Chapter 7, Problem 49GQ

The magnet in the following photo is made from neodymium, iron, and boron.

A magnet mode of on alloy containing the elements Nd, Fe, and B.

1. (a) Write the electron configuration of each of these elements using an orbital box diagram and noble gas notation.
2. (b) Are these elements paramagnetic or diamagnetic?
3. (c) Write the electron configurations of Nd³⁺ and Fe³⁺ using orbital box diagrams and noble gas notation. Are these ions paramagnetic or diamagnetic?

Interpretation Introduction

Interpretation:

The electron configuration of Neodymium, iron, and boron has to be written using orbital box diagram.

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.

Paramagnetic: The Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field.

Diamagnetic properties: In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom.

Answer to Problem 49GQ

Statement (a): The atomic number of Neodymium (Nd) (Z=60) electronic configuration of Neodymium (Nd)= ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^4{\text{6s}}^{\text{2}}$ and Noble gas configuration of $[\text{Xe}]{\text{ 4f}}^4{\text{6s}}^{\text{2}}$.

The electronic configuration of Iron (Fe) =${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{6}4{\text{s}}^2$ and noble gas configuration of $[\text{Ar}]3{\text{d}}^{6}4{\text{s}}^2$.

The electronic configuration of Boron (B) =${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^1$ and noble gas configuration of $[\text{He}]2s^2{\text{2p}}^1$

Statement (b): The all elements ($\text{Nd, Fe and B}$) are obeyed for paramagnetic properties.

Statement (c): The orbital notation of Neodymium (Nd³⁺) ions= ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^3{\text{6s}}^0$ noble gas configuration of Neodymium (Nd³⁺) ions= $[\text{Xe}]{\text{ 4f}}^3{\text{6s}}^0$. The electronic configuration of Iron (Fe³⁺) ions =${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{5}4{\text{s}}^0$ and noble gas configuration of $[\text{Ar}]3{\text{d}}^{5}4{\text{s}}^0$. Than these ions are obeyed for paramagnetic properties.

To determine: The orbital notation method, noble gas configuration methods and Magnetic property should be explained given the different type of Neodymium (Nd), Iron (Fe) and Boron (B) element and its (Nd³⁺, Fe³⁺) ions.

Explanation of Solution

The electron configuration of Neodymium (Nd) element:

The orbital box diagram as follows,

$\begin{array}{l}\text{Atomic}\text{number}\text{of}Neodymium\left(Nd\right)\text{=60}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Nd)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^4{\text{6s}}^{\text{2}}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^4{\text{6s}}^2\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^4{\text{6s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^4{\text{6s}}^2\end{array}$

Hence, the Noble gas configuration of $[\text{Xe}]{\text{ 4f}}^4{\text{6s}}^{\text{2}}$.

The electron configuration of  Iron (Fe):

The orbital box diagram as follows,

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Iron }\left(\text{Fe}\right)\text{=26}\\ spdf\text{with orbtital notation}=\text{Fe}[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{6}4{\text{s}}^2\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow \downarrow }\\ {\text{3p}}^6{\text{3d}}^{6}4{\text{s}}^2\\ spdf\text{with noble gas notation}=[\text{Ar}]3{\text{d}}^{6}4{\text{s}}^2{}^{}\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow \downarrow }\\ {\text{3d}}^{6}4{\text{s}}^2\end{array}$

Hence, the Noble gas configuration of $[\text{Ar}]3{\text{d}}^{6}4{\text{s}}^2$.

The electron configuration of  Boron (B):

The orbital box diagram as follows,

$\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Boron }\left(\text{B}\right)\text{=5}\\ spdf\text{with orbtital notation}=B[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^1]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{}\boxed{}\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^1\\ spdf\text{with noble gas notation}=[\text{He}]{\text{2s}}^{\text{2}}{\text{2p}}^1\\ \text{Orbital}\text{box}\text{notation }\text{= [He]}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{}\boxed{}\\ {\text{2s}}^{\text{2}}{\text{2p}}^1\end{array}$

Hence, the Noble gas configuration of $[\text{He}]2s^2{\text{2p}}^1$.

Interpretation Introduction

Interpretation:

The magnetic property for the elements neodymium, iron, and Boron has to be predicted.

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.

Paramagnetic: The Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field.

Diamagnetic properties: In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom.

Explanation of Solution

Analyzing for Magnetic properties:

The given Neodymium (Nd), Iron (Fe) and Boron (B) elements shows good paramagnetic character, because unpaired electrons are present.

  $\begin{array}{l}Neodymium\left(Nd\right)\text{element}=[\text{Xe}]{\text{ 4f}}^4{\text{6s}}^{\text{2}}[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{= [Xe]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{\uparrow \downarrow }\\ {\text{4f}}^4{\text{6s}}^2\\ \downarrow \\ \text{All}\text{four}\text{electrons}are\text{unpaired}\to \text{Obeyed for paramagantic }\end{array}$

The Iron (Fe) system is we discussed below,

  $\begin{array}{l}\text{Iron }\left(\text{Fe}\right)\text{=26 }[\text{Ar}]3{\text{d}}^{6}4{\text{s}}^2[\because \text{Atomic}\text{number}\text{of}\text{Ar}=18]\\ \text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow \downarrow }\\ {\text{3d}}^{6}4{\text{s}}^2\\ \downarrow \\ \text{All}\text{four}\text{electrons}are\text{unpaired}\to \text{Obeyed for paramagantic }\end{array}$

It has four unpaired electrons which show paramagnetic property.

The Boron (B) system is we discussed below,

  $\begin{array}{l}\text{Boron }\left(\text{B}\right)\text{=5 }[He]2s^2{\text{2p}}^1[\because \text{Atomic}\text{number}\text{of}He=2]\\ \text{= [He]}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{}\boxed{}\\ 2{\text{s}}^2{\text{2p}}^{\text{1}}\\ \downarrow \\ \text{All}\text{one}\text{electron}isa\text{unpaired}\to \text{Obeyed for paramagantic }\end{array}$

It has one unpaired electrons which shows paramagnetic property.

Interpretation Introduction

Interpretation:

The electron configuration of Nd³⁺ and Fe³⁺ using orbital box diagram and its paramagnetic property has to be predicted.

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.

Paramagnetic: The Paramagnetic properties are due to the presence of some unpaired electrons, and from the realignment of the electron paths caused by the external magnetic field.

Diamagnetic properties: In diamagnetic materials all the electron are paired so there is no permanent net magnetic moment per atom.

Explanation of Solution

Electronic configuration of Neodymium (Nd³⁺) system

The Neodymium (Nd) was oxidized to Neodymium (Nd³⁺) ions, it lost for three electrons in outermost (4f and 6s) shells. Than this orbital filling method are shown below.

$\begin{array}{l}\text{Atomic}\text{number}\text{of}Neodymium\left(Nd\right)\text{=60}\\ \text{Complete (spdf)}\text{notation}\text{of}\text{(Nd)}\text{=}{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{4f}}^3{\text{6s}}^0\\ \text{Loss for four electrons in }Neodymium\left(Nd\right)\text{valancy shells=60-3=57}\\ \text{Orbital filling method}=\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^{\text{2}}{\text{3p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{d}}^{10}4{\text{s}}^{2}4{\text{p}}^6\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 4{\text{d}}^{8}5{\text{s}}^2{\text{5p}}^{\text{6}}\\ \boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^3{\text{6s}}^0\\ spdf\text{with noble gas notation}=[\text{Xe}]{\text{ 4f}}^3{\text{6s}}^0[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{Orbital}\text{box}\text{notation }\text{= [Xe]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\\ {\text{4f}}^3{\text{6s}}^0\end{array}$

Electronic configuration of Iron (Fe³⁺) system

The Iron (Fe) was oxidized to Iron (Fe³⁺)  ions, it lost for three electrons in outermost (4f and 6s) shells. Than this orbital filling method are shown below.

$\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Iron }\left(\text{Fe}\right)\text{=26}\\ spdf\text{with orbtital notation}=\text{Fe}[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{5}4{\text{s}}^0]\\ \text{Loss for four electrons in Iron }\left(\text{Fe}\right)\text{valancy shells=26-3=23}{\text{[Here Fe ionto Fe}}^{\text{3+}}\text{ions]}\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ {\text{3p}}^6{\text{3d}}^{5}4{\text{s}}^0\\ spdf\text{with noble gas notation}=[\text{Ar}]3{\text{d}}^{5}4{\text{s}}^0\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ {\text{3d}}^{5}4{\text{s}}^0\end{array}$

Analyzing for Magnetic properties for (Nd³⁺) and (Fe³⁺) systems:

The given Statement (c) Neodymium (Nd³⁺)and Iron (Fe³⁺) elements are very good paramagnetic character, because all electrons in unpaired only see the above orbital notation method. Than the both system electron filling method are shown below.

  $\begin{array}{l}Neodymium\left(Nd\right)\text{element}=[\text{Xe}]{\text{ 4f}}^3{\text{6s}}^0[\because \text{Atomic}\text{number}\text{of}\text{Xe}=54]\\ \text{= [Xe]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}[HerelossforthreeelectronsinNdsystem\text{Nd}\text{into}{\text{Nd}}^{\text{3+}}]\\ {\text{4f}}^3{\text{6s}}^0\\ \downarrow \\ \text{All}\text{three}\text{electrons}are\text{unpaired}\to \text{Obeyed for paramagantic }\end{array}$

The Iron (Fe³⁺) system is we discussed below,

  $\begin{array}{l}\text{Iron }\left(\text{Fe}\right)\text{=26 }[\text{Ar}]3{\text{d}}^{5}4{\text{s}}^0[\because \text{Atomic}\text{number}\text{of}\text{Ar}=18]\\ \text{= [Ar]}\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ {\text{3d}}^{5}4{\text{s}}^0\\ \downarrow \\ \text{All}\text{five}\text{electrons}are\text{unpaired}\to \text{Obeyed for paramagantic }\end{array}$