Question 2

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

(a) K__I___{3}

(b) H_{2}__S___{4}O_{6}

(c) __Fe___{3}O_{4 }

(d) __C__H_{3}__C__H_{2}OH

(e) __C__H_{3}__C__OOH

Answer

(a) K__I___{3 }

Let assume oxidation number of l is x.

In KI_{3}, the oxidation number (O.N.) of K is +1.

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

⇒ +1 +3x = 0

⇒ 3x = -1

⇒ x = -1/3

Hence, the average oxidation number of I is - 1/3

However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI_{3} to find the oxidation states. In a KI_{3 }molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI_{3} molecule, the O.N. of the two I atoms forming the I_{2} molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(b) H_{2}__S___{4}O_{6}

Let assume oxidation number of S is x.

The oxidation number (O.N.) of H is +1.

The oxidation number (O.N.) of O is -2.

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

⇒ 2 + 4x - 12 = 0

⇒ 4x -10 = 0

⇒ 4x = +10

⇒ x = +10/4

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)** Fe_{3}O_{4 }

Let assume oxidation number of Fe is x.

The oxidation number (O.N.) of O is -2.

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

⇒ 3x - 8 = 0

⇒ 3x = 8

⇒ x = 8/3

However, O.N. cannot be fractional.

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

(d) __C__H_{3}__C__H_{2}OH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + 2(+1) + 1(-2) + 1(+1) = 0

⇒ x +3 + x +2 - 2 + 1 = 0

⇒ 2x + 4 = 0

⇒ 2x = -4

⇒ x = -2

Hence, the oxidation number of C is -2.

(e) __C__H_{3}__C__OOH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0

⇒ 2x + 3 - 2 - 2 + 1 = 0

⇒ 2x + 0 = 0

⇒ x = 0

However, 0 is average O.N. of C.

The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH_{3}COOH.

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Deepak rao
2019-05-19 14:46:36

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