Chapter 4, Problem 4.89QP

Indicate which one of the two species in each of the following pairs is smaller: (a) Cl or Cl−; (b) Na or Na⁺; (c) O²⁻ or S²⁻; (d) Mg²⁺ or Al³⁺; (e) Au or Au³⁺.

Interpretation Introduction

Interpretation:

In the given set of species which will be smaller in size has to be identified.

Concept Introduction:

• The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius.  An ion is formed by either loss or gain of electrons from its valence shell.
• As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell.  As we move across the period the ionic radius increases as the electrons are added to the same subshell.
• An anion is formed when an electron is added to the valence shell of an atom.  The anion has a net negative charge in it.  In anion the extra electron added occupies more space and maximizes the shielding.
• Anions will have larger size compared to cations.
• The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
• When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
• A cation is formed when an electron is lost by an atom from its valence shell.  The cation has a net positive charge.  In cation the shielding decreases as the electron is removed from the valence shell.
• If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
• The trend in periodic table can be described as well.  As we move down the group the ionic radius decreases as the electrons are added to a new shell.  But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.

To identify: The species which will have smaller size from the given species.

Answer to Problem 4.89QP

Answer

In (a) $\text{Cl}$ will be smaller in the given set of species

Explanation of Solution

The number of electrons and protons in the given set of species (a) is,

Species	Total number of Electrons	Total number of Protons
$\text{Cl}$	17	17
${\text{Cl}}^{\text{-}}$	18	17

The total number of electrons and protons present for the given species are found out and presented in the above table.  From this we can see that total number of protons in all the given species is same, but the total numbers of electrons are different.

$\text{Cl}$ have smaller size than ${\text{Cl}}^{\text{-}}$ because,

By comparing the total number of protons and electrons in the table given in the previous step and as comparing to the size of atom size of anion is larger. And also here the proton number is lesser than the electron in ${\text{Cl}}^{\text{-}}$ so the size of the ion will be larger due to more shielding

Interpretation Introduction

Interpretation:

In the given set of species which will be smaller in size has to be identified.

Concept Introduction:

• The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius.  An ion is formed by either loss or gain of electrons from its valence shell.
• As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell.  As we move across the period the ionic radius increases as the electrons are added to the same subshell.
• An anion is formed when an electron is added to the valence shell of an atom.  The anion has a net negative charge in it.  In anion the extra electron added occupies more space and maximizes the shielding.
• Anions will have larger size compared to cations.
• The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
• When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
• A cation is formed when an electron is lost by an atom from its valence shell.  The cation has a net positive charge.  In cation the shielding decreases as the electron is removed from the valence shell.
• If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
• The trend in periodic table can be described as well.  As we move down the group the ionic radius decreases as the electrons are added to a new shell.  But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.

To identify: The species which will have smaller size from the given species.

Answer to Problem 4.89QP

Answer

In (b) ${\text{Na}}^{\text{+}}$ will be smaller in the given set of species

Explanation of Solution

The number of electrons and protons in the given set of species (b)

Species	Total number of Electrons	Total number of Protons
$\text{Na}$	11	11
${\text{Na}}^{\text{+}}$	10	11

The total number of electrons and protons present for the given species are found out and presented in the above table.  From this we can see that total numbers of protons in all the given species are same, but the total numbers of electrons are different.

${\text{Na}}^{\text{+}}$ have smaller size than $\text{Na}$, because

By comparing the total number of protons and electrons in the table given in the previous step it is clear that the number of proton is greater than the electron in ${\text{Na}}^{\text{+}}$ so the size of ion will be smaller due to less shielding. Therefore ${\text{Na}}^{\text{+}}$ have smaller size than $\text{Na}$.

Interpretation Introduction

Interpretation:

In the given set of species which will be smaller in size has to be identified.

Concept Introduction:

• The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius.  An ion is formed by either loss or gain of electrons from its valence shell.
• As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell.  As we move across the period the ionic radius increases as the electrons are added to the same subshell.
• An anion is formed when an electron is added to the valence shell of an atom.  The anion has a net negative charge in it.  In anion the extra electron added occupies more space and maximizes the shielding.
• Anions will have larger size compared to cations.
• The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
• When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
• A cation is formed when an electron is lost by an atom from its valence shell.  The cation has a net positive charge.  In cation the shielding decreases as the electron is removed from the valence shell.
• If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
• The trend in periodic table can be described as well.  As we move down the group the ionic radius decreases as the electrons are added to a new shell.  But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.

To identify: The species which will have smaller size from the given species.

Answer to Problem 4.89QP

Answer

In (c) ${\text{O}}^{\text{2-}}$ will be smaller in the given set of species

Explanation of Solution

The number of electrons and protons in the given set of species (c)

Species	Total number of Electrons	Total number of Protons
${\text{O}}^{\text{2-}}$	8	10
${\text{S}}^{\text{2-}}$	16	18

The total number of electrons and protons present for the given species are found out and presented in the above table.

${\text{O}}^{\text{2-}}$ Will be smaller compared to ${\text{S}}^{\text{2-}}$

The given two species belong to group “6A” of periodic table.  The oxygen atom comes before the sulphur atom when we move down the periodic table.  As discussed above, when we move down the group the ionic radius increases because the electrons are added to a new subshell.  Hence, ${\text{O}}^{\text{2-}}$ will be smaller compared to ${\text{S}}^{\text{2-}}$

Interpretation Introduction

Interpretation:

In the given set of species which will be smaller in size has to be identified.

Concept Introduction:

• The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius.  An ion is formed by either loss or gain of electrons from its valence shell.
• As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell.  As we move across the period the ionic radius increases as the electrons are added to the same subshell.
• An anion is formed when an electron is added to the valence shell of an atom.  The anion has a net negative charge in it.  In anion the extra electron added occupies more space and maximizes the shielding.
• Anions will have larger size compared to cations.
• The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
• When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
• A cation is formed when an electron is lost by an atom from its valence shell.  The cation has a net positive charge.  In cation the shielding decreases as the electron is removed from the valence shell.
• If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
• The trend in periodic table can be described as well.  As we move down the group the ionic radius decreases as the electrons are added to a new shell.  But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.

To identify: The species which will have smaller size from the given species.

Answer to Problem 4.89QP

Answer

In (d) ${\text{Al}}^{\text{3+}}$ will be smaller in the given set of species

Explanation of Solution

The number of electrons and protons in the given set of species (d)

Species	Total number of Electrons	Total number of Protons
${\text{Mg}}^{\text{2+}}$	10	12
${\text{Al}}^{\text{3+}}$	10	13

The total number of electrons and protons present for the given species are found out and presented in the above table.  From this we can see that total number of electrons in all the given ions is same, but the total numbers of protons are different.

${\text{Al}}^{\text{3+}}$ have smaller size than ${\text{Mg}}^{\text{2+}}$ because,

The number of proton in ${\text{Al}}^{\text{3+}}$ is greater than the number of protons in the ${\text{Mg}}^{\text{2+}}$. When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. Therefore ${\text{Al}}^{\text{3+}}$ have small size than ${\text{Mg}}^{\text{2+}}$.

Interpretation Introduction

Interpretation:

In the given set of species which will be smaller in size has to be identified.

Concept Introduction:

• The distance between the nucleus and the valence shell of a cation or an anion is known as ionic radius.  An ion is formed by either loss or gain of electrons from its valence shell.
• As we move down the group in periodic table the ionic radius increases as the electrons are added to new subshell.  As we move across the period the ionic radius increases as the electrons are added to the same subshell.
• An anion is formed when an electron is added to the valence shell of an atom.  The anion has a net negative charge in it.  In anion the extra electron added occupies more space and maximizes the shielding.
• Anions will have larger size compared to cations.
• The reduction in the effective nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons in the nucleus is known as shielding effect
• When the proton number is greater than the electron, the size of the ion will be smaller due to less shielding. When the proton number is lesser than the electron, the size of the ion will be larger due to more shielding.
• A cation is formed when an electron is lost by an atom from its valence shell.  The cation has a net positive charge.  In cation the shielding decreases as the electron is removed from the valence shell.
• If the total number of electrons is less than the total number of protons in the ion, then the protons present can effectively attract the valence shell decreasing the size of the ion and vice-versa.
• The trend in periodic table can be described as well.  As we move down the group the ionic radius decreases as the electrons are added to a new shell.  But as we move across a period in periodic table the ionic radius increases as the electrons are added to the same subshell.

To identify: The species which will have smaller size from the given species.

Answer to Problem 4.89QP

Answer

In (e) ${\text{Au}}^{\text{3+}}$ will be smaller in the given set of species

Explanation of Solution

The number of electrons and protons in the given set of species (e)

Species	Total number of Electrons	Total number of Protons
$\text{Au}$	79	79
${\text{Au}}^{\text{3+}}$	76	79

The total number of electrons and protons present for the given species are found out and presented in the above table.  From this we can see that total number of protons in the given species is same, but the total numbers of electrons are different.

${\text{Au}}^{\text{3+}}$ have smaller size than $\text{Au}$ because,

By comparing the total number of protons and electrons in the table given in the previous step it is clear that the number of proton is greater than the electron in ${\text{Au}}^{\text{3+}}$ so the size of ion will be smaller due to less shielding. Therefore ${\text{Au}}^{\text{3+}}$ have smaller size than $\text{Au}$.