8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results 

(a) K\underline{I}_{3} 

(b) H_{2}\underline{S_{4}}O_{6}

(c) \underline{Fe}_{3}O_{4}

(d) \underline{C}H_{3}\underline{C}H_{2}OH

(e) \underline{C}H_{3}\underline{C}OOH

 

 

Answers (1)

Solution-

O.N of potassium (K)= +1

O.N of hydrogen(H)= +1 (In case of metalic hydride, -1)

O.N of Oxygen(O) = -2 ( In case of peroxide and superoxide it wil be different ON)

 

(a) K\underline{I}_{3} 

1*1 + 3*x = 0

x = (-1/3)

average  O. N. Of I is -\frac{1}{3}   . But it is wrong because  O.N cannot be fractional. So lets try with structure of KI_{3}

K^+(I-I\leftarrow I)^{-1}

O.N of I = -1 (because a coordinate bond is formed between I_{2}  and  I^{-}  ion. Hence O. N of three I atoms are 0,0 and -1, O.N 0 in I_{2} moleculeand -1 in  I^{-} ion.)

                

(b) H_{2}\underline{S_{4}}O_{6}

Assume O.N of S is x

2*1 + 4*x + 6*(-2) = 0 \Rightarrow x=2.5

Fractional O.N  is not possible so try with structure -

        

The two S atom present in the middle has zero (0) O.N and other two have (+5) O.N [ -2 from two O atom and -1 from OH]                        

(c) \underline{Fe}_{3}O_{4}

  If you calculate the oxidation number of Fe in Fe_{3}O_{4} it would be 8/3 and however, O.N cannot be in fractional.

     Fe_{3}O_{4} \:\:\: is \:an\:equimolar \:mixture\: of \:(FeO)\: and \:(Fe_{2}O_{3}) Here one iron atom has +2 O.N and the other two are of +3 O.N.

 

(d) \bar{C}H_{3}\bar{C}H_{2}OH

      let assume carbon has x oxidation Number

     So,[ x + 1(3) +x +1(2) +(-2)+1 = 0]

             2x = -4

             x=-2

In this molecule two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O. S  of -3 and -1.

 

(e) \bar{C}H_{3}\bar{C}OOH

suppose the oxidation number of Carbon is x.

If we calculate the O.N of x we get x=0

 However, 0 is average O.N. of C atoms. In this molecule two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O.S  of +3 and 3 in CH3COOH This can be more understood by structure-

Here we can see that at right C, +3 O.N (-1 from OH and -2from O atom) and in left C, -3 O.N(contribution from H atom only)

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