Chapter 4, Problem 4.78QP

For each pair of ions, determine which will have the greater number of unpaired electrons: (a) Fe²⁺, Fe³⁺; (b) P³⁺, P⁵⁺: (c) Cr²⁺, Cr³⁺.

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of ions has to be written.

Concept Introduction:

• Aufbau Principle tells that the orbital with the lower energy is filled with electrons first and then the filling of higher energy orbital follows.  By using this valence orbital diagram can be drawn for any atoms or ions.
• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions, the electronic configuration is written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbital is singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.

To identify: The ion which has greater number of unpaired electrons.

Answer to Problem 4.78QP

Answer

In (a) ${\text{Fe}}^{\text{3+}}$ has more number of unpaired electrons than ${\text{Fe}}^{\text{2+}}$

Explanation of Solution

Electronic configuration of $\text{Fe}$ is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$

The electronic configuration of $\text{Fe}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Fe}$ is 26.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Fe}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^6$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{Fe}}^{\text{2+}}$ and valence orbital diagram can be written as follows,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$

The electronic configuration of ${\text{Fe}}^{\text{2+}}$ is found from the electronic configuration of $\text{Fe}$.  ${\text{Fe}}^{\text{2+}}$ is formed from $\text{Fe}$ when two valence electrons are removed from the 4s orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Fe}}^{\text{2+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{Fe}}^{\text{3+}}$ and valence orbital diagram is,

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$

The electronic configuration of ${\text{Fe}}^{\text{3+}}$ is found from the electronic configuration of $\text{Fe}$.  ${\text{Fe}}^{\text{3+}}$ is formed from $\text{Fe}$ when two valence electrons are removed from the 4s orbital and one electron from 3d orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Fe}}^{\text{3+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$.  The valence orbital diagram is drawn as shown above.

Comparing the unpaired electrons in ${\text{Fe}}^{\text{2+}}$ and ${\text{Fe}}^{\text{3+}}$, finding the ion which has greater number of unpaired electrons,

By looking at the valence-orbital diagram we can see that in ${\text{Fe}}^{\text{2+}}$ and ${\text{Fe}}^{\text{3+}}$ has four unpaired electrons and five unpaired electrons respectively.  The unpaired electrons are highlighted in red colour in the above figure.  ${\text{Fe}}^{\text{3+}}$ Has greater number of unpaired electrons than ${\text{Fe}}^{\text{2+}}$.

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of ions has to be written.

Concept Introduction:

• Aufbau Principle tells that the orbital with the lower energy is filled with electrons first and then the filling of higher energy orbital follows.  By using this valence orbital diagram can be drawn for any atoms or ions.
• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions, the electronic configuration is written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbital is singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.

To identify: The ion which has greater number of unpaired electrons.

Answer to Problem 4.78QP

Answer

In (b) ${\text{P}}^{\text{3+}}$ and ${\text{P}}^{\text{5+}}$ does not have unpaired electrons

Explanation of Solution

Electronic configuration of $\text{P}$

$1s^2 2s^2 2p^6 3s^2 3p^3$

The electronic configuration of $\text{P}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{P}$ is 15.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{P}$ is found as $1s^2 2s^2 2p^6 3s^2 3p^3$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{P}}^{\text{3+}}$ and valence orbital diagram can be written as follows,

$1s^{2}2s^{2}2p^{6}3s^2$

The electronic configuration of ${\text{P}}^{\text{3+}}$ is found from the electronic configuration of $\text{P}$.

${\text{P}}^{\text{3+}}$ is formed from $\text{P}$ when three valence electrons are removed from the 3p orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{P}}^{\text{3+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^2$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{P}}^{\text{5+}}$ and valence orbital diagram is,

$1s^{2}2s^{2}2p^6$

The electronic configuration of ${\text{P}}^{\text{5+}}$ is found from the electronic configuration of $\text{P}$.

${\text{P}}^{\text{5+}}$ is formed from $\text{P}$ when three valence electrons are removed from the 3p orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{P}}^{\text{5+}}$ is found as $1s^{2}2s^{2}2p^6$.  The valence orbital diagram is drawn as shown above.

Comparing the unpaired electrons in ${\text{P}}^{\text{3+}}$ and ${\text{P}}^{\text{5+}}$, finding the ion which has greater number of unpaired electrons,

By looking at the valence-orbital diagram we can see that both electrons in ${\text{P}}^{\text{3+}}$ and ${\text{P}}^{\text{5+}}$ does not have unpaired electrons.

Interpretation Introduction

Interpretation: Ground-state electronic configuration of the given set of ions has to be written.

Concept Introduction:

• Aufbau Principle tells that the orbital with the lower energy is filled with electrons first and then the filling of higher energy orbital follows.  By using this valence orbital diagram can be drawn for any atoms or ions.
• Electronic configuration is the arrangement of the electrons of atoms in the orbital.  For atoms and ions, the electronic configuration is written by using Pauli Exclusion Principle and Hund’s rule.
• According to Pauli Exclusion Principle, no two electrons having the same spin can occupy the same orbital.
• According to Hund’s rule, the orbital in the subshell is filled singly by one electron before the same orbital is doubly filled.  When the orbital is singly filled, all the electrons have same spin.  In a doubly filled orbital, there are two electrons with opposite spin.

To identify: The ion which has greater number of unpaired electrons.

Answer to Problem 4.78QP

Answer

In (c) ${\text{Cr}}^{\text{2+}}$ has more number of unpaired electrons than ${\text{Cr}}^{\text{3+}}$

Explanation of Solution

Electronic configuration of $\text{Cr}$ is,

$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$

The electronic configuration of $\text{Fe}$ is found using the total number of electrons present in the atom.  The total number of electrons present in $\text{Cr}$ is 24.  According to Pauli Exclusion Principle and Hund’s rule, the electronic configuration of $\text{Cr}$ is found as $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{Cr}}^{\text{2+}}$ and valence orbital diagram can be written as follows,

$1s^2 2s^2 2p^6 3s^2 3p^{6}3d^4$

The electronic configuration of ${\text{Cr}}^{\text{2+}}$ is found from the electronic configuration of $\text{Cr}$.  ${\text{Cr}}^{\text{2+}}$ is formed from $\text{Cr}$ when one valence electron is removed from 4s orbital and one electron from 3d orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Cr}}^{\text{2+}}$ is found as $1s^2 2s^2 2p^6 3s^2 3p^{6}3d^4$.  The valence orbital diagram is drawn as shown above.

Electronic configuration of ${\text{Cr}}^{\text{3+}}$ and valence orbital diagram

$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$

The electronic configuration of ${\text{Cr}}^{\text{3+}}$ is found from the electronic configuration of $\text{Cr}$.  ${\text{Cr}}^{\text{3+}}$ is formed from $\text{Cr}$ when one valence electron is removed from the 4s orbital and two electrons from 3d orbital.  According to Pauli Exclusion Principle and Hund’s rule, the ground state electronic configuration of ${\text{Cr}}^{\text{3+}}$ is found as $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$.  The valence orbital diagram is drawn as shown above.

Comparing the unpaired electrons in ${\text{Cr}}^{\text{2+}}$ and ${\text{Cr}}^{\text{3+}}$, finding the ion which has greater number of unpaired electrons.

By looking at the valence-orbital diagram we can see that in ${\text{Cr}}^{\text{2+}}$ and ${\text{Cr}}^{\text{3+}}$ has four unpaired electrons and three unpaired electrons respectively.  The unpaired electrons are highlighted in red colour in the above figure.  ${\text{Cr}}^{\text{2+}}$ has greater number of unpaired electrons than ${\text{Cr}}^{\text{3+}}$.