Chapter 7, Problem 21PS

Using orbital box diagrams, depict an electron configuration for each of the following ions: (a) Mg²⁺, (b) K⁺, (c) Cl⁻, and (d) O²⁻.

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{Mg}}^{\text{2+}}$ ions using orbital box diagram.

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 Magnesium ($M{g}^{2+}$) ions.

Given the Magnesium atom has loss of two electrons from outermost shells.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Magnesium}\text{(Mg)}\text{=12}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^{\text{2}}\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{\uparrow \downarrow }\\ 3{\text{s}}^2\end{array}$

When ($\text{Mg}$) was oxidized to (${\text{Mg}}^{\text{2+}}$) ions, it lost for two electrons from outermost (3s) orbitals, hence this orbital notation method shows below.

   $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Magnesium}\text{(Mg)}\text{=12}\\ spdf\text{with orbtital notation}=\text{Mg}[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^0]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{}\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^0\\ spdf\text{with noble gas notation}=[\text{Ne]}3{\text{s}}^0\\ \text{Orbital}\text{box}\text{notation }\text{= [Ne]}\boxed{}\\ 3{\text{s}}^0\end{array}$

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

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{K}}^{\text{+}}$ ions using orbital box diagram.

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 Potassium ions ($K^{+}$) ions.

The single potassium atoms having (19) electrons in (s, p) orbital shells and its atomic number (Z=19). Moreover the (K) atoms has loss of one electrons in outermost (4s) shells.

Hence we can write oxidation reaction has shown below.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Potassium}\text{(K)}\text{=19}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}3{\text{p}}^{\text{6}}4{\text{s}}^{\text{1}}]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ 3{\text{p}}^{\text{6}}4{\text{s}}^{\text{1}}\\ spdf\text{with noble gas notation}=[\text{Ar]}4{\text{s}}^{\text{1}}\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow }\\ 4{\text{s}}^1\end{array}$

When (K) was oxidized to (K⁺) ions, it lost for one electron in outermost (4s) orbitals, hence this orbital notation method shows below.

   $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Potassium}\text{(K)}\text{=19}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}3{\text{p}}^{\text{6}}4{\text{s}}^0]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{}\\ 3{\text{p}}^{\text{6}}4{\text{s}}^0\\ spdf\text{with noble gas notation}=[\text{Ar]}4{\text{s}}^0\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{}\\ 4{\text{s}}^1\end{array}$

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

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{Cl}}^{-}$ ions using orbital box diagram.

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 Chlorine ions (Cl^-) ions.

The single chlorine atoms having (17) electrons in (s, p) orbital shells and its atomic number (Z=17). Moreover the (Cl) atom has gain of one electron into outermost (3p) shells.

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Chlorine}\text{(Cl)}\text{=17}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^5]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ 3{\text{p}}^5\\ spdf\text{with noble gas notation}=[\text{Ar]}3{\text{s}}^{\text{2}}{\text{3p}}^5\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\\ 3{\text{s}}^{\text{2}}3{\text{p}}^5\end{array}$

When (Cl) was gain to (Cl^-) ions, it gain for one electron into outermost (3s) orbitals, hence this orbital notation method shows below.

   $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Chlorine}\text{(Cl)}\text{=17}\\ spdf\text{with orbtital notation}=[{\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^5]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{6}3{\text{s}}^{\text{2}}\\ \boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{p}}^5\\ spdf\text{with noble gas notation}=[\text{Ar]}3{\text{s}}^{\text{2}}{\text{3p}}^6\\ \text{Orbital}\text{box}\text{notation }\text{= [Ar]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\\ 3{\text{s}}^{\text{2}}3{\text{p}}^6\end{array}$

Hence, the electronic configuration of chlorine ions (Cl^-) = ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^6{\text{3s}}^2{\text{3p}}^6$ and noble gas configuration of $[\text{Ar}]3{\text{s}}^{\text{2}}{\text{3p}}^{\text{6}}$ 

Interpretation Introduction

Interpretation:

The electronic configuration has to be depicted for ${\text{O}}^{2-}$ ions using orbital box diagram.

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

Finally we consider the orbital filling method of Oxygen (II) ions (O^2-) ions.

The oxygen atom (O) is a monoatomic anion one or more electrons added to the valance shell of a non-metal atom so that electronic configuration of the ion is the same as the electronic configuration of the noble gas in the periodic table. Here single (O) atom gains of two electrons and oxygen become oxygen anion, attains electron configuration as the noble gas Neon (Ne).

  $\begin{array}{l}\text{Atomic}\text{number}\text{of}\text{Oxygen}\text{(O)}\text{=}\text{8}\\ spdf\text{with orbtital notation}=[1{\text{s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4]\\ \text{Orbital filling method }\text{= }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\\ {\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^4\\ spdf\text{with noble gas notation}=[\text{He}]2{\text{s}}^{\text{2}}{\text{2p}}^4\\ \text{Orbital}\text{box}\text{notation }\text{= [He]}\boxed{\uparrow \downarrow }\boxed{\uparrow \downarrow }\boxed{\uparrow }\boxed{\uparrow }\\ 2{\text{s}}^{\text{2}}{\text{2p}}^4\end{array}$

When (O) was gain to (O^2-) ions, it gain for two electrons into outermost (2p) orbitals, hence this orbital notation method shows below.

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

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