Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 144

Consider the anion whose charge is \(2^{-}\) and whose electron configuration is identical to that of argon, Ar. What is the symbol for this anion?

Short Answer

Expert verified
The symbol for the anion whose electron configuration is identical to that of argon, Ar, and has a 2- charge is \(S^{2-}\).

Step by step solution

01

Identify the element with an atomic number of 16

Since argon has 18 electrons, we need to find the element that has 16 electrons in its neutral state, thus having an atomic number of 16 (which means 16 protons in its nucleus). A periodic table tells us that the element with an atomic number of 16 is sulfur, S.
02

Determine the electron configuration of the anion

Now that we've identified sulfur as the element to consider, we can determine its electron configuration as an anion with a 2- charge. A neutral sulfur atom has 16 electrons, so our anion would have 18 electrons (16 + 2) due to the 2- charge. This electron configuration would match that of argon.
03

Identify the symbol for the anion

The element that matches the given criteria is sulfur, S. An anion of sulfur with a 2- charge is represented with the symbol S²⁻.

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

When a group IIA element \(X\) reacts with a group VIIA element \(Y\), what will be the formula? Why?

Identify the period 2 element that is described by the ionization data below. \(\mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+} 1 \mathrm{e}^{-} \quad \mathrm{IE}(1)=1.40 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{+}(\mathrm{g}) \rightarrow \mathrm{M}^{2+} 1 \mathrm{e}^{-} \mathrm{IE}(2)=2.86 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{2+}(\mathrm{g}) \rightarrow \mathrm{M}^{3+} 1 \mathrm{e}^{-} \mathrm{IE}(3)=4.58 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{3+}(\mathrm{g}) \rightarrow \mathrm{M}^{4+} 1 \mathrm{e}^{-} \mathrm{IE}(4)=7.48 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{4+}(\mathrm{g}) \rightarrow \mathrm{M}^{5+} 1 \mathrm{e}^{-} \mathrm{IE}(5)=9.44 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{5+}(\mathrm{g}) \rightarrow \mathrm{M}^{6+} 1 \mathrm{e}^{-} \mathrm{IE}(6)=5.33 \times 10^{4} \mathrm{~J} / \mathrm{mol}\) \(\mathrm{M}^{6+}(\mathrm{g}) \rightarrow \mathrm{M}^{7+} 1 \mathrm{e}^{-} \mathrm{IE}(7)=6.44 \times 10^{4} \mathrm{~J} / \mathrm{mol}\)

Draw the Bohr model for a \(\mathrm{Cl}\) atom and for a \(\mathrm{Cl}^{-}\) ion. How many electrons are there in the valence shell in each drawing?

How can you tell how many electrons a representative nonmetal is likely to gain? What, in general, will be the charge of the anion it forms?

Write the ground-state electron configuration for each of the following atoms or ions. Which have a valence-shell octet? (a) \(\overline{\mathrm{Ar}}\) (b) \(\mathrm{Na}^{+}\) (c) \(\mathrm{C}^{2-}\) (d) \(\mathrm{O}^{2-}\) (e) \(\mathrm{Ca}^{2+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic 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