Chapter 6, Problem 64A

Interpretation Introduction

(a)

Interpretation:

The element having the larger ionization energy among Li and N needs to be determined.

Concept introduction:

Ionization energy can be defined as the amount of energy needed to remove an electron form an atom or an ion.

Answer to Problem 64A

N has the largest ionization energy.

Explanation of Solution

The elements with the smaller ionic radius or the elements in the further right in the period have the larger ionization energy.

N has the atomic number of 7 and the electronic configuration [He] 2s²2p³. N is present in Group 15 (A) i.e. in p-block. And Li has an atomic number of 3 and electronic configuration [He] 2s² located in group 1i.e. in s-block.

Therefore, as N is placed right in comparison with Li, Hence, N has the larger ionization energy.

(b)

Interpretation Introduction

Interpretation:

The element that has the larger ionization energy among Kr and Ne needs to be determined.

Concept introduction:

Ionization energy can be defined as the amount of energy needed to remove an electron form an atom or an ion.

Answer to Problem 64A

Ne has the larger ionization energy.

Explanation of Solution

Kr has the atomic number of 36 and the electronic configuration [Ar] 3d¹⁰ 4s²4p⁶. Kr is present in Group 18 i.e. in p-block. And Ne has an atomic number of 10 and electronic configuration [He] 2s²4p⁶, located in group 18i.e. in p-block..

In the periodic table, the size of elements increases and ionization energy decreases down the group.

Due to this, Ne has a smaller size since its electrons are held closer to the nucleus of the atom than that of Kr. Hence, Ne has a larger ionization energy.

(c)

Interpretation Introduction

Interpretation:

The element having the larger ionization energy among Cs and Li needs to be determined.

Concept introduction:

Ionization energy can be defined as the amount of energy needed to remove an electron form an atom or an ion.

Answer to Problem 64A

Li has the larger ionization energy.

Explanation of Solution

Cs has the atomic number of 55 and the electronic configuration [Xe] 6s¹. Ce is present in Group 1 i.e. in s-block. And Li has an atomic number of 3 and electronic configuration [He] 2s²located in group 1 i.e. in s-block.

In the periodic table, the size of elements increases and ionization energy decreases down the group. Due to this, Li has a smaller size since its electrons are held closer to the nucleus of the atom than that of Cs. Hence, Li has a larger ionization energy.