Chapter 4, Problem 4.46QP

(a)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of ${\text{N}}_{\text{2}}$ in ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$

• According to the rules, the alkaline earth metals always having $+2$ oxidation number.
• The sum of oxidation numbers of all the atoms present in the molecule is zero

$\begin{array}{l}\text{3(2+)+2X=0}\\ \text{(6+)+2X=0}\\ \text{2X=-6}\\ \text{X=-3}\end{array}$

• The oxidation number of ${\text{N}}_{\text{2}}$ in ${\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}$ is $-3$

(b)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of ${\text{O}}_{\text{2}}$ in ${\text{CsO}}_{\text{2}}$is

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the earth metals always having $+1$ oxidation number only
• The $\text{Cs}$ in $+1$ oxidation state and $\text{O}$ is,

  $\begin{array}{l}\text{(1+)+2(X)=0}\\ \text{2X=-1}\\ \text{X=-}\raisebox{1ex}{$\text{1}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.\end{array}$

• The oxidation number of $\text{O}$ in ${\text{CsO}}_{\text{2}}$ is $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

(c)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{C}$ in ${\text{CaC}}_{\text{2}}$

• According to the rules, the alkaline earth metals always having $+2$ oxidation number.
• The sum of oxidation numbers of all the atoms present in the molecule is zero
• The $\text{Ca}$ in $+2$ oxidation state and C is,

  $\begin{array}{l}\text{1(2+)+2(X)=0}\\ \text{(2+)+2(X)=0}\\ \text{2X=-2}\\ \text{X=-1}\\ \text{C=-1}\end{array}$

• The oxidation number of $\text{C}$ in ${\text{CaC}}_{\text{2}}$ is $-1$

(d)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{C}$ in ${\text{CO}}_{\text{3}}{}^{\text{2-}}$

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is -2 and the general oxidation state of oxygen is $-2$.
• The $\text{O}$ in $-2$ oxidation state and C is,

  $\begin{array}{l}\text{1(X)+3(-2)=-2}\\ \text{(X)+(-6)=-2}\\ \text{X=}\text{+4}\end{array}$

• The oxidation number of $\text{C}$ in ${\text{CO}}_{\text{3}}{}^{\text{2-}}$ is $+4$

(e)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{C}$ in ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}$

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is -2 and the general oxidation state of oxygen is $-2$.
• The $\text{O}$ in $-2$ oxidation state and C is,

  $\begin{array}{l}\text{4(-2)+2(X)=-2}\\ \text{(-8)+2(X)=0}\\ \text{2X=+6}\\ \text{C=+3}\end{array}$

• The oxidation number of $\text{C}$ in ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}$ is $+3$

(f)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{O}$ in ${\text{ZnO}}_{\text{2}}{}^{\text{2-}}$

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the general oxidation state of Oxygen is $-2$.
• The stable  oxidation state and $\text{Zn}$ in $+2$ is and O is,

  $\begin{array}{l}\text{1(+2)+2(X)=-2}\\ \text{(+2)+2X=-2}\\ \text{2X=}-4\\ \text{X = -2}\end{array}$

• The oxidation number of $\text{O}$ in ${\text{ZnO}}_{\text{2}}{}^{\text{2-}}$ is $-2$

(g)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{B}$ in ${\text{NaBH}}_{\text{4}}$

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and general oxidation number for oxygen is $-2$ and hydrogen is $+1$ but in the hydride compound general oxidation number of hydrogen is $-1$
• The $\text{H}$ in $-1$ and $\text{Na}$ is in $\text{+1}$ oxidation states and $\text{B}$ is,

  $\begin{array}{l}\text{1(+1)+1(X)+4(-1)=0}\\ \text{(+1)+X+(-4)=0}\\ \text{X=+3}\end{array}$

• The oxidation number of $\text{B}$ in ${\text{NaBH}}_{\text{4}}$ is $+3$

(h)

Interpretation Introduction

Interpretation:

The oxidation numbers of underlined atoms in given species are should be given.

Concept introduction:

Rules for assigning the Oxidation number:

• The oxidation number of atom in the elemental form is zero.
• The sum of oxidation numbers of all the atoms present in the molecule is zero.
• The general oxidation number for oxygen is $-2$ and hydrogen is $+1$.
• The combination of any atoms in group 7 the oxidation number is varied.
• The Inter halogen compounds the Fluorine is always having $-1$ oxidation state.
• In the oxohalide compounds oxidation number of halide is in positive.
• The general oxidation state of halides is -1.
• The earth metals always having $+1$ oxidation number only.
• The oxidation state of hydrogen in metal hydride is $-1$.

Explanation of Solution

The oxidation number of $\text{W}$ in ${\text{WO}}_{\text{4}}{}^{\text{2-}}$

• According to the rules, the sum of oxidation numbers of all the atoms present in the molecule is zero and the general oxidation state of Oxygen is $-2$.
• The $\text{O}$ in $-2$ oxidation state and $\text{W}$ is,

  $\begin{array}{l}\text{X+4(-2)=-2}\\ \text{X+(-8)=-2}\\ \text{X=}\text{8+(-2)}\\ \text{W=}\text{+6}\end{array}$

• The oxidation number of $\text{W}$ in ${\text{WO}}_{\text{4}}{}^{\text{2-}}$ is $+6$.