Chapter 20, Problem 23E

(a)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are $\text{s, p, d}$ and $\text{f}$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]{\text{nf}}^{\text{x}}{\text{nd}}^{\text{y}}{\text{ns}}^{\text{z}}$

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of $\text{Ti,}{\text{Ti}}^{2+}$ and ${\text{Ti}}^{4+}$.

Answer to Problem 23E

Answer

The electron configuration of $\text{Ti}$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}$. The electron configuration of ${\text{Ti}}^{2+}$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{2}}$. The electron configuration of ${\text{Ti}}^{4+}$ is $\left[\text{Ar}\right]{\text{3d}}^{\text{0}}$.

Explanation of Solution

Explanation

Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are $\text{s,p,d}$ and $\text{f}$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]{\text{nf}}^{\text{x}}{\text{nd}}^{\text{y}}{\text{ns}}^{\text{z}}$

Where,

• $\text{n}$ is the shell of principal orbital.
• $\text{x}$ is the number of electrons in $\text{f}$ orbital.
• $\text{y}$ is the number of electrons in $\text{d}$ orbital.
• $\text{z}$ is the number of electrons in $\text{s}$ orbital.

Titanium is a group $4$ element. The atomic number of Titanium $\left(\text{T}\text{}\text{i}\right)$ is $\text{22}$ and its electronic configuration is,

$\left[\text{Ar}\right]{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}$

Titanium loses two electrons to acquire two positive charges and becomes ${\text{Ti}}^{2+}$. The electronic configuration of ${\text{Ti}}^{2+}$ is,

${\text{Ti}}^{2+}=\text{Ti}-2{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{T}\text{}\text{i}\right)$ in the above equation.

$\begin{array}{c}{\text{Ti}}^{2+}=\text{Ti}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}-2{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{2}}\end{array}$

Titanium loses four electrons to acquire four positive charges and becomes ${\text{Ti}}^{4+}$. The electronic configuration of ${\text{Ti}}^{4+}$ is,

${\text{Ti}}^{4+}=\text{Ti}-4{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{T}\text{}\text{i}\right)$ in the above equation.

$\begin{array}{c}{\text{Ti}}^{4+}=\text{Ti}-4{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{2}}{\text{4s}}^{\text{2}}-4{\text{e}}^{-}\\ =\left[\text{Ar}\right]{\text{3d}}^{\text{0}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are $\text{s, p, d}$ and $\text{f}$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]{\text{nf}}^{\text{x}}{\text{nd}}^{\text{y}}{\text{ns}}^{\text{z}}$

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of $\text{Re,}{\text{Re}}^{2+}$ and ${\text{Re}}^{3+}$.

Answer to Problem 23E

Answer

The electron configuration of $\text{Re}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{5}}{\text{6s}}^{\text{2}}$. The electron configuration of ${\text{Re}}^{2+}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{5}}$. The electron configuration of ${\text{Re}}^{3+}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{4}}$.

Explanation of Solution

Explanation

Rhenium is a group $7$ element. The atomic number of Rhenium $\left(\text{Re}\right)$ is $\text{75}$ and its electronic configuration is,

$\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{5}}{\text{6s}}^{\text{2}}$

Rhenium loses two electrons to acquire two positive charges and becomes ${\text{Re}}^{2+}$. The electronic configuration of ${\text{Re}}^{2+}$ is,

${\text{Re}}^{2+}=\text{Re}-2{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{Re}\right)$ in the above equation.

$\begin{array}{c}{\text{Re}}^{2+}=\text{Re}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]4{\text{f}}^{\text{14}}{\text{5d}}^{\text{5}}{\text{6s}}^{\text{2}}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]4{\text{f}}^{\text{14}}{\text{5d}}^{\text{5}}\end{array}$

Rhenium loses three electrons to acquire three positive charges and becomes ${\text{Re}}^{3+}$. The electronic configuration of ${\text{Re}}^{3+}$ is,

${\text{Re}}^{3+}=\text{Re}-3{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{Re}\right)$ in the above equation.

$\begin{array}{c}{\text{Re}}^{3+}=\text{Re}-3{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{5}}{\text{6s}}^{\text{2}}-3{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{4}}\end{array}$

(c)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are $\text{s, p, d}$ and $\text{f}$.

The metals from group $\text{3-12}$ are known as first row transition metals. The general form of valence electron configuration of these metals is,

$\left[\text{Nobel}\text{gas}\right]{\text{nf}}^{\text{x}}{\text{nd}}^{\text{y}}{\text{ns}}^{\text{z}}$

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of $\text{Ir,}{\text{Ir}}^{2+}$ and ${\text{Ir}}^{3+}$.

Answer to Problem 23E

Answer

The electron configuration of $\text{Ir}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}{\text{6s}}^{\text{2}}$. The electron configuration of ${\text{Ir}}^{2+}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}$. The electron configuration of ${\text{Ir}}^{3+}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{6}}$.

Explanation of Solution

Explanation

Iridium is a group $9$ element. The atomic number of Iridium $\left(\text{Ir}\right)$ is $\text{77}$ and its electronic configuration is,

$\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}{\text{6s}}^{\text{2}}$

Iridium loses two electrons to acquire two positive charges and becomes ${\text{Ir}}^{2+}$. The electronic configuration of ${\text{Ir}}^{2+}$ is,

${\text{Ir}}^{2+}=\text{Ir}-2{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{Ir}\right)$ in the above equation.

$\begin{array}{c}{\text{Ir}}^{2+}=\text{Ir}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}{\text{6s}}^{\text{2}}-2{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}\end{array}$

Iridium loses three electrons to acquire three positive charges and becomes ${\text{Ir}}^{3+}$. The electronic configuration of ${\text{Ir}}^{3+}$ is,

${\text{Ir}}^{3+}=\text{Ir}-3{\text{e}}^{-}$

Substitute the electron configuration of $\left(\text{Ir}\right)$ in the above equation.

$\begin{array}{c}{\text{Ir}}^{3+}=\text{Ir}-3{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{7}}{\text{6s}}^{\text{2}}-3{\text{e}}^{-}\\ =\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{6}}\end{array}$