Chapter 10, Problem 116SCQ

(a)

Interpretation Introduction

Interpretation:

For the given reaction the under given conditions the valence electrons present in the given compound should be determined.

Concept Introduction:

Valence electrons: The outermost electrons present in the compound other than the inner core electrons are denoted as valence electrons.  The valence electrons are responsible for the bond formation.

There are about 19 valence electrons present in the given compound.

Explanation of Solution

The given compound ${\text{ClO}}_{\text{2}}$ contains two oxygen atoms and one chlorine atom. The atomic number of oxygen is 8 it says that there are about 6 valence electrons present in the outer most shell of each oxygen since arrangement of electrons around atom is as follows.

${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{ and 2p}}^{\text{4}}$.

Similarly the chlorine atom contains 7 valence electrons in its outer most shell which totally comprises of $\left(7+2\times 6\right)19$ valence electrons present in the given compound.

(b)

Interpretation Introduction

Interpretation:

For the given reaction the under given conditions the valence electrons present in the given compound ${\text{ClO}}_{\text{2}}$, the electrons dot structure, hybridization shape around the central atom present in the given compound, bond angle and the mass of ${\text{ClO}}_{\text{2}}$ obtained under given reaction conditions should be determined.

Concept introduction:

Lewis dot structure: It is the representation of chemical substance that how the valence are represented as dot and placed around the each atom present in the substance by considering the number of atoms with their valence electrons and number of bonds present in the given chemical substance.

Answer to Problem 116SCQ

The electron dot structure for the given compound is as follows,

Explanation of Solution

The electron dot structure for the given ${\text{ClO}}_{\text{2}}{}^{-}$ is drawn as follows,

  

There are two oxygen atoms and one chlorine atom present in the given compound.

The oxygen atom contains 8 electrons that is represented as follows, ${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{ and 2p}}^{\text{4}}$ which shows that it contains 6 valence electrons with it.

Similarly, the electrons present in the chlorine is represented as follows

${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^6,3s^{2}and3p^5$ shows that there are about 7 valence electrons with it.

Therefore, the chemical substance totally, has 20 valence electrons with it since addition of valence electrons present in the atom of the given compound says that.  There are 2 O atoms each with 6 valence electrons one chlorine atom with 7 valence electrons and one negative charge which denotes present of one extra electrons which totally equals to 20.

The valence electrons are represented as dot around each atom present in the compound such that each atom tends to exhibit stable configuration that is 8electrons in its valence shell.

(c)

Interpretation Introduction

Interpretation:

For the given reaction the under given conditions the angle and the mass of ${\text{ClO}}_{\text{2}}$ obtained under given reaction conditions should be determined.

Concept introduction:

Hybridization refers to mixing of orbitals.  The orbitals of different atoms overlap with each other and form a set of new orbitals called hybrid orbitals.  The number of orbitals that are combined equal to the number of hybrid orbitals formed by combination.

An isolated atom remains in excited state and comes back to ground state.   In this situation, it is difficult for orbitals to hybridize.

Hybridization is a hypothetical concept.  It refers to mixing of atomic orbitals and the resultant orbitals formed are known as hybrid orbitals.  After hybridization, the orbitals cannot be distinguished individually.  Based on hybridization one can predict the geometry of the molecule though some deviations do exist.

Answer to Problem 116SCQ

The chlorine atom in the given compound is found to be ${\text{sp}}^{\text{3}}$

Explanation of Solution

The given compound contains chlorine with two oxygen atoms results to have two bonds with it.

The electronic configuration of Chlorine and oxygen is as follows,

Electronic configuration of chlorine is ${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{,2p}}^6,3s^{2}and3p^5$ and that of oxygen is

  ${\text{1s}}^{\text{2}}{\text{,2s}}^{\text{2}}{\text{ and 2p}}^{\text{4}}$.

${\text{ClO}}_{\text{2}}$ has chlorine at the center gets bonded with two terminal oxygen atoms.  The neutral Chlorine has one unpaired electron in its valence shell which accommodates one additional electron since there are totally three p-subshells where bonding occurs.  Oxygen can accommodate two additional electrons with it since it has only 4 electrons in its p sub shell.

The available orbitals that is s orbitals, one p orbital from chlorine and two p orbitals from oxygen atom gets hybridized and results to form new hybrid orbitals classified as ${\text{sp}}^{\text{3}}$ and the hybridization around chlorine is found to be ${\text{sp}}^{\text{3}}$.

The central chlorine in compound ${\text{ClO}}_{\text{2}}{}^{\text{-}}$ contains two lone pair electrons and the repulsion between the two lone pairs are high which gives bent shape to the molecule.

(d)

Interpretation Introduction

Interpretation:

For the given reaction the under given conditions the valence electrons present in the given compound ${\text{ClO}}_{\text{2}}$, the electrons dot structure, hybridization shape around the central atom present in the given compound, bond angle and the mass of ${\text{ClO}}_{\text{2}}$ obtained under given reaction conditions should be determined.

Concept introduction:

Bond angle: The angle formed between two bonds that is where two atoms gets bonded with the third central atom present.

Answer to Problem 116SCQ

The species ${\text{O}}_{\text{3}}$ will have larger bond angle compared to the other given compound ${\text{ClO}}_{\text{2}}{}^{\text{-}}$

Explanation of Solution

The central chlorine in compound ${\text{ClO}}_{\text{2}}{}^{\text{-}}$ contains two lone pair electrons whereas the central atom oxygen in ${\text{O}}_{\text{3}}$ one lone pair around which makes ${\text{ClO}}_{\text{2}}{}^{\text{-}}$ bond angle less due to lone pair-lone pair repulsions between the lone pairs in ${\text{ClO}}_{\text{2}}{}^{\text{-}}$ is greater than the repulsions present between the bond pair and lone pair in the other given compound ${\text{O}}_{\text{3}}$ since repulsion follows the following order

$\text{lone pair - lone pair > lone pair - bond pair > bond pair - bond pair}$.

Therefore, ${\text{O}}_{\text{3}}$ will have larger bond angle compared to the other given compound ${\text{ClO}}_{\text{2}}{}^{\text{-}}$

(e)

Interpretation Introduction

Interpretation:

For the given reaction the under given conditions the mass of ${\text{ClO}}_{\text{2}}$ obtained under given reaction conditions should be determined.

Concept introduction:

Ideal gas equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas.  Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties.  At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

        $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 116SCQ

The mass of given compound formed is equal to $11.16g$

Explanation of Solution

Given:

  $\begin{array}{l}{\text{Mass of NaClO}}_{\text{2}}\text{ = 15}\text{.6g}\\ {\text{P}}_{{\text{NaClO}}_{\text{2}}}\text{=}\text{1050 mm Hg = }\frac{1050}{760}=1.382atm\\ \text{V}\text{=}\text{1}\text{.45L}\\ {\text{T = 22}}^{\text{o}}\text{C = 273}\text{.15+22 = 295}\text{.15K}\\ {\text{Mass of ClO}}_{\text{2}}produced=?\end{array}$

First the given mass of $NaCl{O}_2$ is converted into moles as follows,

  $\begin{array}{l}\text{moles =}\frac{\text{mass}}{\text{molar mass}}\\ =\frac{\text{15}\text{.6 g}}{\text{90}\text{.44 g/mol}}\\ =0.1725moles\end{array}$

There are about $0.1725moles$ of $NaCl{O}_2$ present in the given amount.

Then, the amount of $C{l}_2$ gas is calculated as follows,

  $\begin{array}{l}\text{PV = nRT}\\ \text{n = }\frac{\text{PV}}{\text{RT}}=\frac{1.382atm\times \text{1}\text{.45L}}{\text{0}\text{.0821}\times \text{295}\text{.15K}}=0.0827moles\end{array}$

Now, analyzing the given chemical equation it is evident that 2 moles of $NaCl{O}_2$ requires 1 mole of $C{l}_2$ hence one mole of $NaCl{O}_2$ need $\frac{1}{2}\text{moles}\text{of }C{l}_2$. From the given value it is obvious that we have $0.1725moles$ of $NaCl{O}_2$ which means it requires

  $\text{0}\text{.1725×}\frac{1}{2}\text{ = 0}{\text{.09 moles of Cl}}_{\text{2}}$.

Similarly, there are $\text{0}{\text{.0827 moles of Cl}}_{\text{2}}$ the given reaction says that 1 mole of $C{l}_2$ needs ${\text{2 moles of NaClO}}_{\text{2}}$ hence the given amount of $C{l}_2$ requires $\text{0}\text{.0827}\times \text{2 = 0}\text{.1654 moles}$ of $NaCl{O}_2$.

Analyzing the above calculations we have only about $0.0827moles$ of $C{l}_2$ hence it acts as limiting reagent and the formation of $Cl{O}_2$ depends on the given amount of $C{l}_2$.

Examining the given chemical reaction show that 1 mole of $C{l}_2$ results to give 2 moles of $Cl{O}_2$.

Therefore, available amount of $C{l}_2$ gives $\text{0}\text{.0827moles ×2 = 0}{\text{.1654 moles of ClO}}_{\text{2}}$

  $\begin{array}{l}\text{Moles =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Mass}\text{=Moles×}\text{Molar mass}\\ \text{=0}\text{.1654moles}\times 67.45g/mol\\ =11.16g\end{array}$