How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\)

Short Answer

Expert verified
The number of unpaired electrons for each complex ion are: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 unpaired electrons b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 unpaired electrons c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\): 0 unpaired electrons

Step by step solution

01

Identify the central atom and its electron configuration

For each complex ion: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) - Central atom: Ru (Ruthenium) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) - Central atom: Ni (Nickel) c. \(\mathrm{V}(\mathrm{en})_{3}^{3+}\) - Central atom: V (Vanadium) Now we can find the electron configuration of these atoms: a. Ru: [Kr]4d^75s^1 b. Ni: [Ar]3d^84s^2 c. V: [Ar]3d^34s^2
02

Determine the oxidation states of the central atoms

Now we need to find the oxidation states of the central atoms: a. Ru: \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) - Since ammonia (NH3) is a neutral ligand, the oxidation state of Ru is +2. b. Ni: \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) - Since water (H2O) is a neutral ligand, the oxidation state of Ni is +2. c. V: \(\mathrm{V}(\mathrm{en})_{3}^{3+}\) - Since ethylenediamine (en) is a bidentate neutral ligand, the oxidation state of V is +3.
03

Account for the oxidation state and find the electron configuration of the central ion

Now we can find the electron configuration of the central ion with the corresponding oxidation state: a. Ruthenium(II) ion (Ru2+): [Kr]4d^6 b. Nickel(II) ion (Ni2+): [Ar]3d^8 c. Vanadium(III) ion (V3+): [Ar]3d^2
04

Determine the effect of ligands on d-orbital splitting and find the unpaired electrons

To find the number of unpaired electrons, we must consider the effect of ligands on the d-orbital splitting. a. Since we are considering the low-spin case for \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\), the NH3 ligands, being a strong-field ligand, cause the d-electrons to pair. The electron configuration in the low-spin case will be t2g^6 eg^0. Thus: - Unpaired electrons in \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 b. In \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\), the water is a weak-field ligand. The electron configuration will be t2g^6 eg^2. Thus: - Unpaired electrons in \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 c. In \(\mathrm{V}(\mathrm{en})_{3}^{3+}\), the ethylenediamine (en) is a strong-field ligand. The electron configuration will be t2g^2 eg^0. Thus: - Unpaired electrons in \(\mathrm{V}(\mathrm{en})_{3}^{3+}\): 0

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!

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

Iron is present in the earth's crust in many types of minerals. The iron oxide minerals are hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) and magnetite \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right) .\) What is the oxidation state of iron in each mineral? The iron ions in magnetite are a mixture of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions. What is the ratio of \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ions in magnetite? The formula for magnetite is often written as \(\mathrm{FeO} \cdot \mathrm{Fe}_{2} \mathrm{O}_{3} .\) Does this make sense? Explain.

Consider the complex ions \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}, \mathrm{Co}(\mathrm{CN})_{6}^{3-},\) and \(\mathrm{CoF}_{6}^{3-} .\) The wavelengths of absorbed electromagnetic radiation for these compounds (in no specific order) are \(770 \mathrm{nm}\) \(440 \mathrm{nm},\) and \(290 \mathrm{nm} .\) Match the complex ion to the wavelength of absorbed electromagnetic radiation.

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colors of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complex ions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four coordination compounds?

Give formulas for the following complex ions. a. tetrachloroferrate(III) ion b. pentaammineaquaruthenium(III) ion c. tetracarbonyldihydroxochromium(III) ion d. amminetrichloroplatinate(II) ion

What is the electron configuration for the transition metal ion in each of the following compounds? a. \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) b. \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}\) c. \(\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}\) d. \(\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NO}_{2}\right)_{2}\right] \mathrm{I}\)

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