Chapter 5, Problem 116QRT

(a)

Interpretation Introduction

Interpretation:

The period in which sulphur is located in periodic table has to be located.

Concept Introduction:

Period:  Horizontal rows in periodic table are called periods.  In periodic table, seven periods are present, with each one beginning at the far left.

Explanation of Solution

A modern periodic table is given by

Figure 1

Sulphur belongs to p-block element of periodic table.  The outermost electrons present in third shell of the atom.  Therefore, sulphur belongs to third period.

(b)

Interpretation Introduction

Interpretation:

The complete electron configuration of sulphur has to be written.

Concept Introduction:

Electron configuration:  The complete description of the atomic orbitals occupied by all the electrons in an atom or monoatomic ion is known as its electron configuration.

Explanation of Solution

Sulphur belongs to p-block element of periodic table.  It has $16$ electrons and $16$ protons.  According to Aufbau principle, electrons are arranged in increasing order of energy level.  The electron configuration of sulphur is given by

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

(c)

Interpretation Introduction

Interpretation:

The atom of an element involved in given reaction has smallest first ionization energy and smallest radius has to be identified.

Concept Introduction:

Electron configuration:  The complete description of the atomic orbitals occupied by all the electrons in an atom or monoatomic ion is known as its electron configuration.

Atomic radius:  It is defined as distance between nucleus of atom and electron present in outermost valence shell of the atom.

Explanation of Solution

Given,

    ${\text{SO}}_{\text{2}}{}_{}\text{+}{\text{2Cl}}_{\text{2}}\stackrel{}{\to }{\text{SOCl}}_{\text{2}}\text{+}{\text{OCl}}_{\text{2}}$

The smallest first ionization energy of the three atoms is S, and the smallest radius is O atom.  According to periodic trends, elements present in far right of periodic table have highest ionization energy and smallest radius.

(d)

Interpretation Introduction

Interpretation:

The mass of chlorine needed to form $\text{675g}$ of thionyl chloride has to be calculated.

Explanation of Solution

Given,

    ${\text{SO}}_{\text{2}}{}_{}\text{+}{\text{2Cl}}_{\text{2}}\stackrel{}{\to }{\text{SOCl}}_{\text{2}}\text{+}{\text{OCl}}_{\text{2}}$

    $\begin{array}{l}{\text{675 g SOCl}}_{\text{2}}\text{×}\frac{{\text{1 mol SOCl}}_{\text{2}}}{\text{118}{\text{.970 g SOCl}}_{\text{2}}}\text{×}\frac{{\text{2 mol Cl}}_{\text{2}}}{{\text{1 mol SOCl}}_{\text{2}}}\text{×}\frac{\text{70}{\text{.906 g Cl}}_{\text{2}}}{{\text{1 mol Cl}}_{\text{2}}}\\ \\ \text{=}{\text{805 g Cl}}_{\text{2}}\end{array}$

The mass of chlorine needed is ${\text{805 g Cl}}_{\text{2}}$

(e)

Interpretation Introduction

Interpretation:

The theoretical mass of thionyl chloride has to be calculated.

Explanation of Solution

Given,

    ${\text{SO}}_{\text{2}}{}_{}\text{+}{\text{2Cl}}_{\text{2}}\stackrel{}{\to }{\text{SOCl}}_{\text{2}}\text{+}{\text{OCl}}_{\text{2}}$

    $\begin{array}{l}\text{10}{\text{.0 g SO}}_{\text{2}}\text{×}\frac{{\text{1molSO}}_{\text{2}}}{\text{64}{\text{.064 g SO}}_{\text{2}}}\text{×}\frac{\text{1mol}{\text{SOCl}}_{\text{2}}}{\text{1mol}{\text{SO}}_{\text{2}}}\text{=}\text{0}\text{.156mol}{\text{SOCl}}_{\text{2}}\\ \\ \text{20g}{\text{Cl}}_{\text{2}}\text{×}\frac{\text{1mol}{\text{Cl}}_{\text{2}}}{\text{70}{\text{.906 g Cl}}_{\text{2}}}\text{×}\frac{\text{1mol}{\text{SOCl}}_{\text{2}}}{\text{2mol}{\text{Cl}}_{\text{2}}}\text{=}\text{0}{\text{.141 mol SOCl}}_{\text{2}}\\ \\ {\text{Cl}}_{\text{2}}\text{is the limiting reactant}\text{.}\\ \\ \text{0}{\text{.141 mol SOCl}}_{\text{2}}\text{×}\frac{\text{118}{\text{.970 g SOCl}}_{\text{2}}}{\text{1}\text{mol}{\text{SOCl}}_{\text{2}}}\text{=}\text{16}{\text{.8 g SOCl}}_{\text{2}}\end{array}$

The mass of thionyl chloride formed is $\text{16}{\text{.8 g SOCl}}_{\text{2}}.$