Chapter 12, Problem 39QAP

Name the noble gas atom that has the same electron configuration as each of the ions in the following compounds.

msp;  $\begin{array}{l}\text{a}.\text{ barium sulfide},\text{ BaS}\\ \text{b}.\text{ strontium fluoride},{\text{ SrF}}_{\text{2}}\\ \text{c}.\text{ magnesium oxide},\text{ MgO}\\ \text{d}.\text{ aluminum sulfide},{\text{ Al}}_{\text{2}}{\text{S}}_{\text{3}}\end{array}$

Interpretation Introduction

(a)

Interpretation:

The noble gas atom that has the same configuration as each of the ions in the given compound are to be predicted.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron in an orbital is given by the electron configuration of that atom.

Atoms lose or gain electrons to become stable by attaining nearest noble gas configuration. While doing so, they are converted to their respective ions. The positive ion and the negative ion combine to form their corresponding salt.

Answer to Problem 39QAP

${\text{Ba}}^{2+}$ has a similar electron configuration as that of xenon and ${\text{S}}^{2-}$ has a similar electron configuration as that of argon.

Explanation of Solution

The salt given is $\text{BaS}$. It consists of one cation and one anion.

The cation present in $\text{BaS}$ is ${\text{Ba}}^{2+}$.

The electron configuration of $\text{Ba}$ with $\text{Z}=56$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6 3{\text{d}}^{10} 4{\text{s}}^2 4{\text{p}}^6 4{\text{d}}^{10} 5{\text{s}}^2 5{\text{p}}^6 6{\text{s}}^2$.

Barium present in $\text{BaS}$ is in the form of ${\text{Ba}}^{2+}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6 3{\text{d}}^{10} 4{\text{s}}^2 4{\text{p}}^6 4{\text{d}}^{10} 5{\text{s}}^2 5{\text{p}}^6$. This electron configuration corresponds to the electron configuration of xenon. Therefore, the ion, ${\text{Ba}}^{2+}$ present in $\text{BaS}$ has a similar electron configuration of xenon.

The anion present in $\text{BaS}$ is ${\text{S}}^{2-}$.

The electron configuration of $\text{S}$ with $\text{Z}=16$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^4$.

Sulfur present in $\text{BaS}$ is in the form of ${\text{S}}^{2-}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6$. This electron configuration corresponds to the electron configuration of argon. Therefore, the ion, ${\text{S}}^{2-}$ present in $\text{BaS}$ has a similar electron configuration of argon.

Interpretation Introduction

(b)

Interpretation:

The noble gas atom that has the same configuration as each of the ions in the given compound are to be predicted.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron in an orbital is given by the electron configuration of that atom.

Atoms lose or gain electrons to become stable by attaining nearest noble gas configuration. While doing so, they are converted to their respective ions. The positive ion and the negative ion combine to form their corresponding salt.

Answer to Problem 39QAP

${\text{Sr}}^{2+}$ has a similar electron configuration of krypton and ${\text{F}}^{-}$ has a similar electron configuration of neon.

Explanation of Solution

The salt given is ${\text{SrF}}_2$. It consists of one cation and one anion.

The cation present in ${\text{SrF}}_2$ is ${\text{Sr}}^{2+}$.

The electron configuration of $\text{Sr}$ with $\text{Z}=38$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6 3{\text{d}}^{10} 4{\text{s}}^2 4{\text{p}}^6 5{\text{s}}^2$.

Strontium present in ${\text{SrF}}_2$ is in the form of ${\text{Sr}}^{2+}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6 3{\text{d}}^{10} 4{\text{s}}^2 4{\text{p}}^6$. This electron configuration corresponds to the electron configuration of krypton. Therefore, the ion, ${\text{Sr}}^{2+}$ present in ${\text{SrF}}_2$ has a similar electron configuration of krypton.

The anion present in ${\text{SrF}}_2$ is ${\text{F}}^{-}$.

The electron configuration of $\text{F}$ with $\text{Z}=9$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^5$.

Fluorine present in ${\text{SrF}}_2$ is in the form of ${\text{F}}^{-}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6$. This electron configuration corresponds to the electron configuration of neon. Therefore, the ion, ${\text{F}}^{-}$ present in ${\text{SrF}}_2$ has a similar electron configuration of neon.

Interpretation Introduction

(c)

Interpretation:

The noble gas atom that the has same configuration as each of the ions in the given compound are to be predicted.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron in an orbital is given by the electron configuration of that atom.

Atoms lose or gain electrons to become stable by attaining nearest noble gas configuration. While doing so, they are converted to their respective ions. The positive ion and the negative ion combine to form their corresponding salt.

Answer to Problem 39QAP

${\text{Mg}}^{2+}$ has a similar electron configuration of neon and ${\text{O}}^{2-}$ has a similar electron configuration of neon.

Explanation of Solution

The salt given is $\text{MgO}$. It consists of one cation and one anion.

The cation present in $\text{MgO}$ is ${\text{Mg}}^{2+}$.

The electron configuration of $\text{Mg}$ with $\text{Z}=12$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2$.

Magnesium present in $\text{MgO}$ is in the form of ${\text{Mg}}^{2+}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6$. This electron configuration corresponds to the electron configuration of neon. Therefore, the ion, ${\text{Mg}}^{2+}$ present in $\text{MgO}$ has a similar electron configuration of neon.

The anion present in $\text{MgO}$ is ${\text{O}}^{2-}$.

The electron configuration of $\text{O}$ with $\text{Z}=8$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^4$.

Oxygen present in $\text{MgO}$ is in the form of ${\text{O}}^{2-}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6$. This electron configuration corresponds to the electron configuration of neon. Therefore, the ion, ${\text{O}}^{2-}$ present in $\text{MgO}$ has a similar electron configuration of neon.

Interpretation Introduction

(d)

Interpretation:

The noble gas atom that has the same configuration as each of the ions in the given compound are to be predicted.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron in an orbital is given by the electron configuration of that atom.

Atoms lose or gain electrons to become stable by attaining nearest noble gas configuration. While doing so, they are converted to their respective ions. The positive ion and the negative ion combine to form their corresponding salt.

Answer to Problem 39QAP

${\text{Al}}^{3+}$ has a similar electron configuration of neon and ${\text{S}}^{2-}$ has a similar electron configuration of argon.

Explanation of Solution

The salt given is ${\text{Al}}_{\text{2}}{\text{S}}_3$. It consists of one cation and one anion.

The cation present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ is ${\text{Al}}^{3+}$.

The electron configuration of $\text{Al}$ with $\text{Z}=13$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^1$.

Aluminum present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ is in the form of ${\text{Al}}^{3+}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6$. This electron configuration corresponds to the electron configuration of neon. Therefore, the ion, ${\text{Al}}^{3+}$ present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ has a similar electron configuration of neon.

The anion present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ is ${\text{S}}^{2-}$.

The electron configuration of $\text{S}$ with $\text{Z}=16$ is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^4$.

Sulfur present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ is in the form of ${\text{S}}^{2-}$ and the electron configuration of the ion is $1{\text{s}}^{2}2{\text{s}}^2 2{\text{p}}^6 3{\text{s}}^2 3{\text{p}}^6$. This electron configuration corresponds to the electron configuration of argon. Therefore, the ion, ${\text{S}}^{2-}$ present in ${\text{Al}}_{\text{2}}{\text{S}}_3$ has a similar electron configuration of argon.