A metal ion with a net +3 charge has five electrons in the \(3 d\) subshell. Identify the metal.

Short Answer

Expert verified
The metal ion is Vanadium.

Step by step solution

01

Determine Number of Electrons

Since the net charge of this metal ion is +3, it means this metal ion has lost three electrons. Therefore, the neutral atom would have had three additional electrons. given that the 3d subshell has 5 electrons, the total electron count for the neutral atom is 5 (in the 3d subshell) + 3 (the ones that were lost), which equals 8 electrons.
02

Cross-check with Atomic Number

The atomic number of an element is equal to its count of protons. In a neutral atom, the number of protons equals the number of electrons. So, the atomic number of this atom is the same as the total number of electrons, which is 8.
03

Identify the Metal

With the atomic number, you now look up the element on the Periodic Table. An element with atomic number 8 is Oxygen. However, Oxygen is not a metal. This tells you that the metal ion lost not only 3s electrons but 2p electrons as well. In 2p level there are 6 electrons so 6+3 = 9. The element having 9 in 2p3 level is Vanadium (V). After losing three electrons, we have a +3 charged V ion.

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

Write equations representing the following processes: (a) The electron affinity of \(\mathrm{S}^{-}\) (b) The third ionization energy of titanium (c) The electron affinity of \(\mathrm{Mg}^{2+}\) (d) The ionization energy of \(\mathrm{O}^{2-}\)

The ionization energies of sodium (in \(\mathrm{kJ} / \mathrm{mol}\) ), starting with the first and ending with the eleventh, are 495.9,4560,6900,9540,13,400,16,600,20,120 \(25,490,28,930,141,360,170,000 .\) Plot the \(\log\) of ionization energy ( \(y\) axis) versus the number of ionization \((x\) axis \() ;\) for example, log 495.9 is plotted versus 1 (labeled \(I E_{1}\), the first ionization energy), \(\log 4560\) is plotted versus 2 (labeled \(I E_{2}\), the second ionization energy), and so on. (a) Label \(I E_{1}\) through \(I E_{11}\) with the electrons in orbitals such as \(1 s, 2 s, 2 p\) and \(3 s .\) (b) What can you deduce about electron shells from the breaks in the curve?

(a) The formula of the simplest hydrocarbon is \(\mathrm{CH}_{4}\) (methane). Predict the formulas of the simplest compounds formed between hydrogen and the following elements: silicon, germanium, tin, and lead. (b) Sodium hydride (NaH) is an ionic compound. Would you expect rubidium hydride (RbH) to be more or less ionic than NaH? (c) Predict the reaction between radium (Ra) and water. (d) When exposed to air, aluminum forms a tenacious oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) coating that protects the metal from corrosion. Which metal in Group 2 A would you expect to exhibit similar properties? Why?

Explain, in terms of their electron configurations, why \(\mathrm{Fe}^{2+}\) is more easily oxidized to \(\mathrm{Fe}^{3+}\) than \(\mathrm{Mn}^{2+}\) is to \(\mathrm{Mn}^{3+}\)

A student is given samples of three elements, \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), which could be an alkali metal, a member of Group \(4 \mathrm{~A},\) and a member of Group \(5 \mathrm{~A} .\) She makes the following observations: Element X has a metallic luster and conducts electricity. It reacts slowly with hydrochloric acid to produce hydrogen gas. Element \(Y\) is a light-yellow solid that does not conduct electricity. Element \(Z\) has a metallic luster and conducts electricity. When exposed to air, it slowly forms a white powder. A solution of the white powder in water is basic. What can you conclude about the elements from these observations?

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