Chapter 2, Problem 2.62P

(a)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{F}}^{-}$ 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 ${\text{F}}^{-}$.  As this is a negatively charged ion, it is an anion.  The symbol present in the ion represent the element fluorine.  Atomic number of fluorine is $9$.  This means it contains $9$ protons and $9$ electrons.  The ion given has one negative charge.  This means it is formed when one electron is added to the fluorine atom.  This is represented as shown below.

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

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

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

Nitrogen has an atomic number of $7$.  Gain of three electron results in formation of ${\text{N}}^{3-}$ 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 $10$.

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 $10$.

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

(b)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{Se}}^{2-}$ ion has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Se}}^{2-}$.  As this is a negatively charged ion, it is an anion.  The symbol present in the ion represent the element selenium.  Atomic number of selenium is $34$.  This means it contains $34$ protons and $34$ electrons.  The ion given has two negative charges.  This means it is formed when two electrons are added to the selenium atom.  This is represented as shown below.

    $\text{Se}+2{\text{e}}^{-}\to {\text{Se}}^{2-}$

Therefore, the number of electrons present in ${\text{Se}}^{2-}$ is $36$.

Isoelectronic ions that have $36$ electrons like ${\text{Se}}^{2-}$ are given below.

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 $36$.

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

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

(c)

Interpretation Introduction

Interpretation:

Three ions that are isoelectronic with ${\text{Ba}}^{2+}$ ion has to be given.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{Ba}}^{2+}$.  As this is a positively charged ion, it is a cation.  The symbol present in the ion represent the element barium.  Atomic number of barium is $56$.  This means it contains $56$ protons and $56$ electrons.  The ion given has two positive charges.  This means it is formed when two electrons are removed from the barium atom.  This is represented as shown below.

    $\text{Ba}\to {\text{Ba}}^{\text{2+}}+2{\text{e}}^{-}$

Therefore, the number of electrons present in ${\text{Ba}}^{\text{2+}}$ is $54$.

Isoelectronic ions that have $54$ electrons like ${\text{Ba}}^{\text{2+}}$ are given below.

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 $54$.

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 $54$.

Iodine has an atomic number of $53$.  Gain of one electron results in formation of ${\text{I}}^{-}$ ion with total number of electrons as $54$.

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

(d)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer part (a).

Explanation of Solution

Given ion is ${\text{La}}^{3+}$.  As this is a positively charged ion, it is a cation.  The symbol present in the ion represent the element lanthanum.  Atomic number of lanthanum is $57$.  This means it contains $57$ protons and $57$ electrons.  The ion given has three positive charges.  This means it is formed when three electrons are removed from the lanthanum atom.  This is represented as shown below.

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

Therefore, the number of electrons present in ${\text{La}}^{3+}$ is $54$.

Isoelectronic ions that have $54$ electrons like ${\text{La}}^{\text{3+}}$ are given below.

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 $54$.

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 $54$.

Iodine has an atomic number of $53$.  Gain of one electron results in formation of ${\text{I}}^{-}$ ion with total number of electrons as $54$.

Therefore, the ions that are isoelectronic with ${\text{La}}^{\text{3+}}$ are ${\text{Cs}}^{+}$, ${\text{Te}}^{2-}$, and ${\text{I}}^{-}$.