Chapter 4, Problem 4.73QP

(a)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of metal ions has to be written.

Concept Introduction:

• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions the electronic configuration are written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbitals are singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.
• Half-filled orbitals are comparatively stable as completely filled orbitals.  Therefore if there is a possibility of forming half filled orbital then the electron will be moved to the respective orbitals giving rise to more stability.
• When cation is formed it means the electrons are removed from the outermost orbital of atom.  If anion is formed means then the electrons are added to the atom in its outermost orbital.

Answer to Problem 4.73QP

Answer

The ground-state electronic configuration of (a) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$

To write: Ground-state electronic configuration of ${\text{Ni}}^{\text{2+}}$ .

Explanation of Solution

Electronic configuration of $\text{Ni}$ is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^8$

The electronic configuration of $\text{Ni}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Ni}$ is 28.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Ni}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^8$.

Electronic configuration of ${\text{Ni}}^{\text{2+}}$ is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$

The electronic configuration of ${\text{Ni}}^{\text{2+}}$ is found from the electronic configuration of $\text{Ni}$.  ${\text{Ni}}^{\text{2+}}$ is formed from $\text{Ni}$ when two valence electrons are removed from the outermost orbitals.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Ni}}^{\text{2+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^8$.

(b)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of metal ions has to be written.

Concept Introduction:

• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions the electronic configuration are written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbitals are singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.
• Half-filled orbitals are comparatively stable as completely filled orbitals.  Therefore if there is a possibility of forming half filled orbital then the electron will be moved to the respective orbitals giving rise to more stability.
• When cation is formed it means the electrons are removed from the outermost orbital of atom.  If anion is formed means then the electrons are added to the atom in its outermost orbital.

To write: Ground-state electronic configuration of ${\text{Cu}}^{\text{+}}$ .

Answer to Problem 4.73QP

Answer

The ground-state electronic configuration of (b) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$

Explanation of Solution

Electronic configuration of $\text{Cu}$ is,

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

The electronic configuration of $\text{Cu}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Cu}$ is 29.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Cu}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10}$.

Electronic configuration of ${\text{Cu}}^{\text{+}}$ is,

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

The electronic configuration of ${\text{Cu}}^{\text{+}}$ is found from the electronic configuration of $\text{Cu}$.  ${\text{Cu}}^{\text{+}}$ is formed from $\text{Cu}$ when one electron is removed from the outermost orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Cu}}^{\text{+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}$.

(c)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of metal ions has to be written.

Concept Introduction:

• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions the electronic configuration are written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbitals are singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.
• Half-filled orbitals are comparatively stable as completely filled orbitals.  Therefore if there is a possibility of forming half filled orbital then the electron will be moved to the respective orbitals giving rise to more stability.
• When cation is formed it means the electrons are removed from the outermost orbital of atom.  If anion is formed means then the electrons are added to the atom in its outermost orbital

To write: Ground-state electronic configuration of ${\text{Ag}}^{\text{+}}$ .

Answer to Problem 4.73QP

Answer

The ground-state electronic configuration of (c) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}$

Explanation of Solution

Electronic configuration of $\text{Ag}$

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

The electronic configuration of $\text{Ag}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Ag}$ is 47.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Ag}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{1}4d^{10}$.

Electronic configuration of ${\text{Ag}}^{\text{+}}$ is,

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

The electronic configuration of ${\text{Ag}}^{\text{+}}$ is found from the electronic configuration of $\text{Ag}$.  ${\text{Ag}}^{\text{+}}$ is formed from $\text{Ag}$ when one electron is removed from the outermost orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Ag}}^{\text{+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}4d^{10}$.

(d)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of metal ions has to be written.

Concept Introduction:

• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions the electronic configuration are written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbitals are singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.
• Half-filled orbitals are comparatively stable as completely filled orbitals.  Therefore if there is a possibility of forming half filled orbital then the electron will be moved to the respective orbitals giving rise to more stability.
• When cation is formed it means the electrons are removed from the outermost orbital of atom.  If anion is formed means then the electrons are added to the atom in its outermost orbital.

To write: Ground-state electronic configuration of ${\text{Au}}^{\text{+}}$ .

Answer to Problem 4.73QP

Answer

The ground-state electronic configuration of (d) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^{10}$

Explanation of Solution

Electronic configuration of $\text{Au}$ is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{1}4f^{14}5d^{10}$

The electronic configuration of $\text{Au}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Au}$ is 24.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Au}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{1}4f^{14}5d^{10}$.

Electronic configuration of ${\text{Au}}^{\text{+}}$ is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^{10}$

The electronic configuration of ${\text{Au}}^{\text{+}}$ is found from the electronic configuration of $\text{Au}$.  ${\text{Au}}^{\text{+}}$ Is formed from $\text{Au}$ when three electrons are removed from the outermost orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Au}}^{\text{+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^{10}$.

(e)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of metal ions has to be written.

Concept Introduction:

• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions the electronic configuration are written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbitals are singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.
• Half-filled orbitals are comparatively stable as completely filled orbitals.  Therefore if there is a possibility of forming half filled orbital then the electron will be moved to the respective orbitals giving rise to more stability.
• When cation is formed it means the electrons are removed from the outermost orbital of atom.  If anion is formed means then the electrons are added to the atom in its outermost orbital.

To write: Ground-state electronic configuration of ${\text{Au}}^{\text{3+}}$ .

Answer to Problem 4.73QP

Answer

The ground-state electronic configuration of (e) is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^8$

Explanation of Solution

Electronic configuration of $\text{Au}$ is,

                    $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{1}4f^{14}5d^{10}$

The electronic configuration of $\text{Au}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Au}$ is 79.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Au}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{1}4f^{14}5d^{10}$.

Electronic configuration of ${\text{Au}}^{\text{3+}}$ is,

        $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^8$

The electronic configuration of ${\text{Au}}^{\text{3+}}$ is found from the electronic configuration of $\text{Au}$.  ${\text{Au}}^{\text{3+}}$ is formed from $\text{Au}$ when two electrons are removed from the outermost orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Au}}^{\text{3+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}4f^{14}5d^8$.