Chapter 5, Problem 5.37P

(a)

Interpretation Introduction

Interpretation:

Number of unpaired electrons present in ground state electronic configuration of ${\text{Cl}}^{-}$ has to be given.

Concept Introduction:

Electrons of an atom are arranged in orbitals by the order of increasing energy.  This arrangement is known as electronic configuration of atom.  This can be represented using noble-gas shorthand notation also.

Ions are formed from neutral atom either by removal or addition of electrons from the valence shell.

Each orbital has two electrons and they both are in opposite spin.  Electrons are filled up in the orbitals of a sub-shell by following Hund’s rule.  This rule tells that all the orbitals are singly filled in a sub-shell and then the pairing occurs.

Explanation of Solution

Given ion is ${\text{Cl}}^{-}$.  Atomic number of chlorine is $17$.  Ground state electronic configuration of chlorine is shown below.

    $\text{Cl}=1s^2 2s^2 2p^6 3s^2 3p^5$

The given ${\text{Cl}}^{-}$ ion is formed when one electron is added to the valence shell.  In this case the one electron is added to $3p$ orbital.  Therefore, the electronic configuration of ${\text{Cl}}^{-}$ ion is written as shown below.

    ${\text{Cl}}^{-}=1s^2 2s^2 2p^6 3s^2 3p^6$

According to Hund’s rule, the electrons are added to the orbitals only after singly filling all the orbitals.  After this takes place only the pairing of electrons is done.  Therefore, the expansion of orbitals in the outermost shell is given as shown below.

    ${\text{Cl}}^{-}=1s^2 2s^2 2p^6 3s^2 3p_x{}^{2}3p_y{}^{2}3p_z{}^2$

Therefore, there are no unpaired electrons in ${\text{Cl}}^{-}$ ion in ground state electronic configuration.

(b)

Interpretation Introduction

Interpretation:

Number of unpaired electrons present in ground state electronic configuration of ${\text{O}}^{+}$ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{O}}^{+}$.  Atomic number of oxygen is $8$.  Ground state electronic configuration of oxygen is shown below.

    $\text{O}=1s^2 2s^2 2p^4$

The given ${\text{O}}^{+}$ ion is formed when one electron is removed from the valence shell.  In this case the one electron is removed $2p$ orbital.  Therefore, the electronic configuration of ${\text{O}}^{+}$ ion is written as shown below.

    ${\text{O}}^{\text{+}}=1s^2 2s^2 2p^3$

According to Hund’s rule, the electrons are added to the orbitals only after singly filling all the orbitals.  After this takes place only the pairing of electrons is done.  Therefore, the expansion of orbitals in the outermost shell is given as shown below.

    ${\text{O}}^{+}=1s^2 2s^2 2p_x{}^{1}2p_y{}^{1}2p_z{}^1$

Therefore, there are three unpaired electrons in ${\text{O}}^{+}$ ion in ground state electronic configuration.

(c)

Interpretation Introduction

Interpretation:

Number of unpaired electrons present in ground state electronic configuration of ${\text{Al}}^{3+}$ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Al}}^{3+}$.  Atomic number of aluminium is $13$.  Ground state electronic configuration of aluminium is shown below.

    $\text{Al}=1s^2 2s^2 2p^6 3s^2 3p^1$

The given ${\text{Al}}^{3+}$ ion is formed when three electrons are removed from the valence shell.  In this case the three electrons are removed from $3$ shell.  Therefore, the electronic configuration of ${\text{Al}}^{3+}$ ion is written as shown below.

    ${\text{Al}}^{\text{3+}}=1s^2 2s^2 2p^6$

According to Hund’s rule, the electrons are added to the orbitals only after singly filling all the orbitals.  After this takes place only the pairing of electrons is done.  Therefore, the expansion of orbitals in the outermost shell is given as shown below.

    ${\text{Al}}^{3+}=1s^2 2s^2 2p_x{}^{2}2p_y{}^{2}2p_z{}^2$

Therefore, there are no unpaired electrons in ${\text{Al}}^{3+}$ ion in ground state electronic configuration.

(d)

Interpretation Introduction

Interpretation:

Number of unpaired electrons present in ground state electronic configuration of ${\text{Xe}}^{+}$ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Xe}}^{+}$.  Atomic number of xenon is $54$.  Ground state electronic configuration of xenon is shown below.

    $\text{Xe}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6$

The given ${\text{Xe}}^{+}$ ion is formed when one electron is removed from the valence shell.  In this case the one electron is removed $p$ orbital.  Therefore, the electronic configuration of ${\text{Xe}}^{+}$ ion is written as shown below.

    ${\text{Xe}}^{\text{+}}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^5$

According to Hund’s rule, the electrons are added to the orbitals only after singly filling all the orbitals.  After this takes place only the pairing of electrons is done.  Therefore, the expansion of orbitals in the outermost shell is given as shown below.

    ${\text{Xe}}^{+}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p_x{}^{2}5p_y{}^{2}5p_z{}^1$

Therefore, there is one unpaired electron in ${\text{Xe}}^{+}$ ion in ground state electronic configuration.

(e)

Interpretation Introduction

Interpretation:

Number of unpaired electrons present in ground state electronic configuration of ${\text{K}}^{+}$ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{K}}^{+}$.  Atomic number of potassium is $19$.  Ground state electronic configuration of potassium is shown below.

    $\text{K}=1s^2 2s^2 2p^6 3s^2 3p^{6}4s^1$

The given ${\text{K}}^{+}$ ion is formed when one electron is removed from the valence shell.  In this case the one electron is removed from $s$ orbital.  Therefore, the electronic configuration of ${\text{K}}^{+}$ ion is written as shown below.

    ${\text{K}}^{\text{+}}=1s^2 2s^2 2p^{6}3s^{2}3p^6$

According to Hund’s rule, the electrons are added to the orbitals only after singly filling all the orbitals.  After this takes place only the pairing of electrons is done.  Therefore, the expansion of orbitals in the outermost shell is given as shown below.

    ${\text{K}}^{\text{+}}=1s^2 2s^2 2p^{6}3s^{2}3p_x{}^{2}3p_y{}^{2}3p_z{}^2$

Therefore, there are no unpaired electrons in ${\text{K}}^{\text{+}}$ ion in ground state electronic configuration.