Oxidation number of each element in CHF 3 has to be determined. Concept Introduction: Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements. 1. Elements present in their free state have zero oxidation number. 2. Oxidation number of hydrogen is generally + 1 , except for metal hydrides. 3. Oxidation number of oxygen is − 2 , except for peroxides. 4. Metals have positive oxidation numbers. 5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds. 6. Sum of oxidation numbers of different elements in neutral atom is zero. 7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Identify the amino acids by name. Illustrate a titration curve for this tetrapeptide indicating the pKa's for each ionizable groups and identify the pI for this tetrapeptide. please help

↓ ina xSign x Sign X labs X Intro X Cop Xa chat X My Cx Grac X Laur x Laur xash learning.com/ihub/assessment/f188d950-dd73-11e0-9572-0800200c9a66/d591b3f2-d5f7-4983-843c-0d00c1c0340b/f2b47861-07c4-4d1b-a1ee-e7db2 +949 pts /3400 K Question 16 of 34 > © Macmillan Learning Draw the major E2 reaction product formed when cis-1-chloro-2-ethylcyclohexane (shown) reacts with hydroxide ion in DMSO. H CH2CH3 H H HO- H H H Cl DMSO H H C Select Draw Templates More C H 0 2 Erase

A common buffer for stabilizing antibodies is 100 mM Histidine at pH 7.0. Describe the preparation of this buffer beginning with L-Histidine monohydrochloride monohydrate and 1 M NaOH. Be certain to show the buffering reaction that includes the conjugate acid and base.

Answer 2

Question

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Answer 6

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

Answer 7

Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CHF3$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$ , except for metal hydrides.

3. Oxidation number of oxygen is $−2$ , except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

Answer 8

(a)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen group, its oxidation number is $−1$ . Since $CHF3$ is not hydride compound, oxidation state of $H$ is $+1$ .

Expression for oxidation number in $CHF3$ is as follows:

$[Oxidation number of C+Oxidation number of H+3(Oxidation number of F)]=0$ (1)

Rearrange equation (1) for oxidation number of $C$ .

$Oxidation number of C=[−Oxidation number of H−3(Oxidation number of F)]$ (2)

Substitute $−1$ for oxidation number of $F$ and $+1$ for oxidation number of $H$ in equation (2).

$Oxidation number of C=[−(+1)−3(−1)]=−1+3=+2$

Hence, oxidation number of $C$ is $+2$ , that of $H$ is $+1$ and that of $F$ is $−1$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

Since $P2O5$ is not peroxide, oxidation number of $O$ is $−2$ .

Expression for oxidation number in $P2O5$ is as follows:

$[2(Oxidation number of P)+5(Oxidation number of O)]=0$ (3)

Rearrange equation (3) for oxidation number of $P$ .

$Oxidation number of P=[−5(Oxidation number of O)2]$ (4)

Substitute $−2$ for oxidation number of $O$ in equation (4).

$Oxidation number of P=[−5(−2)2]=+102=+5$

Hence, oxidation number of $P$ is $+5$ and that of $O$ is $−2$ .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 20

(c)

Expert Solution

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $F$ is $−1$ .

Expression for oxidation number in $SF6$ is as follows:

$[Oxidation number of S+6(Oxidation number of F)]=0$ (5)

Rearrange equation (5) for oxidation number of $S$ .

$Oxidation number of S=−6(Oxidation number of F)$ (6)

Substitute $−1$ for oxidation number of $F$ in equation (6).

$Oxidation number of S=−6(−1)=+6$

Hence, oxidation number of $F$ is $−1$ and that of $S$ is $+6$ .

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

Answer 25

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 26

(d)

Expert Solution

Explanation of Solution

Since $SnSO4$ is not peroxide compound, oxidation state of $O$ is $−2$ . Since chlorine is member of halogen group, its oxidation number is $−1$ . $Sn$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$ .

Expression for oxidation number in $SnSO4$ is as follows:

$[Oxidation number of Sn+Oxidation number of S+4(Oxidation number of O)]=0$ (7)

Rearrange equation (7) for oxidation number of $S$ .

$Oxidation number of S=[−Oxidation number of Sn−4(Oxidation number of O)]$ (8)

Substitute $−2$ for oxidation number of $O$ and $+2$ for oxidation number of $Sn$ in equation (8).

$Oxidation number of S=[−(+2)−4(−2)]=−2+8=+6$

Hence, oxidation number of $Sn$ is $+2$ , that of $S$ is $+6$ and that of $O$ is $−2$ .

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

Answer 31

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 32

(e)

Expert Solution

Explanation of Solution

Since $CH3OH$ is not peroxide, oxidation number of $O$ is $−2$ . It is not hydride compound so oxidation number of $H$ is $+1$ .

Expression for oxidation number in $CH3OH$ is as follows:

$[Oxidation number of C+4(Oxidation number of H)+Oxidation number of O]=0$ (9)

Rearrange equation (9) for oxidation number of $C$ .

$Oxidation number of C=[−4(Oxidation number of H)−Oxidation number of O]$ (10)

Substitute $−2$ for oxidation number of $O$ and $+1$ for oxidation number of $H$ in equation (10).

$Oxidation number of C=[−4(+1)−(−2)]=−4+2=−2$

Hence, oxidation state of $O$ is $−2$ , that of $H$ is $+1$ and that of $C$ is $−2$ .

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Answer 37

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 38

(f)

Expert Solution

Explanation of Solution

Since $H3PO4$ is neither hydride nor peroxide compound, oxidation number of $H$ is $+1$ and that of $O$ is $−2$ .

Expression for oxidation number in $H3PO4$ is as follows:

$[3(Oxidation number of H)+Oxidation number of P+4(Oxidation number of O)]=0$ (11)

Rearrange equation (11) for oxidation number of $P$ .

$Oxidation number of P=[−3(Oxidation number of H)−4(Oxidation number of O)]$ (12)

Substitute $+1$ for oxidation number of $H$ and $−2$ for oxidation number of $O$ in equation (12).

$Oxidation number of P=[−3(+1)−4(−2)]=−3+8=+5$

Hence, oxidation number of $H$ is $+1$ , that of $P$ is $+5$ and that of $O$ is $−2$ .

Answer 39

(f)

Expert Solution

Answer 40

(f)

Expert Solution

Answer 41

Expert Solution

Answer 42

Ch. 17.1 - Prob. 17.1P Ch. 17.1 - Prob. 17.2P Ch. 17.1 - Prob. 17.3P Ch. 17.1 - Prob. 17.4P Ch. 17.2 - Prob. 17.5P Ch. 17.3 - Prob. 17.6P Ch. 17.3 - Prob. 17.7P Ch. 17.4 - Prob. 17.8P Ch. 17.5 - Prob. 17.9P Ch. 17 - Prob. 1RQ

Videos

Explanation of Solution

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Want to see more full solutions like this?

Chapter 17 Solutions

Videos

Explanation of Solution

Interpretation: Oxidation number of each element in P2O5<m:math><m:mrow><m:msub><m:mtext>P</m:mtext><m:mn>2</m:mn></m:msub><m:msub><m:mtext>O</m:mtext><m:mn>5</m:mn></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Oxidation number of each element in SF6<m:math><m:mrow><m:msub><m:mrow><m:mtext>SF</m:mtext></m:mrow><m:mn>6</m:mn></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Oxidation number of each element in SnSO4<m:math><m:mrow><m:msub><m:mrow><m:mtext>SnSO</m:mtext></m:mrow><m:mn>4</m:mn></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Oxidation number of each element in CH3OH<m:math><m:mrow><m:msub><m:mrow><m:mtext>CH</m:mtext></m:mrow><m:mtext>3</m:mtext></m:msub><m:mtext>OH</m:mtext></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Oxidation number of each element in H3PO4<m:math><m:mrow><m:msub><m:mtext>H</m:mtext><m:mtext>3</m:mtext></m:msub><m:msub><m:mrow><m:mtext>PO</m:mtext></m:mrow><m:mtext>4</m:mtext></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Want to see more full solutions like this?

Chapter 17 Solutions

Interpretation:

Oxidation number of each element in $P2O5$ has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

Oxidation number of each element in $SF6$ has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

Oxidation number of each element in $SnSO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

Oxidation number of each element in $CH3OH$ has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

Oxidation number of each element in $H3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).