Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 24

Predict the powerful oxidizing agent in the 3 rd period: (a) sulphur (b) sodium (c) chlorine (d) bromine

Short Answer

Expert verified
Answer: (c) Chlorine

Step by step solution

01

Understanding oxidizing agents

An oxidizing agent is a substance that can accept electrons from another substance. The strength of an oxidizing agent depends on its ability to accept electrons, which is usually determined by its electronegativity and the change in oxidation state during the reaction.
02

Determining the electronegativity and oxidation state of given elements

Let's determine the electronegativity and common oxidation states of the given elements. You can find these values in a periodic table. (a) Sulphur: Electronegativity = 2.58, Oxidation states = -2, +4, +6 (b) Sodium: Electronegativity = 0.93, Oxidation states = +1 (c) Chlorine: Electronegativity = 3.16, Oxidation states = -1, +1, +3, +5, +7 (d) Bromine: Electronegativity = 2.96, Oxidation states = -1, +1, +3, +4, +5
03

Comparing the oxidizing abilities of given elements

A strong oxidizing agent will have higher electronegativity and a greater increase in its oxidation state during the reaction. The greater electronegativity makes it easier for the atom to attract electrons, while the increase in oxidation state ensures that it gains electrons. After comparing the periods, we can see that chlorine has the highest electronegativity among the given elements and can undergo the greatest increase in oxidation state (from -1 to +7). Thus, it has the strongest oxidizing ability.
04

Concluding which element is the powerful oxidizing agent

Based on the comparison, it is clear that chlorine (\(Cl\)) is the most powerful oxidizing agent among the given elements in the 3rd period. The correct option is (c) chlorine.

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

The number of valence electrons that can be present in the second element of any period is (a) 1 (b) 2 (c) 5 (d) 7

Which of the following sequence of explanation is appropriate for explaining the reason for the periodicity of reducing property in a period or group? (1) Tendency to undergo oxidation decreases in a period and increases in a group. (2) In a period ionization energy increases and in a group it decreases. (3) In a period the atomic size decreases and in a group it increases. (4) The elements present in the left side of the periodic table have strong reducing property. (a) 3214 (b) 3421 (c) 4312 (d) 4213

Which of the following triads does not follow Dobereiner's law of triads? (a) \(\mathrm{Li}, \mathrm{Na}, \mathrm{K}\) (b) \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) (c) Be, \(\mathrm{Mg}\), \(\mathrm{Ca}\) (d) \(\mathrm{Cu}, \mathrm{Ag}, \mathrm{Au}\)

An element with atomic number " 32 " belongs to (a) 4 th period, VIA group (b) 3 rd period, IVA group (c) 4 th period, IVA group (d) 5 th period, VA group

Assertion (A): First ionization energy of beryllium is greater than that of boron. Reason (R): Boron has larger size than beryllium. (a) Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is not the correct explanation of \(A\). (c) \(\mathrm{A}\) is true, \(\mathrm{R}\) is false (d) \(\mathrm{A}\) is false, \(\mathrm{R}\) is true
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

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