Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $\left({\text{Ce}}^{\text{3+}}\right)$ has to be predicted using an orbital box diagram and noble gas configuration.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons (e-) of an given molecule or respective atoms in atomic or molecular orbital’s.

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) =${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^1{\text{5d}}^1{\text{6s}}^{\text{2}}$ and Noble gas configuration of $[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^{\text{1}}{\text{6s}}^{\text{2}}$. The electronic configuration of Cerium (Ce³⁺) ions=

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^1{\text{5d}}^0{\text{6s}}^0$ noble gas configuration of Chromium (Cr³⁺) ions=$[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^0{\text{6s}}^0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) =${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{11}{\text{6s}}^{\text{2}}$ and Noble gas configuration of $[\text{Xe}]{\text{ 4f}}^{11}{\text{6s}}^{\text{2}}$. The electronic configuration of Holmium (Ho³⁺) ions= 

${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{10}{\text{6s}}^0$ noble gas configuration of Holmium (Ho³⁺) ions=$[\text{Xe}]{\text{ 4f}}^{10}{\text{6s}}^0$

To determine: The orbital notation method, noble gas configuration methods and should be explained given the cerium (Ce), Holmium (Ho) elements and its (Ce³⁺), (Ho³⁺) ions.

Explanation of Solution

• Electronic configuration for Cerium (Ce) system:

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Cerium}\text{(Ce)}\text{=58}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Ce)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^1{\text{5d}}^1{\text{6s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^6{\text{5d}}^0{\text{6s}}^2\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^{\text{1}}{\text{6s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^1{\text{5d}}^1{\text{6s}}^2\end{array}$

Hence, the electron configuration of cerium is $[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^{\text{1}}{\text{6s}}^{\text{2}}$

• Electronic configuration for Cerium (Ce³⁺) system:

When Cerium (Ce) was oxidized into (Ce³⁺) ions, it loses three electrons from outermost (5d, 6s) shells. Than this orbital filling method are shown below.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Cerium}\text{(Ce)}\text{= 58}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Ce)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^1{\text{5d}}^0{\text{6s}}^0\\ \text{Loss for three electrons in Cerium }\left(\text{Ce}\right)\text{valancy shells = 58-3 = 55}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^6{\text{5d}}^0{\text{6s}}^0\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^0{\text{6s}}^0[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^1{\text{5d}}^0{\text{6s}}^0\end{array}$

Hence, the electron configuration of cerium is $[\text{Xe}]{\text{ 4f}}^1{\text{5d}}^0{\text{6s}}^0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $\left({\text{Ho}}^{\text{3+}}\right)$ has to be predicted using an orbital box diagram and noble gas configuration.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons (e-) of an given molecule or respective atoms in atomic or molecular orbital’s.

Aufbau principle: This rule statues that ground state of an atom or ions electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. If consider the 1s shell is filled the 2s subshell is occupied.

Hund's Rule: The every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Pauli exclusion rule: an atomic orbital may describe at most two electrons, each with opposite spin direction.

Explanation of Solution

• The electron configuration of Cerium:

The Holmium (Ho) has 13 electrons in outermost shells the 11 electrons in (4f) shells, two electron in (6s) orbitals, the electrons filling method presented below.

$\begin{array}{l}\text{Atomic}\text{number}\text{of}Holmium\left(Ho\right)\text{=67}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Ho)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{11}{\text{6s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^{11}{\text{5d}}^0{\text{6s}}^2\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^{11}{\text{6s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^{11}{\text{5d}}^0{\text{6s}}^2\\ (\text{or)}\\ \text{=[Xe]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow \downarrow }\\ {\text{4f}}^{11}{\text{6s}}^2\end{array}$

Hence, the electron configuration of Holmium is $[\text{Xe}]{\text{ 4f}}^{11}{\text{6s}}^{\text{2}}$

• The electron configuration of Ce³⁺:

When Holmium (Ho) was oxidized into (Ho³⁺) ions, it loses for three electrons from outermost (4f and 6s) shells. Than this orbital filling method are shown below.

$\begin{array}{l}\text{Atomic}\text{number}\text{of}Holmium\left(Ho\right)\text{=67}\\ \text{Loss for four electrons in Holmium }\left(\text{Ho}\right)\text{valancy shells=67-3=64}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Ho)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^{10}{\text{6s}}^0\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^{10}{\text{5d}}^0{\text{6s}}^0\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^{10}{\text{6s}}^0[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^{10}{\text{5d}}^0{\text{6s}}^0\\ (\text{or)}\\ \text{=[Xe]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\\ {\text{4f}}^{10}{\text{6s}}^0\end{array}$

Hence, the electron configuration of Holmium (III) is $[\text{Xe}]{\text{ 4f}}^{10}{\text{6s}}^0$