Chapter 4, Problem 4.69QP

(a)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Rb}$

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

The electronic configuration of $\text{Rb}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Rb}$ is 37.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Rb}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^1$

Electronic configuration of ${\text{Rb}}^{\text{+}}$

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

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

(b)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Sr}$

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

The electronic configuration of $\text{Sr}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Sr}$ is 38.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Sr}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^2$

Electronic configuration of ${\text{Sr}}^{\text{2+}}$

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

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

(c)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Sn}$

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

The electronic configuration of $\text{Sn}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Sn}$ is 50.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Sn}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^2$

Electronic configuration of ${\text{Sn}}^{\text{2+}}$

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

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

(d)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Te}$

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

The electronic configuration of $\text{Te}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Te}$ is 52.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Te}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^4$

Electronic configuration of ${\text{Te}}^{\text{2-}}$

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

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

(e)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Ba}$

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

The electronic configuration of $\text{Ba}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Ba}$ is 56.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Ba}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^2$

Electronic configuration of ${\text{Ba}}^{\text{2+}}$

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

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

(f)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

The ground-state electronic configuration of (f) 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 $\text{In}$

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

The electronic configuration of $\text{In}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{In}$ is 49.  According to Pauli Exclusion Principle, the electronic configuration of $\text{In}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^1$

Electronic configuration of ${\text{In}}^{\text{3+}}$

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

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

(g)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

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

Explanation of Solution

Electronic configuration of $\text{Tl}$

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

The electronic configuration of $\text{Tl}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Tl}$ is 81.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Tl}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^1$

Electronic configuration of ${\text{Tl}}^{\text{+}}$

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

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

(h)

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of 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 is 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 ions are formed from the atoms the electrons are added or removed from the outermost orbital.

To write: Ground-state electronic configuration for the given ions,

Answer to Problem 4.69QP

Answer

The ground-state electronic configuration of (h) 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 $\text{Tl}$

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

The electronic configuration of $\text{Tl}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Tl}$ is 81.  According to Pauli Exclusion Principle, the electronic configuration of $\text{Se}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^{10}6p^1$

Electronic configuration of ${\text{Tl}}^{\text{3+}}$

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

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