Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Identify each of the following coordination complexes as either diamagnetic or paramagnetic: (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) (b) square planar \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{\mathrm{s}}\right]^{2+}\) (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\)

Short Answer

Expert verified
The coordination complexes are classified as follows: (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is diamagnetic, (b) square planar \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is paramagnetic, (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{\mathrm{s}}\right]^{2+}\) is paramagnetic, and (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) is paramagnetic.

Step by step solution

01

Determine the oxidation state and electron configuration of the central metal ion

For each coordination complex, determine the oxidation state of the central metal ion. Then, find the electron configuration of the central metal ion in its respective oxidation state. (a) The oxidation state of Ag in \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is +1. Therefore, the electron configuration of Ag(I) is [Kr]4d10. (b) Cu has an oxidation state of +2 in the square planar complex \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\). The electron configuration of Cu(II) is [Ar]3d9. (c) In \(\left[\mathrm{Ru}(\mathrm{bipy})_{\mathrm{s}}\right]^{2+}\), Ru has an oxidation state of +2. The electron configuration of Ru(II) is [Kr]4d6. (d) Co has an oxidation state of +2 in \(\left[\mathrm{CoCl}_{4}\right]^{2-}\). The electron configuration of Co(II) is [Ar]3d7.
02

Count the unpaired electrons in each central metal ion

For each electron configuration determined in step 1, count the number of unpaired electrons. (a) Ag(I) has the electron configuration [Kr]4d10, which means it has 0 unpaired electrons. (b) Cu(II) has the electron configuration [Ar]3d9, which means it has 1 unpaired electron. (c) Ru(II) has the electron configuration [Kr]4d6, which means it has 2 unpaired electrons. (d) Co(II) has the electron configuration [Ar]3d7, which means it has 3 unpaired electrons.
03

Determine if each coordination complex is diamagnetic or paramagnetic

Based on the number of unpaired electrons counted in step 2, classify each coordination complex as diamagnetic or paramagnetic. (a) \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is diamagnetic, as it has 0 unpaired electrons. (b) Square planar \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is paramagnetic, as it has 1 unpaired electron. (c) \(\left[\mathrm{Ru}(\mathrm{bipy})_{\mathrm{s}}\right]^{2+}\) is paramagnetic, as it has 2 unpaired electrons. (d) \(\left[\mathrm{CoCl}_{4}\right]^{2-}\) is paramagnetic, as it has 3 unpaired electrons.

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.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation State
Understanding the oxidation state of a metal in a coordination complex helps us know how many electrons the metal has lost. The loss of electrons affects its chemical behavior and interactions with ligands.
In coordination chemistry, the oxidation state is usually determined by examining the charges and bonding of ligands and the metal itself. For example:

• In \([\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\), silver (Ag) has an oxidation state of +1.
• The oxidation state of copper (Cu) in \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is +2.

Determining oxidation states is often the first step in analyzing the properties of coordination complexes, as it directly influences electron configurations.
Electron Configuration
The electron configuration of a metal ion in a complex describes the distribution of electrons across its atomic orbitals. A useful tool, it helps us understand its magnetic properties and reactivity.
To find the electron configuration for a specific oxidation state, remove electrons starting from the outermost shell. Consider these examples:

• Ag(I) in \([\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) has the configuration [Kr]4d10.
• Cu(II) in \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) has [Ar]3d9.

This knowledge helps to count the unpaired electrons, which is key to predicting whether the complex is diamagnetic or paramagnetic.
Diamagnetic and Paramagnetic
Whether a coordination complex is diamagnetic or paramagnetic depends on the presence of unpaired electrons in its metal ion. This influences how the complex behaves in a magnetic field.

• **Diamagnetic** substances have all paired electrons and are slightly repelled by magnets. For instance, \([\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\) is diamagnetic, as it has 0 unpaired electrons.

• **Paramagnetic** substances have one or more unpaired electrons, making them attracted to magnets. For example:
- \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is paramagnetic with 1 unpaired electron.
- \([\mathrm{Ru}(\mathrm{bipy})_{\mathrm{s}}\right]^{2+}\) has 2 unpaired electrons and is also paramagnetic.

Recognizing these properties is essential in coordination chemistry.
Coordination Chemistry
Coordination chemistry involves studying the structures and functions of complex compounds with central metal atoms bonded to surrounding molecules or ions, called ligands.
In these complexes, the central metal and ligands form a specific geometrical shape, often influencing the complex's chemical properties and reactivity.

• Square planar geometry, as seen in \([\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\), typically affects magnetic and optical behaviors due to its specific bonding angles.
• Ligands like NH₃ (ammonia) or bipy (2,2'-bipyridine) coordinate to the metal, providing electron pairs that complete the metal’s electron shell.

This coordination creates unique chemical behaviors and plays a pivotal role in applications ranging from catalysis to medicine.

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 red color of ruby is due to the presence of \(\mathrm{Cr}(\mathrm{III})\) ions at octahedral sites in the dose-packed exide lattice of \(\mathrm{Al}_{2} \mathrm{O}_{2}\). Draw the crystal-field splitting diagram for \(\mathrm{Cr}\) (III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.

Which type of substance is attracted by a magnetic field, a diamagnetic substance or a paramagnetic substance?

Sketch all the possible stereoisomers of (a) tetrahedral $\left[\mathrm{Cd}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{2}\right],(\mathbf{b})\( square-planar \)\left[\operatorname{Ir} \mathrm{Cl}_{2}\left(\mathrm{PH}_{3}\right)_{2}\right]^{-},$ (c) octahedral $\left[\mathrm{Fe}(\sigma \text { -phen })_{2} \mathrm{Cl}_{2}\right]^{+}$

The molecule dimethylphosphinoethane \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{PCH}_{2} \mathrm{CH}_{2}\right.\) \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{2}\), which is abbreviated dmpel is used as a ligand for some complexes that serve as catalysts. A complex that contains this ligand is \(\mathrm{Mo}(\mathrm{CO})_{4}\) (dmpe). (a) Draw the Lewis structure for dmpe, and compare it with ethylenediamine as a coordinating ligand. (b) What is the oxidation state of Mo in \(\mathrm{Na}_{2}\left[\mathrm{Mo}(\mathrm{CN})_{2}(\mathrm{CO})_{2}\right.\) (dmpe)]? (c) Sketch the structure of the \(\left[\mathrm{Mo}(\mathrm{CN})_{2}(\mathrm{CO})_{2}(\text { dmpe })\right]^{2-}\) ion, including all the possible isomers.

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 \(\mathrm{MnO}_{4}^{-}\)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 \(\mathrm{MnO}_{4}^{-}\)? Explain. (e) The \(\mathrm{VO}_{4}^{3-}\) 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.

See all solutions

Recommended explanations on Chemistry Textbooks

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