Chapter 5, Problem 5.40P

(a)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with ${\text{O}}^{2-}$ 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 ${\text{O}}^{2-}$.  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 ${\text{O}}^{2-}$ ion is formed when two electrons are added to the valence shell.  In this case the two electrons are added to $2p$ orbital.  Therefore, the electronic configuration of ${\text{O}}^{2-}$ ion is written as shown below.

    ${\text{O}}^{2-}=1s^2 2s^2 2p^6$

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

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

Total number of electrons that is present in ${\text{O}}^{2-}$ ion is $10$.  The neutral atom that has eighteen electrons is found to be neon.  Therefore, the ${\text{O}}^{2-}$ ion is found to be isoelectronic with neon atom.

(b)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with ${\text{Ca}}^{\text{+}}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Ca}}^{+}$.  Atomic number of calcium is $20$.  Ground state electronic configuration of calcium is shown below.

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

The given ${\text{Ca}}^{+}$ ion is formed when one electron is removed from the valence shell.  In this case the one electron is removed $4s$ orbital.  Therefore, the electronic configuration of ${\text{Ca}}^{+}$ ion is written as shown below.

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

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

    $\begin{array}{c}\text{Total}\text{number}\text{of}\text{electrons}=2+2+6+2+6+1\\ =19\end{array}$

Total number of electrons that is present in ${\text{Ca}}^{\text{+}}$ ion is $19$.  The neutral atom that has nineteen electrons is found to be potassium.  Therefore, the ${\text{Ca}}^{\text{+}}$ ion is found to be isoelectronic with potassium atom.

(c)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with ${\text{He}}^{\text{+}}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

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

    $\text{He}=1s^2$

The given ${\text{He}}^{+}$ ion is formed when one electron is removed from the valence shell.  In this case the one electrons are removed from the $1s$ orbital.  Therefore, the electronic configuration of ${\text{He}}^{+}$ ion is written as shown below.

    ${\text{He}}^{+}=1s^1$

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

    $\text{Total}\text{number}\text{of}\text{electrons}=1$

Total number of electrons that is present in ${\text{He}}^{\text{+}}$ ion is $1$.  The neutral atom that has one electron is found to be hydrogen.  Therefore, the ${\text{He}}^{\text{+}}$ ion is found to be isoelectronic with hydrogen atom.

(d)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with ${\text{Pb}}^{2+}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Pb}}^{2+}$.  Atomic number of lead is $82$.  Ground state electronic configuration of lead is shown below.

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

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

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

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

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

Total number of electrons that is present in ${\text{Pb}}^{\text{2+}}$ ion is $80$.  The neutral atom that has eighty electrons is found to be mercury.  Therefore, the ${\text{Pb}}^{\text{2+}}$ ion is found to be isoelectronic with mercury atom.

(e)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with ${\text{N}}^{3-}$ has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{N}}^{3-}$.  Atomic number of nitrogen is $7$.  Ground state electronic configuration of nitrogen is shown below.

    $\text{N}=1s^2 2s^2 2p^3$

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

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

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

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

Total number of electrons that is present in ${\text{N}}^{3-}$ ion is $10$.  The neutral atom that has eighteen electrons is found to be neon.  Therefore, the ${\text{N}}^{3-}$ ion is found to be isoelectronic with neon atom.