Chapter 21, Problem 17PS

(a)

Interpretation Introduction

Interpretation: The complete balanced equation should 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 17PS

The complete balanced equation for the reaction of sodium with bromine is:

    $2Na\left(s\right)+B{r}_2\left(l\right)\to 2NaBr\left(s\right)$

Explanation of Solution

Sodium 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.

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

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

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

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

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 bromine is present as ${\text{Br}}_{\text{2}}$ in the reaction, hence the stoichiometric coefficient of $\text{NaBr}$ is $2$ due to which sodium also has a stoichiometric coefficient of $2$.

Thus, the overall balanced equation is:

    $\text{2Na}\left(\text{s}\right){\text{+Br}}_{\text{2}}\left(\text{l}\right)\to \text{2NaBr}\left(\text{s}\right)$

(b)

Interpretation Introduction

Interpretation: The complete balanced equation should 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 17PS

The complete balanced equation for the reaction of magnesium with oxygen is:

  $2Mg\left(s\right)+O_2\left(g\right)\to 2MgO\left(s\right)$

Explanation of Solution

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

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

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. Magnesium has a charge of $+2$ and oxygen has a charge of $-2$. Thus, the formula of the product formed is $\text{MgO}$.

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 $\text{MgO}$ is $2$ due to which magnesium also has a stoichiometric coefficient of $2$.

Thus, the overall balanced equation is:

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

(c)

Interpretation Introduction

Interpretation: The complete balanced equation should 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 17PS

The complete balanced equation for the reaction of Aluminium with fluorine is:

$2Al\left(s\right)+3F_2\left(g\right)\to 2Al{F}_3\left(s\right)$

Explanation of Solution

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

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

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

  ${\text{F}}_2+e^{-}\to 2{\text{F}}^{-}$

The number of electrons is not same in both the equations. Aluminium bear charge and fluorine bear charge. Thus, the formula of the product is ${\text{AlF}}_{\text{3}}$.

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 there is ${\text{F}}_{\text{2}}$ in the reactant side but aluminium forms the compound with three fluorine atoms, hence the stoichiometric coefficients of $\text{Al}$ is $2$, ${\text{F}}_{\text{2}}$ is $3$ and ${\text{AlF}}_3$ is $2$.

Thus, the overall balanced equation is:

    $\text{2Al}\left(\text{s}\right){\text{+3F}}_{\text{2}}\left(\text{g}\right)\to {\text{2AlF}}_{\text{3}}\left(\text{s}\right)$

(d)

Interpretation Introduction

Interpretation: The complete balanced equation should 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 17PS

The complete balanced equation for the reaction of carbon with oxygen is:

    $C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$

Explanation of Solution

Carbon belongs to group $\text{4A}$ of the periodic table and has four electrons in the outermost shell. The electronic configuration of carbon is ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{2}}$. Carbon 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, carbon forms covalent compounds.

The electronegativity difference between carbon and oxygen is less than $2$. Thus, the compound formed is covalent compound by sharing of electrons between carbon and oxygen.

An excess of oxygen is assumed for the reaction. Thus, the product formed is ${\text{CO}}_{\text{2}}$.

Thus, the overall balanced equation is:

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