# 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$

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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