Chapter 2, Problem 2.61P

(a)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{K}}^{\text{+}}$ ion has to be given.

Concept Introduction:

Neutral atom contains equal number of protons and electrons.  When an electron is added or removed from a neutral atom, charged species known as ion is formed.  If electrons are removed, then the formed ion is known as cation and if electrons are added, the formed ion is known as anion.  Atom that has lost electrons will have positive charge and atoms that have gained electrons will have negative charge.

Isoelectronic species are the ones that contain same number of electrons but different number of protons.  They have different chemical and physical properties.

Explanation of Solution

Given ion is ${\text{K}}^{\text{+}}$.  As this is a positively charged ion, it is a cation.  The symbol present in the ion represent the element potassium.  Atomic number of potassium is $19$.  This means it contains $19$ protons and $19$ electrons.  The ion given has one positive charge.  This means it is formed when one electron is removed from the potassium atom.  This is represented as shown below.

    $\text{K}\to {\text{K}}^{\text{+}}+{\text{e}}^{-}$

Therefore, the number of electrons present in ${\text{K}}^{\text{+}}$ is $18$.

Isoelectronic ions that have $18$ electrons like ${\text{K}}^{\text{+}}$ are given below.

Chlorine has an atomic number of $17$.  Gain of one electron results in formation of ${\text{Cl}}^{-}$ ion with total number of electrons as $18$.

Calcium has an atomic number of $20$.  Loss of two electrons results in formation of ${\text{Ca}}^{\text{2+}}$ ion with total number of electrons as $18$.

Scandium has an atomic number of $21$.  Loss of three electrons results in formation of ${\text{Sc}}^{\text{3+}}$ ion with total number of electrons as $18$.

Therefore, the ions that are isoelectronic with ${\text{K}}^{\text{+}}$ are ${\text{Cl}}^{-}$, ${\text{Ca}}^{\text{2+}}$, and ${\text{Sc}}^{\text{3+}}$.

(b)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with $\text{Kr}$ has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given species is $\text{Kr}$.  Atomic number of krypton is $36$.  This means it contains $36$ protons.  Therefore, the number of electrons present in $\text{Kr}$ is $36$.

Isoelectronic ions that have $36$ electrons like $\text{Kr}$ are given below.

Bromine has an atomic number of $35$.  Gain of one electron results in formation of ${\text{Br}}^{-}$ ion with total number of electrons as $36$.

Rubidium has an atomic number of $37$.  Loss of one electron results in formation of ${\text{Rb}}^{\text{+}}$ ion with total number of electrons as $36$.

Strontium has an atomic number of $38$.  Loss of two electrons results in formation of ${\text{Sr}}^{\text{2+}}$ ion with total number of electrons as $36$.

Therefore, the ions that are isoelectronic with $\text{Kr}$ are ${\text{Br}}^{-}$, ${\text{Rb}}^{\text{+}}$, and ${\text{Sr}}^{\text{2+}}$.

(c)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{N}}^{3-}$ ion has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{N}}^{3-}$.  As this is a negatively charged ion, it is an anion.  The symbol present in the ion represent the element nitrogen.  Atomic number of nitrogen is $7$.  This means it contains $7$ protons and $7$ electrons.  The ion given has three negative charges.  This means it is formed when three electrons are added to the nitrogen atom.  This is represented as shown below.

    $\text{N}+3{\text{e}}^{-}\to {\text{N}}^{3-}$

Therefore, the number of electrons present in ${\text{N}}^{3-}$ is $10$.

Isoelectronic ions that have $10$ electrons like ${\text{N}}^{3-}$ are given below.

Fluorine has an atomic number of $9$.  Gain of one electron results in formation of ${\text{F}}^{-}$ ion with total number of electrons as $10$.

Sodium has an atomic number of $11$.  Loss of one electron results in formation of ${\text{Na}}^{\text{+}}$ ion with total number of electrons as $10$.

Magnesium has an atomic number of $12$.  Loss of two electrons results in formation of ${\text{Mg}}^{\text{2+}}$ ion with total number of electrons as $10$.

Therefore, the ions that are isoelectronic with ${\text{N}}^{3-}$ are ${\text{F}}^{-}$, ${\text{Na}}^{\text{+}}$, and ${\text{Mg}}^{\text{2+}}$.

(d)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{I}}^{-}$ ion has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{I}}^{-}$.  As this is a negatively charged ion, it is an anion.  The symbol present in the ion represent the element iodine.  Atomic number of iodine is $53$.  This means it contains $53$ protons and $53$ electrons.  The ion given has one negative charge.  This means it is formed when one electron is added to the iodine atom.  This is represented as shown below.

    $\text{I}+{\text{e}}^{-}\to {\text{I}}^{-}$

Therefore, the number of electrons present in ${\text{I}}^{-}$ is $54$.

Isoelectronic ions that have $54$ electrons like ${\text{I}}^{-}$ are given below.

Tellurium has an atomic number of $52$.  Gain of two electrons results in formation of ${\text{Te}}^{2-}$ ion with total number of electrons as $54$.

Cesium has an atomic number of $55$.  Loss of one electron results in formation of ${\text{Cs}}^{\text{+}}$ ion with total number of electrons as $54$.

Barium has an atomic number of $56$.  Loss of two electrons results in formation of ${\text{Ba}}^{\text{2+}}$ ion with total number of electrons as $54$.

Therefore, the ions that are isoelectronic with ${\text{I}}^{-}$ are ${\text{Te}}^{2-}$, ${\text{Cs}}^{\text{+}}$, and ${\text{Ba}}^{\text{2+}}$.