The electron configuration of Cerium and ( Ce 3+ ) 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.

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Answer 1

Question

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

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Answer 2

Question

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

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Answer 3

Question

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

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Answer 4

Question

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

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Answer 5

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

Answer 6

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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.

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

Answer 7

Chapter 7, Problem 40GQ

a)

Interpretation Introduction

Interpretation:

The electron configuration of Cerium and $(Ce3+)$ 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 8

a)

Expert Solution

Answer to Problem 40GQ

Statement (a): The atomic number of Cerium (Ce) (Z=58) electronic configuration of Cerium (Ce) = $1s22s22p63s23p63d104s24p64d105s25p64f15d16s2$ and Noble gas configuration of $[Xe] 4f15d16s2$ . The electronic configuration of Cerium (Ce³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f15d06s0$ noble gas configuration of Chromium (Cr³⁺) ions= $[Xe] 4f15d06s0$

Statement (b): The atomic number of Holmium (Ho) (Z=67) electronic configuration of Holmium (Ho) = $1s22s22p63s23p63d104s24p64d105s25p64f116s2$ and Noble gas configuration of $[Xe] 4f116s2$ . The electronic configuration of Holmium (Ho³⁺) ions=

$1s22s22p63s23p63d104s24p64d105s25p64f106s0$ noble gas configuration of Holmium (Ho³⁺) ions= $[Xe] 4f106s0$

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.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Electronic configuration for Cerium (Ce) system:

$Atomic number of Cerium (Ce) =58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d16s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ ↑ ↑↓ 4f6 5d0 6s2spdf with noble gas notation = [Xe] 4f15d16s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ ↑ ↑↓ 4f1 5d1 6s2$

Hence, the electron configuration of cerium is $[Xe] 4f15d16s2$

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.

$Atomic number of Cerium (Ce) = 58Complete (spdf) notation of (Ce) = 1s22s22p63s23p63d104s24p64d105s25p64f15d06s0Loss for three electrons in Cerium (Ce)valancy shells = 58-3 = 55Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑ 4f6 5d0 6s0spdf with noble gas notation = [Xe] 4f15d06s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑ 4f1 5d0 6s0$

Hence, the electron configuration of cerium is $[Xe] 4f15d06s0$

Answer 13

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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.

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$

Hence, the electron configuration of Holmium (III) is $[Xe] 4f106s0$

Answer 14

b)

Interpretation Introduction

Interpretation:

The electron configuration of Holmium and $(Ho3+)$ 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 15

b)

Expert Solution

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.

$Atomic number of Holmium (Ho) =67Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f116s2Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2spdf with noble gas notation = [Xe] 4f116s2 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 5d0 6s2 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑↑ ↑↓ 4f11 6s2$

Hence, the electron configuration of Holmium is $[Xe] 4f116s2$

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.

$Atomic number of Holmium (Ho) =67Loss for four electrons in Holmium (Ho)valancy shells=67-3=64Complete (spdf) notation of (Ho) = 1s22s22p63s23p63d104s24p64d105s25p64f106s0Orbital filling method =↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s2 2s2 2p6 3s2 3p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 3d10 4s2 4p6 ↑↓↑↓↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 4d8 5s2 5p6 ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0spdf with noble gas notation = [Xe] 4f106s0 [∵ Atomic number of Xe=54]Orbital box notation = [Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 5d0 6s0 (or) =[Xe] ↑↓↑↓↑↓↑↓↑↑ 4f10 6s0$