Chapter 4, Problem 4.28EP

Fill in the blanks in each line in the following table. The first line is already completed as an example.

Interpretation Introduction

Interpretation:

Filling of each blank in the following table has to be done:

$\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ \_& \text{13P}& \text{10e}& \_\end{array}$

Concept Introduction:

Atoms are composed of three types of particles called subatomic particles.  They are as follows:

• Protons: Positively charged particles in an atom.
• Neutrons: Neutral charged particles in an atom.
• Electrons: Negatively charged particles in an atom.

The neutral atom has equal number of protons and electrons.  Gaining or loosing of electrons of an atom forms ion.

Negative charged ions are formed by gaining one or more electrons and it has more electrons than protons.

Positive charged ions are formed by losing one or more electrons and it has more protons than electrons.

Answer to Problem 4.28EP

Complete table is shown below:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ {\text{Al}}^{\text{3+}}& \text{13P}& \text{10e}& \text{3e lost}\end{array}$

Explanation of Solution

The charge on an ion and its magnitude is equal to the number of protons minus the number of electrons.

The given ion has $\text{13}$ protons and $\text{10}$ electrons.  Aluminium element has $\text{13}$ protons, three electrons are lesser than the total number of protons.  Thus, it lost three electrons and then the charge of the aluminium ion is ${\text{Al}}^{\text{3+}}$.

Hence, the symbol of ion is ${\text{Al}}^{\text{3+}}$, $\text{13}$ has protons, $\text{10}$ electrons and lost three electrons.

Interpretation Introduction

Interpretation:

Filling of each blank in the following table has to be done:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ {\text{Ca}}^{\text{2+}}& \text{\_}& \text{\_}& \text{\_}\end{array}$

Concept Introduction:

Atoms are composed of three types of particles called subatomic particles.  They are as follows:

• Protons: Positively charged particles in an atom.
• Neutrons: Neutral charged particles in an atom.
• Electrons: Negatively charged particles in an atom.

The neutral atom has equal number of protons and electrons.  Gaining or loosing of electrons of an atom forms ion.

Negative charged ions are formed by gaining one or more electrons and it has more electrons than protons.

Positive charged ions are formed by losing one or more electrons and it has more protons than electrons.

Answer to Problem 4.28EP

Complete table is shown below:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ {\text{Ca}}^{\text{2+}}& \text{20P}& \text{18e}& \text{2e lost}\end{array}$

Explanation of Solution

The charge on an ion and its magnitude is equal to the number of protons minus the number of electrons.

The atomic number of calcium is $\text{20}$ thus the number of protons in calcium atom is $\text{20 }$ and it has charge $\text{+2}$ means it lost two electrons.

  $\begin{array}{c}\text{The number of electrons = Total numer of protons - number of electrons lost}\\ \text{= 20 - 2 = 18e}\end{array}$

Hence, the number of protons are $\text{20}$, number of electrons are $\text{18}$ , lost two electrons and the symbol of the calcium ion is ${\text{Ca}}^{\text{2+}}$.

Interpretation Introduction

Interpretation:

Filling of each blank in the following table has to be done:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ \_& \text{16P}& \_& \text{2e gained}\end{array}$

Concept Introduction:

Atoms are composed of three types of particles called subatomic particles.  They are as follows:

• Protons: Positively charged particles in an atom.
• Neutrons: Neutral charged particles in an atom.
• Electrons: Negatively charged particles in an atom.

The neutral atom has equal number of protons and electrons.  Gaining or loosing of electrons of an atom forms ion.

Negative charged ions are formed by gaining one or more electrons and it has more electrons than protons.

Positive charged ions are formed by losing one or more electrons and it has more protons than electrons.

Answer to Problem 4.28EP

Complete table is shown below:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ {\text{S}}^{\text{2-}}& \text{16P}& \text{18}& \text{2e gained}\end{array}$

Explanation of Solution

The charge on an ion and its magnitude is equal to the number of protons minus the number of electrons.

The ion has $\text{16}$ protons, the element is sulphur and it gained two electrons then the charge of the sulphur ion is ${\text{S}}^{\text{2-}}$.

  $\begin{array}{c}\text{Number of electrons = Number of protons + electrons gained}\\ \text{ = 16+2 = 18e}\end{array}$

Hence, ion is ${\text{S}}^{\text{2-}}$, it has $\text{16}$ protons, $\text{18}$ electrons and it gain two electrons.

Interpretation Introduction

Interpretation:

Filling of each blank in the following table has to be done:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ \_& \_& \text{18e}& \text{1e lost}\end{array}$

Concept Introduction:

Atoms are composed of three types of particles called subatomic particles. They are as follows:

• Protons: Positively charged particles in an atom.
• Neutrons: Neutral charged particles in an atom.
• Electrons: Negatively charged particles in an atom.

The neutral atom has equal number of protons and electrons.  Gaining or loosing of electrons of an atom form ion.

Negative charged ions are formed by gaining one or more electrons and it has more electrons than protons.

Positive charged ions are formed by losing one or more electrons and it has more protons than electrons.

Answer to Problem 4.28EP

Complete table is shown below:

  $\begin{array}{cccc}\text{Ion}& \begin{array}{l}\text{Number of }\\ \text{protons}\end{array}& \begin{array}{l}\text{Number of }\\ \text{electrons}\end{array}& \begin{array}{l}\text{Electrons }\\ \text{Lost/Gained}\end{array}\\ {\text{K}}^{\text{+}}& \text{19P}& \text{18e}& \text{1e lost}\end{array}$

Explanation of Solution

The charge on an ion and its magnitude is equal to the number of protons minus the number of electrons.

The number of electrons in ion is $\text{18}$ and it lost one electron.

  $\begin{array}{c}\text{Number of protons = Number of electrons + number of electrons lost}\\ \text{= 18+1 = 19p}\end{array}$

Potassium has $19$ protons and it lost one electron then the symbol of the ion is ${\text{K}}^{\text{+}}$.

Hence, ion is ${\text{K}}^{\text{+}}$, it has $19$ protons, $18$ electrons and it lost one electron.