Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 38

Which of the following elements acts as the best reducing agent? (a) \(\mathrm{Na}\) (b) \(\mathrm{Cl}\) (c) \(\mathrm{Mg}\) (d) \(\mathrm{F}\)

Short Answer

Expert verified
Answer: (a) Na (Sodium)

Step by step solution

01

Determine the electron loss tendencies

We need to look at the periodic table and see where these elements are placed. This way, we can get an idea of their willingness to lose or gain electrons and act as reducing agents. The elements are as follows: (a) Na (Sodium) - Group 1, Alkali metal (b) Cl (Chlorine) - Group 17, Halogen (c) Mg (Magnesium) - Group 2, Alkaline Earth metal (d) F (Fluorine) - Group 17, Halogen
02

Identify the characteristics of the groups

In general, elements in Group 1 and Group 2 have a greater tendency to lose electrons compared to other groups due to their low ionization energies, making them good reducing agents. On the other hand, elements from Group 17 have high electron affinity and gain electrons easily, making them poor reducing agents.
03

Compare elements within the same group

Since Na and Mg belong to Group 1 and Group 2 respectively, we now need to compare their electron loss tendencies. In general, elements in Group 1 have a higher tendency to lose electrons (greater reducing ability) compared to Group 2 elements.
04

Determine the best reducing agent

Based on the comparison of elements and their electron loss tendencies from their respective groups, we can infer that \(\mathrm{Na}\) (Sodium) from Group 1 has the highest tendency to lose electrons and hence, acts as the best reducing agent among the given elements. So, the correct answer is (a) \(\mathrm{Na}\).

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

An element with atomic number " \(32 "\) "belongs to (a) \(4^{\text {th }}\) period, VIA group or \(16^{\text {th }}\) group (b) \(3^{\text {rd }}\) period, IVA group or \(14^{\text {th }}\) group (c) \(4^{\text {th }}\) period, IVA group or \(14^{\text {th }}\) group (d) \(5^{\text {th }}\) period, VA group or \(15^{\text {th }}\) group

Fill in the blanks. The ascending order of the first ionization potential of \(\mathrm{C}, \mathrm{N}, \mathrm{O}\) and \(\mathrm{F}\) is ____.

In a period, from the left to right the electron affinity increases, but alkaline earth metals have lower electron affinity than alkali metals because (a) alkaline earth metals have lesser atomic radius than alkali metals. (b) alkaline earth have higher electronegativity than alkali metals. (c) alkaline earth metals have completely filled "s-" orbitals. (d) alkaline earth metals have lesser electronegativity than alkali metals.

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

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

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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