610 \(\mathrm{nm} ?\) (c) What is the energy of this absorption in \(\mathrm{kJ} / \mathrm{mol}\) ?

Four-coordinate metals can have either a tetrahedral or a square-planar geometry; both possibilities are shown here for \(\left[\mathrm{PtCl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right] .\) (a) \(\mathrm{What}\) is the name of this molecule? (b) Would the tetrahedral molecule have a geometric isomer? (c) Would the tetrahedral molecule be diamagnetic or paramagnetic? (d) Would the square-planar molecule have a geometric isomer? (e) Would the square-planar molecule be diamagnetic or paramagnetic? (f) Would determining the number of geometric isomers help you distinguish between the tetrahedral and square-planar geometries? (g) Would measuring the molecule's response to a magnetic field help you distinguish between the two geometries? [Sections 23.4-23.6 ]

Solutions of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\) both octahedral \()\) and \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less that that in octahedral ones, assign a color to each complex.

A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains \(10.0 \% \mathrm{Mn}, 28.6 \%\) potassium, \(8.8\%\) carbon, and 29.2\(\%\) bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] .\) Write the formula of the compound, using brackets to denote the manganese and its coordination sphere.

If the lobes of a given \(d\)-orbital point directly at the ligands, will an electron in that orbital have a higher or lower energy than an electron in a \(d\)-orbital whose lobes do not point directly at the ligands?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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