Chapter 6, Problem 41A

(a)

Interpretation Introduction

Interpretation: The given elements are arranged in increasing order of ionization energy.

Concept introduction:

The first ionization energy is the energy needed for the removal of the outermost electron from the neutral gaseous atom.

  $X\left(g\right)+I{E}_1\to X^{+}\left(g\right)+e^{-}$

Answer to Problem 41A

  $\text{Sr}<\text{Mg}<\text{Be}$

Explanation of Solution

The ionization energy is inversely related to the atomic radius. The higher the atomic radius, the lower will be the ionization energy.

Down the group, atomic radius increases. Strontium, magnesium, and beryllium are placed in the same group (group 2). Beryllium is placed above magnesium and strontium. The magnesium is placed between beryllium and strontium. Thus, the order of atomic radii is,

  $\text{Be}<\text{Mg}<\text{Sr}$

Thus, the increasing order of ionization energy is as follows:

  $\text{Sr}<\text{Mg}<\text{Be}$

(b)

Interpretation Introduction

Interpretation: The given elements are arranged in increasing order of ionization energy.

Concept introduction:

The first ionization energy is the energy needed for the removal of the outermost electron from the neutral gaseous atom.

  $X\left(g\right)+I{E}_1\to X^{+}\left(g\right)+e^{-}$

Answer to Problem 41A

  $\text{Cs}<\text{Ba}<\text{Bi}$

Explanation of Solution

Across the period or from left to right, the first ionization energy increases because of a decrease in atomic radius. Cs is present in group 1, Ba is present in group 2, and Bi is present in group 15. So, the increasing order of ionization energy is as follows:

  $\text{Cs}<\text{Ba}<\text{Bi}$

(c)

Interpretation Introduction

Interpretation: The given elements are arranged in increasing order of ionization energy.

Concept introduction:

The first ionization energy is the energy needed for the removal of the outermost electron from the neutral gaseous atom.

  $X\left(g\right)+I{E}_1\to X^{+}\left(g\right)+e^{-}$

Answer to Problem 41A

  $\text{Na}<\text{Al}<\text{S}$

Explanation of Solution

Across the period or from left to right, the first ionization energy increases because of a decrease in atomic radius. Na is present in group 1, Al is present in group 13, and S is present in group 16. So, the increasing order of ionization energy is as follows:

  $\text{Na}<\text{Al}<\text{S}$