Chapter 2, Problem 2.19E

(a)

Interpretation Introduction

Interpretation:

Chemical formulas of two compounds formed between Thallium and oxygen has to be

determined.

Concept Introduction:

The formula to determine moles from mass percentages is as follows:

  $\text{No of Moles}=\frac{Mass\text{percent}}{Atomicmass}$

Explanation of Solution

The formula to determine amount of $Tl$ is as follows:

  $AmountofTl=\frac{Mass\text{percent}ofTl}{Atomicmass}$        (1)

Substitute $89.49\text{ g}$ for mass percent of $Tl$ and $204.38\text{ g/mol}$ for atomic mass in equation (1) to calculate amount of $Tl$ in compound II.

  $\begin{array}{c}AmountofTl=\frac{89.49\text{ g}}{204.38\text{ g/mol}}\\ =0.4378\text{ mol}\end{array}$

Substitute $\text{96}\text{.23 g}$ for mass percent of $Tl$ and $204.38\text{ g/mol}$ for atomic mass in equation (1) to calculate amount of $Tl$ in compound II.

  $\begin{array}{c}AmountofTl=\frac{\text{96}\text{.23 g}}{204.38\text{ g/mol}}\\ =0.4708\text{ mol}\end{array}$

The formula to determine amount of $\text{O}$$Tl$ is as follows:

  $Amountof\text{O}=\frac{Mass\text{percent}of\text{O}}{Atomicmass}$        (2)

Substitute $\text{10}\text{.51 g}$ for mass percent of $\text{O}$ and $204.38\text{ g/mol}$ for atomic mass in equation (2) to calculate amount of $\text{O}$ in compound II.

  $\begin{array}{c}Amountof\text{O}=\frac{10.51\text{ g}}{16\text{ g/mol}}\\ =0.6568\text{ mol}\end{array}$

Substitute $\text{10}\text{.51 g}$ for the value of mass percent and $204.38\text{ g/mol}$ for atomic mass of $\text{O}$ in equation (2) to calculate amount of $\text{O}$ in compound II.

  $\begin{array}{c}Amountof\text{O}=\frac{\text{3}\text{.77 g}}{16\text{ g/mol}}\\ =0.2356\text{ mol}\end{array}$

Thus the amount can be represented as;

  $T{l}_{0.4378}{\text{O}}_{0.6568}\to T{l}_{\frac{0.4378}{0.4378}}{\text{O}}_{\frac{0.6568}{0.4378}}\to T{l}_1{\text{O}}_{1.5}$

Hence the chemical formula for compound I is $T{l}_2{\text{O}}_3$.

Similarly, for compound II the amount can be represented as;

  $T{l}_{0.4708}{\text{O}}_{0.2356}\to T{l}_{1.99}{\text{O}}_1\to T{l}_2\text{O}$

Hence the chemical formula for compound II is $T{l}_2\text{O}$.

(b)

Interpretation Introduction

Interpretation:

Oxidation number of thallium in each compound has to be determined.

Concept Introduction:

The sum of the individual oxidation states of various species present in compound is equal to the charge possessed by overall compound. If the molecule is neutral than the charge is sum is equated to zero.

Explanation of Solution

The formula to determine oxidation number for $T{l}_2{\text{O}}_3$ is as follows:

  $2\left(\text{Oxidation Number of Tl}\right)+3\left(\text{Oxidation Number of O}\right)=0$        (3)

Substitute $-2$ for the oxidation number of $\text{O}$ in equation (3) to calculate oxidation number of $Tl$.

  $\begin{array}{c}2\left(OxidationNumberofTl\right)+3\left(-2\right)=0\\ OxidationNumberofTl=\frac{6}{2}\\ =3\end{array}$

The formula to determine oxidation number for $T{l}_2\text{O}$ is as follows:

  $2\left(\text{Oxidation Number of Tl}\right)+1\left(\text{Oxidation Number of O}\right)=0$        (4)

Substitute $-2$ for the oxidation number of $\text{O}$ in equation (4) to calculate oxidation number of $Tl$.

  $\begin{array}{c}2\left(OxidationNumberofTl\right)+1\left(-2\right)=0\\ OxidationNumberofTl=\frac{2}{2}\\ =1\end{array}$

Therefore $Tl$ has an oxidation number of $+3$ in $T{l}_2{\text{O}}_3$ and an oxidation number of $+1$ in

$T{l}_2\text{O}$.

(c)

Interpretation Introduction

Interpretation:

Electron configuration for each thallium ion has to be written.

Concept Introduction:

In chemical reactions only outer electrons are involved thus the electronic configuration is simplified when the core electrons are represented by the closest noble gases followed by valence electrons. For instance, nitrogen with atomic number seven is written as $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^3$.

The convention followed to remove or add electrons is electrons of largest principal quantum number are lost first. In case of subshells of the same $\text{n}$ value, electrons are lost from the subshell with highest $\text{l}$ value.

Explanation of Solution

$Tl$ has an oxidation number of $+3$ in $T{l}_2{\text{O}}_3$ and an oxidation number of $+1$ in $T{l}_2\text{O}$.

$Tl$ is a p-block metal with atomic number of 81 that gives rise to electronic configuration $\left[\text{Xe}\right]{\text{6s}}^{\text{2}}{\text{4f}}^{14}{\text{5d}}^{10}{\text{6p}}^1$. To form ${\text{Tl}}^{3+}$ ion it loses three electrons; two from $\text{6s}$ and one from $\text{6p}$. This results in $\left[\text{Xe}\right]{\text{4f}}^{14}{\text{5d}}^{10}$ electronic configuration for ${\text{Tl}}^{3+}$.

Similarly in case of $T{l}_2\text{O}$ it loses one electrons; from $\text{6p}$. This results in $\left[\text{Xe}\right]{\text{4f}}^{14}{\text{5d}}^{10}$ electronic configuration $\left[\text{Xe}\right]{\text{4f}}^{14}{\text{5d}}^{10}{\text{6s}}^{\text{2}}$ for ${\text{Tl}}^{+}$.

(d)

Interpretation Introduction

Interpretation:

The melting point data has to be used to determine the more covalent compound.

Concept Introduction:

In any polar covalent bond, the difference in electronegativities gives rise to partial positive and negative charges respectively on the cation and anions.

The covalent character is determined with respect to the relative extent of distortion of anionic cloud of anion. For instance, highly charged small alkali metal cation such as ${\text{Li}}^{+}$ is capable to cause polarization in large-sized halides such as iodide ion. Thus there is considerable covalent character in $\text{LiI}$.

In general, greater is the charge on the metal cation or smaller is the size of cation more is its polarizing power to distort the electron cloud of anion and thus more is the covalent character.

Explanation of Solution

$Tl$ has an oxidation number of $+3$ in $T{l}_2{\text{O}}_3$ and an oxidation number of $+1$ in $T{l}_2\text{O}$.

$T{l}_2{\text{O}}_3$ has very high melting point relative to $T{l}_2\text{O}$. Since greater energy is required to overcome the strong interionic forces of attraction in ionic solids so it can be concluded that $T{l}_2{\text{O}}_3$ must be an ionic compound. Hence $T{l}_2\text{O}$ has more covalent character.

This answer is not consistent with prediction based on polarizing abilities of the two cations. Since $T{l}_2{\text{O}}_3$ has ${\text{Tl}}^{3+}$ that is smaller in size thus it should have more polarizing ability. However, the melting pint data predict $T{l}_2\text{O}$ with ${\text{Tl}}^{+}$ as more covalent.