To write the balanced chemical equation for the formation of N a B r . Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group 1 and group 2 and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds. Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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. M → 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. X + n e − → X n − The metals of group IA form + 1 ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group IA elements is + 1 . Similarly, group IIA 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side. For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Answer 1

Question

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Answer 2

Question

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Answer 3

Question

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Answer 4

Question

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

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Answer 5

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 6

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 7

Chapter 21.2, Problem 1CYU

(a)

Interpretation Introduction

Interpretation: To write the balanced chemical equation for the formation of $NaBr.$

Concept Introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 8

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 9

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $CaSe$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $1A$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 10

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 11

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $PbO$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $1A$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 12

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 13

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation for the formation of $AlCl3$ should be written.

Concept introduction: Main group elements are categorised as s-block and p-block elements. The s-block elements includes metals belonging to group $1$ and group $2$ and elements from group 13 to 18 are referred to as p-block elements. The reaction of themetals of main group elements with the non metals forms ionic compounds.

Ionic compounds are formed by the loss of electrons from the metal which are gained by the non metals. 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.

$M→Mn++ne−$

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

$X+ne−→Xn−$

The metals of group $IA$ form $+1$ ions because the highest oxidation number is always equal to the group number of that element. Thus, the charge on group $IA$ elements is $+1$ . Similarly, group $IIA$ 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 stoichometric coefficents are multiplied with the compounds or species in the chemical equation to have equal number of atoms on both side of the equation. The equation is known as a balanced chemical equation if there are equal number of atoms of each element in both the product and reactant side.

For the naming of an ionic compound, the cation is always named first followed by the anion and the anion ends with the “-ide”.

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 21.2 - Prob. 1CYU Ch. 21.2 - Write the formula for each of the following (a)...Ch. 21.2 - Prob. 3CYU Ch. 21.2 - Prob. 4CYU Ch. 21.4 - Prob. 3RC Ch. 21.5 - Prob. 1Q Ch. 21.5 - Prob. 2Q Ch. 21.8 - Prob. 1Q Ch. 21.8 - Prob. 2Q Ch. 21.8 - Prob. 3Q

Solution Summary: The author explains the balanced chemical equation for the formation of NaBr.

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