Chapter 4, Problem 4.84QP

Identify the atomic ground-state electron configurations that do not exist. For those that do exist, identify the element.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration exist for (a) and the element is identified as vanadium.

Explanation of Solution

Ground-state electronic configuration of the given element (a) is,

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete argon configuration along with two electrons in 4s orbital and three electrons in 3d subshell.  This is derived as shown above.  As the 3d subshell is filled singly, Hund’s rule is followed.  Hence this atomic ground-state configuration exists.

Element with the given atomic ground-state configuration is

Argon is a noble gas and has a complete octet electronic configuration as $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.  After the noble gas configuration, the given atomic ground-state configuration has two electrons in 4s orbital and three electrons in 3d subshell.  The filling of orbital in the subshell is done by following Hund’s rule and it complies with it.  The total number of electrons present in the given atomic ground-state configuration is 23.  The element with the atomic number 23 is identified as Vanadium.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration does not exist for (b).

Explanation of Solution

Ground-state electronic configuration of (b)

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete neon configuration along with two electrons in 3s subshell and two electrons in 3p subshell.  This is derived as shown above.  From the above drawn configuration we can find that the all the orbitals in 3p subshell is not singly filled before pairing.  As this does not comply with Hund’s rule, the given atomic ground-state configuration does not exist.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration does not exist for (c).

Explanation of Solution

Ground-state electronic configuration of the given element (c)

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete krypton configuration along with two electrons in “s” subshell and three electrons in “p” subshell.  This is derived as shown above.  From the above drawn configuration we can find that the filling of orbitals does not take place according to the increasing order of energy.  After filling of 5s orbital, 4d orbital has to be filled first before 5p orbital is getting filled.  But in this case 5p subshell is filled with electrons.  It does not comply with Hund’s rule and hence the given atomic ground-state configuration does not exist.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration exists for (d) and the element is identified as phosphorous.

Explanation of Solution

Ground-state electronic configuration of the given element (d) is,

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete argon configuration along with two electrons in 4s orbital and three electrons in 3d orbital.  This is derived as shown above.  As the 3d subshell is filled singly, Hund’s rule is followed.  Hence this atomic ground-state configuration exists.

Element with the given atomic ground-state configuration

Neon is a noble gas and has a complete octet electronic configuration as $1s^{2}2s^{2}2p^6$.  After the noble gas configuration, the given atomic ground-state configuration has two electrons in 3s orbital and three electrons in 3p subshell.  The filling of orbital in the subshell is done by following Hund’s rule and it complies with it.  The total number of electrons present in the given atomic ground-state configuration is 15.  The element with the atomic number 15 is identified as Phosphorous.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration does not exist for (e).

Explanation of Solution

Electronic configuration of given element (e) is,

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete argon configuration along with two electrons in 4s orbital and four electrons in 3d subshell.  This is derived as shown above.  As the 3d subshell is filled singly, Hund’s rule is followed.  Hence this configuration exists.

Atomic ground-state configuration

Argon is a noble gas and has a complete octet electronic configuration as $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.  After the noble gas configuration, the given atomic ground-state configuration has two electrons in 4s orbital and four electrons in 3d subshell.  The filling of orbital in the subshell is done by following Hund’s rule and it complies with it, that is shown above.  But in ground-state the filled subshell is less stable compared to the half filled subshell as highlighted in red colour.  Eventhough Hund’s rule is followed in filling of orbitals, the given atomic ground-state configuration in the problem statement does not exist because of the less stability of incompletely filled orbital compared to half filled orbitals.

Interpretation Introduction

Interpretation: The atomic ground-state configuration that do not exist for the given sets need to be identified and the elements for atomic ground-state configuration that exists need to be identified.

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.  The electrons are filled from lowest energy orbital first.
• 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 instead of being incompletely filled orbital.
• For simpler representation of ions or atoms, the electronic configuration of the completed octet noble gas configuration is considered and the remaining orbital alone is shown explicitly.  The ground-state configuration of the noble gases are given below,

$\begin{array}{l}\text{[Ne] = }1s^2 2s^2 2p^6\\ \\ \text{[Ar] = }1s^2 2s^2 2p^6 3s^2 3p^6\\ \\ \text{[Kr] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\\ \\ \text{[Xe] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6\\ \\ \text{[Rn] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6\\ \\ \text{[Og] = }1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}5p^6 6s^2 4f^{14} 5d^{10} 6p^6 7s^2 5f^{14} 6d^{10} 7p^6\end{array}$

To identify: The given atomic ground-state configuration that do not exist and if it exists the element with the atomic ground-state configuration to be identified.

Answer to Problem 4.84QP

Answer

Atomic ground-state configuration exists for (f) and the element is identified as silver.

Explanation of Solution

Ground-state electronic configuration of given element (f) is,

The atomic ground-state configuration is given in the problem statement.  From this we can find that this has a complete Krypton configuration along with one electrons in 5s orbital and ten electrons in 4d subshell.  This is derived as shown above.  As the filling of subshells follows the Hund’s rule, the given atomic ground-state configuration exist.

Element with the given atomic ground-state configuration

Argon is a noble gas and has a complete octet electronic configuration as $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$.  After the noble gas configuration, the given atomic ground-state configuration has one electron in 5s orbital and ten electrons in 4d subshell.  The filling of orbital in the subshell is done by following Hund’s rule and it complies with it.  The total number of electrons present in the given atomic ground-state configuration is 47.  The element with the atomic number 47 is identified as Silver.