Chapter 7, Problem 63A

(a)

Interpretation Introduction

Interpretation: To determine the electronic configuration of nitride anion.

Concept Introduction: The placement of electrons inside orbitals around a nuclear nucleus is known as electronic configuration, also known as electronic structure as well as electron configuration.

Answer to Problem 63A

The electronic configuration of the nitride ion will be ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

Explanation of Solution

• The atomic number of nitrogen atom = $7$ .
• fter gaining three electrons, the number of electrons in the nitrogen atom = $7+3=10$
• The electronic configuration of nitride ion = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

(b)

Interpretation Introduction

Interpretation: To determine the electronic configuration of oxide anion.

Concept Introduction: The placement of electrons inside orbitals around a nuclear nucleus is known as electronic configuration, also known as electronic structure as well as electron configuration.

Answer to Problem 63A

The electronic configuration of oxide ion ( ${\text{O}}^{\text{2-}}$ ) will be ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

Explanation of Solution

• The atomic number of O(oxygen) atom = 8
• After losing two electrons, the number of electrons in O (oxygen) = $8+2=10$
• The electronic configuration of oxide ion ( ${\text{O}}^{\text{2-}}$ ) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

(c)

Interpretation Introduction

Interpretation: To determine the electronic configuration of fluoride ions.

Concept Introduction: The placement of electrons inside orbitals around a nuclear nucleus is known as electronic configuration, also known as electronic structure as well as electron configuration.

Answer to Problem 63A

The electronic configuration of fluoride ion will be ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

Explanation of Solution

• The atomic number of F(fluorine) atom =9.
• After losing one electron, the number of electrons in the fluorine atom = $9+1=10$
• The electronic configuration of fluoride ions. = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

(d)

Interpretation Introduction

Interpretation: To determine the electronic configuration of fluoride ions.

Concept Introduction: The placement of electrons inside orbitals around a nuclear nucleus is known as electronic configuration, also known as electronic structure as well as electron configuration.

Answer to Problem 63A

The electronic configuration of Ne atom will be ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .

Explanation of Solution

• The atomic number of Neon (Ne) atom =10.
• The electronic configuration of Ne atom = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$ .