Chapter 7, Problem 22PS

(a)

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{Na}}^{\text{+}}$ using orbital box diagram and noble gas electronic configuration method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of an given molecule or respective atoms in atomic or molecular orbitals.

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

Let us consider the orbital filling method of Sodium (Na⁺) ions.

Given the Sodium atom has loss of one electron from outermost shells.

 $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{sodium}\text{(Na)}\text{=11}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^1]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^1\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^1\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{\uparrow }\\ 3{\text{s}}^1\end{array}$

When ($\text{Na}$) was oxidized to (${\text{Na}}^{\text{+}}$) ions, it lose one electron from outermost (3s) orbitals, hence this orbital notation method shows below.

   $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{sodium}\text{(Na)}\text{=11}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^0]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{}\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{5}3{\text{s}}^0\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^0\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{}\\ 3{\text{s}}^0\end{array}$

Hence, the electronic configuration of Sodium ions (${\text{Na}}^{\text{+}}$) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^0$ and noble gas configuration of $[\text{Ne}]3{\text{s}}^0$ 

(b)

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{Al}}^{\text{3+}}$ using orbital box diagram and noble gas electronic configuration method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of an given molecule or respective atoms in atomic or molecular orbitals.

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

Let us consider the orbital filling method of Aluminium ions (${\text{Al}}^{\text{3+}}$) ions.

The single Aluminium toms having (13) electrons in (s, p, d) orbital shells and its atomic number (Z=13). Moreover the (Ge) atoms has loss of one electrons in outermost (4s, 4p) shells.

Hence we can write oxidation reaction has shown below.

               $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Aluminium}\text{(Al)}\text{=13}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}3{\text{p}}^1]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow }\boxed{}\boxed{}\\ 3{\text{p}}^1\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^{\text{2}}3{\text{p}}^1\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{}\boxed{}\\ 3{\text{s}}^{\text{2}}3{\text{p}}^1\end{array}$

When (Al) was oxidized to (Al³⁺) ions, it lose three electrons from outermost (3s and 3p) orbitals, hence this orbital notation method shows below.

              $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Aluminium}\text{(Al)}\text{=13}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{0}3{\text{p}}^0]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{}\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^0\\ \boxed{}\boxed{}\boxed{}\\ 3{\text{p}}^0\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^{0}3{\text{p}}^0\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{}\boxed{}\boxed{}\boxed{}\\ 3{\text{s}}^{0}3{\text{p}}^0\end{array}$

Hence, the electronic configuration of Aluminium (III) ions (${\text{Al}}^{\text{3+}}$) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^0{\text{3p}}^0$ and noble gas configuration of $[\text{Ne}]{\text{3s}}^0{\text{3p}}^0$ 

(c)

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{Ge}}^{\text{2+}}$ using orbital box diagram and noble gas electronic configuration method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of an given molecule or respective atoms in atomic or molecular orbitals.

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

 Let us consider the orbital filling method of Germanium ions (${\text{Ge}}^{\text{2+}}$) ions.

The single Ge atoms having (32) electrons in (s, p) orbital shells and its atomic number (Z=32). Moreover the (Ge) atoms has loss of two electrons in outermost (3p, 3s) shells.

Hence we can write gains of electron (Oxidation method) process are presented below.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Germanium}\text{(Ge)}\text{=32}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^2]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{p}}^6{\text{3d}}^{10}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ 4{\text{s}}^{\text{2}}4{\text{p}}^2\\ spdf\text{with noble gas notation}=[{\text{Ar]3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^2\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\boxed{}\\ {\text{3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^2\end{array}$

When (Ge) was oxidized to (Ge²⁺) ions, it lost for two electrons in outermost (4s and 4p) orbitals, hence this orbital notation method shows below.

   $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Germanium}\text{(Ge)}\text{=32}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^0]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{p}}^6{\text{3d}}^{10}\\ \boxed{\uparrow \downarrow }\boxed{}\boxed{}\boxed{}\\ 4{\text{s}}^{\text{2}}4{\text{p}}^0\\ spdf\text{with noble gas notation}=[{\text{Ar]3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^2\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{}\boxed{}\boxed{}\\ {\text{3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^0\end{array}$

Hence, the electronic configuration of germanium ions (${\text{Ge}}^{\text{2+}}$) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6{\text{3d}}^{10}4{\text{s}}^{\text{2}}4{\text{p}}^0$ and noble gas configuration of $[\text{Ar}]4{\text{s}}^{\text{2}}4{\text{p}}^0$ 

(d)

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{F}}^{\text{-}}$ ion using orbital box diagram and noble gas electronic configuration method.

Concept Introduction:

Electronic configuration: The electronic configuration is the distribution of electrons of an given molecule or respective atoms in atomic or molecular orbitals.

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

 Let us consider the orbital filling method of Florine ions (F^-) ions.

The single chlorine atoms having (9) electrons in (s, p) orbital shells and its atomic number (Z=9). Moreover the (F) atom has gain of one electron in outermost (2p) shells.

Hence we can write gains of electron (Reduction method) process are presented below.

                $\begin{array}{l}\text{Atomic}\text{number}\text{of}Fluorine\text{(F)}\text{=9}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^5\\ \\ spdf\text{with noble gas notation}=[He{\text{]2s}}^{\text{2}}{\text{2p}}^5\\ \text{Orbital}\text{box}\text{notation }\text{= [He]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ 2{\text{s}}^{\text{2}}2{\text{p}}^5\end{array}$

When (F) was gain to (F^-) ions, it gain one electron to outermost (2p) orbitals, hence this orbital notation method shows below.

         $\begin{array}{l}\text{Atomic}\text{number}\text{of}Flurine\text{(F)}\text{= 9}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6\\ \\ spdf\text{with noble gas notation}=[He{\text{]2s}}^{\text{2}}{\text{2p}}^6\\ \text{Orbital}\text{box}\text{notation }\text{= [He]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 2{\text{s}}^{\text{2}}2{\text{p}}^6\end{array}$

Hence, the electronic configuration of fluorine ions (${\text{F}}^{\text{-}}$) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6$ and noble gas configuration of $[\text{He}]2{\text{s}}^{\text{2}}{\text{2p}}^6$