Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 11

Which of the following statements is true regarding sodium and chlorine? (A) Sodium has greater electronegativity and a larger first ionization energy. (B) Sodium has a larger first ionization energy and a larger atomic radius. (C) Chlorine has a larger atomic radius and a greater electronegativity. (D) Chlorine has greater electronegativity and a larger first ionization energy.

Short Answer

Expert verified
The correct answer is D) Chlorine has greater electronegativity and a larger first ionization energy.

Step by step solution

01

Analysis of Properties

First of all, it is necessary to understand the meaning of three basic properties of elements in the periodic table: atomic radius, electronegativity, and ionization energy. The atomic radius is the distance from the center of an atom's nucleus to its outermost electron. Electronegativity is a measure of the force of an atom's attraction for the electrons it shares in a chemical bond with another atom. The first ionization energy is the energy required to remove the most loosely bound electron from a neutral atom.
02

Trends in the Periodic Table

Moving from left to right across a period in the periodic table, the atomic radius decreases, the electronegativity increases, and the ionization energy increases. When moving down a group in the periodic table, the atomic radius increases, while the electronegativity and ionization energy decrease.
03

Analyzing the Options Given

According to the trends observed, Sodium, sitting to the left of Chlorine in the periodic table, should have a larger atomic radius and lower ionization energy and electronegativity. Therefore, amongst the given options, only option D, which says 'Chlorine has greater electronegativity and a larger first ionization energy' aligns correctly with the properties.

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

If the solubility of \(\mathrm{BaF}_{2}\), is equal to \(x,\) which of the following expressions is equal to the solubility product, \(K_{s p}\) , for \(\operatorname{BaF}_{2} ?\) (A) \(x^{2}\) (B) 2\(x^{2}\) (C) 2\(x^{3}\) (D) 4\(x^{3}\)

$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following best describes the activity in the salt bridge as the reaction progresses? (A) Electrons flow through the salt bridge from the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell. (B) \(\quad \mathrm{Pb}^{2+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell. (C) \(\mathrm{Na}^{+}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half-cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half- cell. (D) \(\quad \mathrm{Na}^{+}\) flows to the \(\mathrm{Pb} / \mathrm{Pb}^{2}\) half- cell, and \(\mathrm{Cl}^{-}\) flows to the \(\mathrm{Cr} / \mathrm{Cr}^{3+}\) half- cell.

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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