Chapter 12, Problem 22A

(a)

Interpretation Introduction

Interpretation:

The formula of ionic compound formed by pair of atom is to be written.

Concept introduction:

Electronic configuration represents the placement of electrons in various shells, subshells and orbitals of a particular atom.

Answer to Problem 22A

Electronic configuration of aluminium is:

  $\left[Ne\right]3s^{2}3p^1$

Electronic configuration of sulphur is:

  $\left[Ne\right]3s^{2}3p^4$

Formula of ionic compound: $A{l}_2{S}_3$

Explanation of Solution

Aluminium has three electrons in valence shell and hence it has tendency to loss three electrons where sulphur has six electrons in valence shell and hence it has tendency to gain two electrons. Therefore two atom of aluminium will form ionic bond with three atoms of sulphur and the chemical formula is $A{l}_2{S}_3$

(b)

Interpretation Introduction

Interpretation:

The formula of ionic compound formed by pair of atom is to be written.

Concept introduction:

Electronic configuration represents the placement of electrons in various shells, subshells and orbitals of a particular atom.

Answer to Problem 22A

Electronic configuration of radium is:

  $\left[Rn\right]7s^2$

Electronic configuration of oxygen is:

  $1s^{2}2s^{2}2p^4$

Formula of ionic compound: $RaO$

Explanation of Solution

Radium has two electrons in valence shell and hence it has tendency to loss two electrons where oxygen has six electrons in valence shell and hence it has tendency to gain two electrons. Therefore one atom of radium will form ionic bond with one atom of oxygen and the chemical formula is $RaO$

(c)

Interpretation Introduction

Interpretation:

The formula of ionic compound formed by pair of atom is to be written.

Concept introduction:

Electronic configuration represents the placement of electrons in various shells, subshells and orbitals of a particular atom.

Answer to Problem 22A

Electronic configuration of calcium is:

  $\left[Ar\right]4s^2$

Electronic configuration of fluorine is:

  $1s^{2}2s^{2}2p^5$

Formula of ionic compound: $Ca{F}_2$

Explanation of Solution

Calcium has two electrons in valence shell and hence it has tendency to loss two electrons where fluorine has seven electrons in valence shell and hence it has tendency to gain one electros. Therefore one atom of calcium will form ionic bond with two atoms of fluorine and the chemical formula is $Ca{F}_2$

(d)

Interpretation Introduction

Interpretation:

The formula of ionic compound formed by pair of atom is to be written.

Concept introduction:

Electronic configuration represents the placement of electrons in various shells, subshells and orbitals of a particular atom.

Answer to Problem 22A

Electronic configuration of cesium is:

  $\left[Xe\right]6s^1$

Electronic configuration of nitrogen is:

  $1s^{2}2s^{2}2p^3$

Formula of ionic compound: $C{s}_{3}N$

Explanation of Solution

Cesium has one electron in valence shell and hence it has tendency to loss one electron where nitrogen has five electrons in valence shell and hence it has tendency to gain three electrons. Therefore three atoms of cesium will form ionic bond with one atom of nitrogen and the chemical formula is $C{s}_{3}N$

(e)

Interpretation Introduction

Interpretation:

The formula of ionic compound formed by pair of atom is to be written.

Concept introduction:

Electronic configuration represents the placement of electrons in various shells, subshells and orbitals of a particular atom.

Answer to Problem 22A

Electronic configuration of rubidium is:

  $\left[Kr\right]5s^1$

Electronic configuration of phosphorus is:

  $\left[Ne\right]3s^{2}3p^3$

Formula of ionic compound: $R{b}_{3}P$

Explanation of Solution

Rubidium has one electron in valence shell and hence it has tendency to loss one electron where phosphorus has five electrons in valence shell and hence it has tendency to gain three electrons. Therefore three atoms of rubidium will form ionic bond with one atom of phosphorus and the chemical formula is $R{b}_{3}P$