Chapter 8, Problem 49E

Give the formula of a negative ion that would have the same number of electrons as each of the following positive ions.

a. Na⁺

b. Ca²⁺

c. Al³⁺

d. Rb⁺

Interpretation Introduction

Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.

Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.

To determine: The formula of negative ions that have the same number of electrons as in ${\text{Na}}^{+}$.

Answer to Problem 49E

The formula of negative ions is ${\text{F}}^{-},{\text{O}}^{2-}$ and ${\text{N}}^{3-}$.

Explanation of Solution

The electronic configuration of ${\text{Na}}^{+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

It contains 10 electrons.

The atomic number of oxygen $\left(\text{O}\right)$ is 8 which means it has 8 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

It gains two electrons to get 10 electrons and form ${\text{O}}^{2-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

The atomic number of fluorine $\left(\text{F}\right)$ is 9 which means it has 9 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

It gains an electron to get 10 electrons and form ${\text{F}}^{-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

The atomic number of nitrogen $\left(\text{N}\right)$ is 7 which means it has 7 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

It gains three electrons to get 10 electrons and form ${\text{N}}^{3-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

Interpretation Introduction

Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.

Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.

To determine: The formula of negative ions that have the same number of electrons as in ${\text{Ca}}^{2+}$.

Answer to Problem 49E

The formula of negative ions is ${\text{Cl}}^{-},{\text{S}}^{2-}\text{and}{\text{P}}^{3-}$.

Explanation of Solution

The electronic configuration of ${\text{Ca}}^{2+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

It contains 18 electrons.

The atomic number of chlorine $\left(\text{Cl}\right)$ is 17 which means it has 17 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^5$

It gains an electron to get 18 electrons and form ${\text{Cl}}^{-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

The atomic number of sulfur $\left(\text{S}\right)$ is 16 which means it has 16 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^4$

It gains two electrons to get 18 electrons and form ${\text{S}}^{2-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

The atomic number of phosphorous $\left(\text{P}\right)$ is 15 which means it has 15 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^3$

It gains three electrons to get 18 electrons and form ${\text{P}}^{3-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}$

Interpretation Introduction

Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.

Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.

To determine: The formula of negative ions that have the same number of electrons as in ${\text{Al}}^{3+}$.

Answer to Problem 49E

The formula of negative ions is ${\text{F}}^{-},{\text{O}}^{2-}\text{and}{\text{N}}^{3-}$.

Explanation of Solution

The electronic configuration of ${\text{Al}}^{3+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^1$

It contains 10 electrons.

The atomic number of fluorine $\left(\text{F}\right)$ is 9 which means it has 9 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

It gains an electron to get 10 electrons and form ${\text{F}}^{-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

The atomic number of oxygen $\left(\text{O}\right)$ is 8 which means it has 8 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4$

It gains two electrons to get 10 electrons and form ${\text{O}}^{2-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

The atomic number of nitrogen $\left(\text{N}\right)$ is 7 which means it has 7 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5$

It gains three electrons to get 10 electrons and form ${\text{N}}^{3-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}$

Interpretation Introduction

Interpretation: The formula of negative ions that have the same number of electrons as each of the given positive ions is to be written.

Concept introduction: The ionic compound can be separated into ions. Ions are of two types. Positively charged ions are called cation and negatively charged ions are called anions.

To determine: The formula of negative ions that have the same number of electrons as in ${\text{Rb}}^{+}$.

Answer to Problem 49E

The formula of negative ions is ${\text{Br}}^{-},{\text{Se}}^{2-}\text{and}{\text{As}}^{3-}$.

Explanation of Solution

The electronic configuration of ${\text{Rb}}^{+}$ is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$

It contains 36 electrons.

The atomic number of bromine $\left(\text{Br}\right)$ is 35 which means it has 35 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^5$

It gains an electron to get 36 electrons and form ${\text{Br}}^{-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$

The atomic number of selenium $\left(\text{Se}\right)$ is 34 which means it has 34 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{4}}$

It gains two electrons to get 36 electrons and form ${\text{Se}}^{2-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$

The atomic number of arsenic $\left(\text{As}\right)$ is 33 which means it has 33 electrons and its electronic configuration is,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^3$

It gains three electrons to get 36 electrons and form ${\text{As}}^{3-}$ and its electronic configuration becomes,

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}$