Chapter 21, Problem 14PS

(a)

Interpretation Introduction

Interpretation: The complete balanced equation has to be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group $\text{1A}$ and group $\text{2A}$ and elements from group $\text{3A}$ to $\text{8A}$ are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  $\text{M}\to {\text{M}}^{n+}+n{e}^{-}$

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    $\text{X}+n{e}^{-}\to {\text{X}}^{n-}$

The metals of group $\text{1A}$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $\text{1A}$ elements is $+1$. Similarly, group $\text{2A}$ elements form $+2$ ions by losing two electrons and have an oxidation number of $+2$. The non-metal gains these electrons to form anions with $-1$ and $-2$ charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than $2$. If the difference in electronegativity is more than $2$ then ionic compounds are formed.

Answer to Problem 14PS

The complete balanced equation for the reaction of potassium and iodine is:

    $2K\left(s\right)+I_2\left(g\right)\to 2KI\left(s\right)$

Explanation of Solution

Potassium belongs to group $\text{1A}$ of periodic table and has an oxidation number of $+1$. Thus, it loses one electron to attain noble gas configuration.

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

This electron is gained by iodine to form an anion with one negative charge. Iodine belongs to halogen family and it has the oxidation number of $-1$.

    ${\text{I}}_{\text{2}}+e^{\_}\to 2{\text{I}}^{-}$

The number of electrons in both the equations is same. Thus an ionic compound is formed in which potassium has $+1$ charge and iodine bears $-1$ charge. Thus, the formula of the product is $\text{KI}$.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation Since iodine is present as ${\text{I}}_{\text{2}}$ in the reaction, hence the stoichiometric coefficient of $\text{KI}$ is $2$ due to which potassium also has a stoichiometric coefficient of $2$.

Thus, the overall balanced equation is:

    $\text{2K}\left(\text{s}\right){\text{+I}}_{\text{2}}\left(\text{g}\right)\to \text{2KI}\left(\text{s}\right)$

(b)

Interpretation Introduction

Interpretation: The complete balanced equation has to be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group $\text{1A}$ and group $\text{2A}$ and elements from group $\text{3A}$ to $\text{8A}$ are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  $\text{M}\to {\text{M}}^{n+}+n{e}^{-}$

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    $\text{X}+n{e}^{-}\to {\text{X}}^{n-}$

The metals of group $\text{1A}$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $\text{1A}$ elements is $+1$. Similarly, group $\text{2A}$ elements form $+2$ ions by losing two electrons and have an oxidation number of $+2$. The non-metal gains these electrons to form anions with $-1$ and $-2$ charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than $2$. If the difference in electronegativity is more than $2$ then ionic compounds are formed.

Answer to Problem 14PS

The complete balanced equation for the reaction of barium and oxygen is:

    $2Ba\left(s\right)+O_2\left(g\right)\to 2BaO\left(s\right)$

Explanation of Solution

Barium belongs to group $\text{2A}$ of periodic table and thus has $+2$ oxidation number. Barium loses two electrons to attain noble gas configuration.

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

These two electrons are gained by the oxygen leading to the formation of an ionic compound. Oxygen belongs to the sulfur family and exists in -2 oxidation number.

  ${\text{O}}_{\text{2}}\text{+2}{e}^{-}\to 2{\text{O}}^{2-}$

The number of electrons in both the equations is same. Barium has a charge of $+2$ and oxygen has a charge of $-2$. Thus, the formula of the product formed is $\text{BaO}$.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since oxygen is present as ${\text{O}}_{\text{2}}$ in the reaction, hence the stoichiometric coefficient of $\text{BaO}$ is $2$ due to which barium also has a stoichiometric coefficient of $2$.

Thus, the overall balanced equation is:

    $\text{2Ba}\left(\text{s}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\to \text{2BaO}\left(\text{s}\right)$

(c)

Interpretation Introduction

Interpretation: The complete balanced equation has to be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group $\text{1A}$ and group $\text{2A}$ and elements from group $\text{3A}$ to $\text{8A}$ are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  $\text{M}\to {\text{M}}^{n+}+n{e}^{-}$

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    $\text{X}+n{e}^{-}\to {\text{X}}^{n-}$

The metals of group $\text{1A}$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $\text{1A}$ elements is $+1$. Similarly, group $\text{2A}$ elements form $+2$ ions by losing two electrons and have an oxidation number of $+2$. The non-metal gains these electrons to form anions with $-1$ and $-2$ charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than $2$. If the difference in electronegativity is more than $2$ then ionic compounds are formed.

Answer to Problem 14PS

The complete balanced equation for the reaction of aluminium with sulfur is:

    $16Al\left(s\right)+3S_8\left(s\right)\to 8A{l}_2{S}_3\left(s\right)$

Explanation of Solution

Aluminium belongs to group $\text{3A}$ of the periodic table and has a highest oxidation number of $+3$. It loses three electrons to attain noble gas configuration of filled s and p subshells.

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

Sulphur belongs to oxygen family and exists in -2 oxidation number. These two electrons are gained by the sulphur leading to the formation of a product compound.

    ${\text{S}}_8+2e^{\_}\to 8{\text{S}}^{2-}$

The numbers of electrons are not same in both the equations. Aluminium bear charge and sulfur bears $-2$ charge.

The stoichiometric coefficients are multiplied with species to have an equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since sulphur is present as ${\text{S}}_{\text{8}}$ in the reaction and aluminium forms the compound with three sulphur atoms, hence the stoichiometric coefficients of $\text{Al}$ is $16$, ${\text{S}}_{\text{8}}$ is $3$ and ${\text{Al}}_{\text{2}}{\text{S}}_{\text{3}}$ is $8$.

Thus, the overall balanced equation is:

    $\text{16Al}\left(\text{s}\right){\text{+3S}}_{\text{8}}\left(\text{s}\right)\to {\text{8Al}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

(d)

Interpretation Introduction

Interpretation: The complete balanced equation has to be written for the given reaction.

Concept introduction: Main group elements are categorized as s-block and p-block elements. The s-block elements include metals belonging to group $\text{1A}$ and group $\text{2A}$ and elements from group $\text{3A}$ to $\text{8A}$ are referred to as p-block elements. The reaction of the metals with the nonmetals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which is gained by the nonmetals. The metal gets positively charge and the non-metal attains a negative charge thus forming cations and anions respectively. They do so to attain a noble gas configuration or to attain stability.

  $\text{M}\to {\text{M}}^{n+}+n{e}^{-}$

Here, by losing electrons metal M achieve the noble gas configuration. These electrons are gained by the non-metals X as shown below.

    $\text{X}+n{e}^{-}\to {\text{X}}^{n-}$

The metals of group $\text{1A}$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $\text{1A}$ elements is $+1$. Similarly, group $\text{2A}$ elements form $+2$ ions by losing two electrons and have an oxidation number of $+2$. The non-metal gains these electrons to form anions with $-1$ and $-2$ charge.

The compounds of non-metals with oxides and hydrides are covalent compounds. This is because non-metals are less electropositive and the difference in electronegativity between two elements is less than $2$. If the difference in electronegativity is more than $2$ then ionic compounds are formed.

Answer to Problem 14PS

The complete balanced equation for the reaction of silicon with chlorine is:

    $Si\left(s\right)+2C{l}_2\left(g\right)\to SiC{l}_4\left(l\right)$

Explanation of Solution

Silicon belongs to group $\text{4A}$ of the periodic table and has four electrons in the outermost shell. The electronic configuration of silicon is $\left[\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{2}}$. Silicon can either gain four electrons or can lose its four electrons to form an ionic compound. However, a very high energy is required in both the cases. Thus, silicon forms covalent compounds.

The electronegativity difference between silicon and chlorine is less than $2$. Thus, the compound formed is covalent compound by sharing of electrons between silicon and chlorine. The oxidation number of silicon is $+4$. The oxidation number of chlorine is $-1$. Thus, the compound formed is ${\text{SiCl}}_{\text{4}}$.

The stoichiometric coefficients are multiplied with species to have equal number of atoms on both the reactant and product side, for a balanced chemical equation. Since chlorine is present as ${\text{Cl}}_{\text{2}}$ in the reaction and silicon forms compound with four chlorine atoms, thus the stoichiometric coefficient of ${\text{Cl}}_{\text{2}}$ is $2$.

Thus, the overall balanced equation is:

    $\text{Si}\left(\text{s}\right){\text{+2Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{SiCl}}_{\text{4}}\left(\text{l}\right)$