Chapter 9, Problem 9.20P

(a)

Interpretation Introduction

Interpretation:

The condensed electron configurations and Lewis symbols to depict the formation of ions formed from atoms $\text{Ba}$ and $\text{Cl}$ is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

Answer to Problem 9.20P

The condensed electronic configuration of ${\text{Ba}}^{2+}$ is $\left[\text{Xe}\right]$.

The condensed electronic configuration of ${\text{Cl}}^{-}$ is $\left[\text{Ne}\right]3s^{2}3p^6$.

The Lewis orbital diagram is as follows:

The formula of the compound formed is ${\text{BaCl}}_2$.

Explanation of Solution

The condensed electronic configuration of a barium atom $\left(\text{Ba}\right)$ is $\left[\text{Xe}\right]6s^2$.

The condensed electronic configuration of the chlorine atom $\left(\text{Cl}\right)$ is $\left[\text{Ne}\right]3s^{2}3p^5$.

Barium atom loses two electrons to form ${\text{Ba}}^{2+}$ and to attain noble gas configuration. The two electrons are gained by two chlorine atoms to form two ${\text{Cl}}^{-}$ ions and attain noble gas configuration.

The condensed electronic configuration of ${\text{Ba}}^{2+}$ is $\left[\text{Xe}\right]$.

The condensed electronic configuration of ${\text{Cl}}^{-}$ is $\left[\text{Ne}\right]3s^{2}3p^6$.

Barium atom loses two electrons to form ${\text{Ba}}^{2+}$ and then react with two ${\text{Cl}}^{-}$ ions formed from $\text{Cl}$ to form barium chloride $\left({\text{BaCl}}_2\right)$.

The Lewis orbital diagram is as follows:

Conclusion

$\text{Ba}$ belongs to Group 2A(2) and it loses two electrons to form ${\text{Ba}}^{2+}$. $\text{Cl}$ belongs to Group 7A(17) and it gains one electron to form ${\text{Cl}}^{-}$.

(b)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms $\text{Sr}$ and $\text{O}$ is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

Answer to Problem 9.20P

The condensed electronic configuration of ${\text{Sr}}^{2+}$ is $\left[\text{Kr}\right]$.

The condensed electronic configuration of ${\text{O}}^{\text{2}}{}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

The Lewis orbital diagram is as follows:

The formula of the compound formed is $\text{SrO}$.

Explanation of Solution

The condensed electronic configuration of a strontium atom $\left(\text{Sr}\right)$ is $\left[\text{Kr}\right]5s^2$.

The condensed electronic configuration of oxygen atom $\left(\text{O}\right)$ is $\left[\text{He}\right]2s^{2}2p^4$.

Strontium atom loses two electrons to form ${\text{Sr}}^{2+}$ and to attain noble gas configuration. The two electrons are gained by the oxygen atoms to form ${\text{O}}^{2-}$ ions and to attain noble gas configuration.

The condensed electronic configuration of ${\text{Sr}}^{2+}$ is $\left[\text{Kr}\right]$.

The condensed electronic configuration of ${\text{O}}^{\text{2}}{}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

Strontium atom loses two electrons to form ${\text{Sr}}^{2+}$ and then react with ${\text{O}}^{\text{2}}{}^{-}$ ions formed from $\text{O}$ to form strontium oxide $\left(\text{SrO}\right)$.

The Lewis orbital diagram is as follows:

Conclusion

$\text{Sr}$ belongs to Group 2A(2) and it loses two electrons to form ${\text{Sr}}^{2+}$. $\text{O}$ belongs to Group 6A(16) and it gains two electrons to form ${\text{O}}^{\text{2}}{}^{-}$.

(c)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms $\text{Al}$ and $\text{F}$ is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

Answer to Problem 9.20P

The condensed electronic configuration of ${\text{Al}}^{3+}$ is $\left[\text{Ne}\right]$.

The condensed electronic configuration of ${\text{F}}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

The Lewis orbital diagram is as follows:

The formula of the compound formed is ${\text{AlF}}_3$.

Explanation of Solution

The condensed electronic configuration of an aluminium atom $\left(\text{Al}\right)$ is $\left[\text{Ne}\right]3s^{2}3p^1$.

The condensed electronic configuration of the fluorine atom $\left(\text{F}\right)$ is $\left[\text{He}\right]2s^{2}2p^5$.

Aluminium atom loses three electrons to form ${\text{Al}}^{3+}$ and to attain noble gas configuration. The three electrons are gained by the three fluorine atoms to form ${\text{F}}^{-}$ ions and to attain noble gas configuration.

The condensed electronic configuration of ${\text{Al}}^{3+}$ is $\left[\text{Ne}\right]$.

The condensed electronic configuration of ${\text{F}}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

Aluminium atom loses three electrons to form ${\text{Al}}^{3+}$ and then react with three ${\text{F}}^{-}$ ions formed from $\text{F}$ to form aluminium fluoride $\left({\text{AlF}}_3\right)$.

The Lewis orbital diagram is as follows:

Conclusion

$\text{Al}$ belongs to Group 3A(13) and it loses three electrons to form ${\text{Al}}^{3+}$. $\text{F}$ belongs to Group 7A(17) and it gains one electron to form ${\text{F}}^{-}$.

(d)

Interpretation Introduction

Interpretation:

The condensed electron configurations, partial orbital diagrams, and Lewis symbols to depict the formation of ions formed from atoms $\text{Rb}$ and $\text{O}$ is to be determined. Also, the formula of the compound formed is to be given.

Concept introduction:

The electronic configuration tells about the distribution of electrons in various atomic orbitals. The condensed electronic configuration is a way to write the electronic configuration where the inner shell configurations are compressed to the nearest noble gas configuration and only the valence shell configuration is written in the expanded form.

The partial orbital diagram is a pictorial representation of the electrons present in an orbital. Each orbital can occupy only two electrons of opposite spin.

Lewis electron-dot symbol is a representation employed to donate the valence electron present in the atom. It includes atom symbol to represent inner electrons and nucleus and the dots represent the valence present in the atom.

Answer to Problem 9.20P

The condensed electronic configuration of ${\text{Rb}}^{+}$ is $\left[\text{Kr}\right]$.

The condensed electronic configuration of ${\text{O}}^{\text{2}}{}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

The Lewis orbital diagram is as follows:

The formula of the compound formed is ${\text{Rb}}_2\text{O}$.

Explanation of Solution

The condensed electronic configuration of a rubidium atom $\left(\text{Rb}\right)$ is $\left[\text{Kr}\right]5s^1$.

The condensed electronic configuration of oxygen atom $\left(\text{O}\right)$ is $\left[\text{He}\right]2s^{2}2p^4$.

Two rubidium atoms lose one electron respectively to form ${\text{Rb}}^{+}$ and to attain noble gas configuration. The two electrons lost by $\text{Rb}$ atoms are gained by oxygen atoms to form ${\text{O}}^{2-}$ ions and to attain noble gas configuration.

The condensed electronic configuration of ${\text{Rb}}^{+}$ is $\left[\text{Kr}\right]$.

The condensed electronic configuration of ${\text{O}}^{\text{2}}{}^{-}$ is $\left[\text{He}\right]2s^{2}2p^6$.

Two rubidium atoms lose one electron respectively to form ${\text{Rb}}^{+}$ and then react with ${\text{O}}^{\text{2}}{}^{-}$ ions formed from $\text{O}$ to form rubidium oxide $\left({\text{Rb}}_2\text{O}\right)$.

The Lewis orbital diagram is as follows:

Conclusion

$\text{Rb}$ belongs to Group 1A(1) and it loses one electron to form ${\text{Rb}}^{+}$. $\text{O}$ belongs to Group 6A(16) and it gains two electrons to form ${\text{O}}^{\text{2}}{}^{-}$.