Chapter 12, Problem 40QAP

Atoms form ions so as to achieve electron configurations similar to those of the noble gases. For the following pairs of noble gas configurations, give the formulas of two simple ionic compounds that would have comparable electron configurations.

msp;  $\begin{array}{l}\text{a}.\left[\text{He}\right]\text{ and }\left[\text{Ne}\right] \text{c}.\left[\text{He}\right]\text{ and }\left[\text{Ar}\right]\\ \text{b}.\left[\text{Ne}\right]\text{ and }\left[\text{Ne}\right] \text{d}.\left[\text{Ne}\right]\text{ and }\left[\text{Ar}\right]\end{array}$

Interpretation Introduction

(a)

Interpretation:

For the given pair of noble gas configurations, the formula of two simple ionic compounds that would have comparable electron configuration is to be stated.

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 40QAP

The ionic compounds that have comparable electron configuration to the given pair of noble gas configurations are $\text{NaH}$ and $\text{LiF}$.

Explanation of Solution

The given noble gas configurations are $\left[\text{He}\right]\text{and}\left[\text{Ne}\right]$.

Ionic compound 1:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{He}$ with atomic number, $\text{Z}=2$ is $1{\text{s}}^2$. Consider hydrogen with atomic number, $\text{Z}=\text{1}$ has an electron configuration of $1{\text{s}}^1$ and if it gains $1$ electron, it will achieve the configuration of helium. Hence, the anion is ${\text{H}}^{-}$.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider sodium with atomic number, $\text{Z}=\text{11}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of neon. Hence, the cation is ${\text{Na}}^{+}$.

The ionic compound that has comparable electron configuration of the given pair of noble gas configurations is $\text{NaH}$.

Ionic compound 2:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{He}$ with atomic number, $\text{Z}=2$ is $1{\text{s}}^2$. Consider lithium with atomic number, $\text{Z}=3$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of helium. Hence, the cation is ${\text{Li}}^{+}$.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider fluorine with atomic number, $\text{Z}=\text{9}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^5$ and if it gains $1$ electron, it will achieve the configuration of neon. Hence, the anion is ${\text{F}}^{-}$.

The ionic compound that has comparable electron configuration to the given pair of noble gas configurations is $\text{LiF}$.

Interpretation Introduction

(b)

Interpretation:

For the given pair of noble gas configurations, the formula of two simple ionic compounds that would have comparable electron configuration is to be stated.

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 40QAP

The ionic compounds that have comparable electron configuration the given pair of noble gas configurations are $\text{NaF}$ and $\text{MgO}$.

Explanation of Solution

The given noble gas configurations are $\left[\text{Ne}\right]\text{and}\left[\text{Ne}\right]$.

Ionic compound 1:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider fluorine with atomic number, $\text{Z}=\text{9}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^5$ and if it gains $1$ electron, it will achieve the configuration of neon. Hence, the anion is ${\text{F}}^{-}$.

Consider sodium with atomic number, $\text{Z}=\text{11}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of neon. Hence, the cation is ${\text{Na}}^{+}$.

The ionic compound that has comparable electron configuration to the given pair of noble gas configurations is $\text{NaF}$.

Ionic compound 2:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider magnesium with atomic number, $\text{Z}=\text{12}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^2$ and if it loses $2$ electrons, it will achieve the configuration of neon. Hence, the cation is ${\text{Mg}}^{2+}$

Consider oxygen with atomic number, $\text{Z}=\text{8}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^4$ and if it gains $2$ electrons, it will achieve the configuration of neon. Hence, the anion is ${\text{O}}^{2-}$.

The ionic compound that has comparable electron configuration the given pair of noble gas configurations is $\text{MgO}$.

Interpretation Introduction

(c)

Interpretation:

For the given pair of noble gas configurations, the formula of two simple ionic compounds that would have comparable electron configuration is to be stated.

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 40QAP

The ionic compounds that have comparable electron configuration the given pair of noble gas configurations are $\text{KH}$ and $\text{LiCl}$.

Explanation of Solution

The given noble gas configurations are $\left[\text{He}\right]\text{and}\left[\text{Ar}\right]$.

Ionic compound 1:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{He}$ with atomic number, $\text{Z}=2$ is $1{\text{s}}^2$. Consider hydrogen with atomic number, $\text{Z}=\text{1}$ has an electron configuration of $1{\text{s}}^1$ and if it gains $1$ electron, it will achieve the configuration of helium. Hence, the anion is ${\text{H}}^{-}$

The electron configuration of $\text{Ar}$ with atomic number, $\text{Z}=18$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$. Consider potassium with atomic number, $\text{Z}=\text{19}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of argon. Hence, the cation is ${\text{K}}^{+}$

So, the ionic compound that have comparable electron configuration the given pair of noble gas configurations is $\text{KH}$

Ionic compound 2:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{He}$ with atomic number, $\text{Z}=2$ is $1{\text{s}}^2$. Consider lithium with atomic number, $\text{Z}=3$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of helium. Hence, the cation is ${\text{Li}}^{+}$

The electron configuration of $\text{Ar}$ with atomic number, $\text{Z}=18$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$. Consider chlorine with atomic number, $\text{Z}=17$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{5}3{\text{s}}^{2}3{\text{p}}^5$ and if it gains $1$ electron, it will achieve the configuration of argon. Hence, the anion is ${\text{Cl}}^{-}$

So, the ionic compound that have comparable electron configuration the given pair of noble gas configurations is $\text{LiCl}$.

Interpretation Introduction

(d)

Interpretation:

For the given pair of noble gas configurations, the formula of two simple ionic compounds that would have comparable electron configuration is to be stated.

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 40QAP

The ionic compounds that have comparable electron configuration the given pair of noble gas configurations are $\text{NaCl}$ and $\text{KF}$.

Explanation of Solution

The given noble gas configurations are $\left[\text{Ne}\right]\text{and}\left[\text{Ar}\right]$.

Ionic compound 1:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{Ar}$ with atomic number, $\text{Z}=18$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$. Consider chlorine with atomic number, $\text{Z}=17$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{5}3{\text{s}}^{2}3{\text{p}}^5$ and if it gains $1$ electron, it will achieve the configuration of argon. Hence, the anion is ${\text{Cl}}^{-}$.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider sodium with atomic number, $\text{Z}=\text{11}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of neon. Hence, the cation is ${\text{Na}}^{+}$.

The ionic compound that has comparable electron configuration to the given pair of noble gas configurations is $\text{NaCl}$.

Ionic compound 2:

An ionic compound consists of both cations and anions.

The electron configuration of $\text{Ar}$ with atomic number, $\text{Z}=18$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^6$. Consider potassium with atomic number, $\text{Z}=\text{19}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^{6}3{\text{s}}^{2}3{\text{p}}^{6}4{\text{s}}^1$ and if it loses $1$ electron, it will achieve the configuration of argon. Hence, the cation is ${\text{K}}^{+}$.

The electron configuration of $\text{Ne}$ with atomic number, $\text{Z}=10$ is $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^6$. Consider fluorine with atomic number, $\text{Z}=\text{9}$ has an electron configuration of $1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^5$ and if it gains $1$ electron, it will achieve the configuration of neon. Hence, the anion is ${\text{F}}^{-}$.

The ionic compound that have comparable electron configuration the given pair of noble gas configurations is $\text{KF}$.