Chapter 18, Problem 49A

(a)

Interpretation Introduction

Interpretation: The oxidation state of all of the atoms in ${\text{MnO}}_{\text{2}}$ is needs to be explained.

Concept Introduction: In a compound, the charge assigned to an atom is called its oxidation state. Oxidation is resulted by an increase in oxidation state and reduction is results by decreasing oxidation state.

Answer to Problem 49A

In ${\text{MnO}}_{\text{2}}$ , oxidation state of Mn is $+4$ and O is $-2$ .

Explanation of Solution

The electronegativity value of O is higher than Mn. Charge of O as anion is $-2$ . Hence, oxidation state of O is $-2$ . The charge of two O in ${\text{MnO}}_{\text{2}}$ is $-4\left(2\times -2\right)$ . The overall charge of ${\text{MnO}}_{\text{2}}$ is zero. Hence, oxidation state of Mn is $+4$ .

(b)

Interpretation Introduction

Interpretation: The oxidation state of all of the atoms in ${\text{BaCrO}}_{\text{4}}$ is needs to be explained.

Concept Introduction: In a compound, the charge assigned to an atom is called oxidation state. Oxidation is resulted by an increase in oxidation state and reduction is results by decreasing oxidation state.

Answer to Problem 49A

In ${\text{BaCrO}}_{\text{4}}$ , oxidation state of Cr is $+6$ , O is $-2$ and Ba is +2.

Explanation of Solution

The electronegativity of O is higher than chromium and barium. Charge of O as anion is $-2$ . Hence, oxidation state of O is $-2$ . The total charge of four O in ${\text{BaCrO}}_{\text{4}}$ is $-8\left(4\times -2\right)$ . Barium belongs to group 2 elements so, it forms +2 cations. The sum of the oxidation state of an electrically neutral compound is zero. Hence, the oxidation state of Cr is $+6$ .

(c)

Interpretation Introduction

Interpretation: The oxidation state of all of the atoms in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ is needs to be explained.

Concept Introduction: In a compound, the charge assigned to an atom is called oxidation state. Oxidation is resulted by an increase in oxidation state and reduction is results by decreasing oxidation state.

Answer to Problem 49A

In ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ , oxidation state of H is $+1$ , O is $-2$ and S is +4.

Explanation of Solution

As electronegativity of O is higher than S and H. Charge of O as anion is $-2$ . Hence, oxidation state of O is $-2$ . The charge of three O in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ is $-6\left(3\times -2\right)$ . The oxidation state of one H is +1. The sum of the oxidation state of electrically neutral compound is zero. Hence, oxidation state of S is $+4$ .

(d)

Interpretation Introduction

Interpretation: The oxidation state of all of the atoms in ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ is needs to be explained.

Concept Introduction: In a compound, the charge assigned to an atom is called oxidation state. Oxidation is resulted by an increase in oxidation state and reduction is results by decreasing oxidation state.

Answer to Problem 49A

In ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ , oxidation state of Ca is $+2$ , O is $-2$ and P is $+5$ .

Explanation of Solution

The electronegativity of O is higher than Ca and P. Charge of O as anion is $-2$ . Hence, oxidation state of O is $-2$ . Calcium is a group 2 elements so, it forms +2 cations. Its oxidation state is +2. The sum of the oxidation state of electrically neutral compound is zero. Hence, the oxidation state of P is $+5$ .