Chapter 5, Problem 5.66P

(a)

Interpretation Introduction

Interpretation:

Number of unpaired electrons and the neutral atom that is isoelectronic with ${\text{F}}^{\text{+}}$ has to be written.

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.

Isoelectronic species are the ones that have different number of protons but same number of electrons.  No two neutral atom can be isoelectronic.  A neutral atom can be isoelectronic with an ion.

Explanation of Solution

Given ion is ${\text{F}}^{\text{+}}$.  Atomic number of fluorine is $9$.  Ground state electronic configuration of fluorine is shown below.

    $\text{F}=1s^2 2s^2 2p^5$

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

    ${\text{F}}^{\text{+}}=1s^2 2s^2 2p^4$

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.

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

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

The total number of electrons present in ${\text{F}}^{\text{+}}$ ion is found by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}=2+2+4\\ =8\end{array}$

Total number of electrons that is present in ${\text{F}}^{\text{+}}$ ion is $8$.  The neutral atom that has eight electrons is found to be oxygen.  Therefore, the ${\text{F}}^{\text{+}}$ ion is found to be isoelectronic with oxygen atom.

(b)

Interpretation Introduction

Interpretation:

Number of unpaired electrons and the neutral atom that is isoelectronic with ${\text{Sn}}^{\text{2+}}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Sn}}^{\text{2+}}$.  Atomic number of tin is $50$.  Ground state electronic configuration of tin is shown below.

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

The given ${\text{Sn}}^{2+}$ ion is formed when two electrons removed from the valence shell.  In this case the two electrons are removed from $5p$ orbital.  Therefore, the electronic configuration of ${\text{Sn}}^{\text{2+}}$ ion is written as shown below.

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

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.

    ${\text{Sn}}^{2+}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d_{xy}{}^{2}4d_{yz}{}^{2}4d_{zx}{}^{2}4d_{x^2-y^2}{}^{2}4d_{y^2}{}^2$

Therefore, there are no unpaired electrons in ${\text{Sn}}^{\text{2+}}$ ion in ground state electronic configuration.

The total number of electrons present in ${\text{Sn}}^{\text{2+}}$ ion is found by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}=2+2+6+2+6+2+10+6+2+10\\ =48\end{array}$

Total number of electrons that is present in ${\text{Sn}}^{\text{2+}}$ ion is $48$.  The neutral atom that has forty eight electrons is found to be cadmium.  Therefore, the ${\text{Sn}}^{\text{2+}}$ ion is found to be isoelectronic with cadmium atom.

(c)

Interpretation Introduction

Interpretation:

Number of unpaired electrons and the neutral atom that is isoelectronic with ${\text{Bi}}^{\text{3+}}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Bi}}^{\text{3+}}$.  Atomic number of bismuth is $83$.  Ground state electronic configuration of bismuth is shown below.

    $\text{Bi}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^3$

The given ${\text{Bi}}^{\text{3+}}$ ion is formed when three electrons are removed from the valence shell.  In this case the three electrons are removed from the $5p$ orbital.  Therefore, the electronic configuration of ${\text{Bi}}^{3+}$ ion is written as shown below.

    ${\text{Bi}}^{\text{3+}}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10}$

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.

${\text{Bi}}^{3+}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d_{xy}{}^{2}5d_{yz}{}^{2}5d_{zx}{}^{2}5d_{x^2-y^2}{}^{2}5d_{y^2}{}^2$

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

The total number of electrons present in ${\text{Bi}}^{\text{3+}}$ ion is found by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}=2+2+6+2+6+2+10+6+2+10+6+2+14+10\\ =80\end{array}$

Total number of electrons that is present in ${\text{Bi}}^{\text{3+}}$ ion is $80$.  The neutral atom that has eighty electrons is found to be mercury.  Therefore, the ${\text{Bi}}^{\text{3+}}$ ion is found to be isoelectronic with mercury atom.

(d)

Interpretation Introduction

Interpretation:

Number of unpaired electrons and the neutral atom that is isoelectronic with ${\text{Ar}}^{\text{+}}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Ar}}^{\text{+}}$.  Atomic number of argon is $18$.  Ground state electronic configuration of argon is shown below.

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

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

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

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.

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

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

The total number of electrons present in ${\text{Ar}}^{\text{+}}$ ion is found by summing up the superscript.

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}=2+2+6+2+5\\ =17\end{array}$

Total number of electrons that is present in ${\text{Ar}}^{\text{+}}$ ion is $17$.  The neutral atom that has seventeen electrons is found to be chlorine.  Therefore, the ${\text{Ar}}^{\text{+}}$ ion is found to be isoelectronic with chlorine atom.