What are the oxidation numbers 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) $$\underline{C}H_{3}\underline{C}OOH$$

Asked by Abhisek | 1 year ago |  80

##### Solution :-

(a) $$K\underline{I}_{3}$$

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

1(+1) + 3(x) = 0

1 + 3x = 0

x = $$-\dfrac{1}{3}$$

Oxidation no. cannot be fractional. Hence, consider the structure of $$KI_{3}$$​.

In $$KI_{3}$$​ molecule, an iodine atom forms coordinate covalent bond with an iodine molecule. Therefore, in $$KI_{3}$$​ molecule, the oxidation no. of I atoms forming the molecule $$I_{2}$$​ is 0, while the oxidation no. of I atom which is forming coordinate bond is -1. Two of the four S atoms have an O.N. of +5, whereas the other two have an O.N. of 0

(b) $$H_{2}\underline{S}_{4}O_{6}$$

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 4(x) + 6(-2) = 0

2 + 4x -12 = 0

4x = 10

x = $$+2\dfrac{1}{2}$$

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule. The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) $$\underline{Fe}_{3}O_{4}$$

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

3(x) + 4(-2) = 0

3x -8 = 0

x = $$\dfrac{8}{3}$$

Oxidation no. cannot be fractional.

One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.

$$FeO^{+2} Fe_2O_3$$

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

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 1(-2) = 0

2x + 4 -2 = 0

x = -2

Therefore, oxidation no. of C is -2.

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

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 2(-2) = 0

2x + 4 -4 = 0

x = 0

Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in $$CH_{3}COOH$$

Answered by Abhisek | 1 year ago

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