21 Chemistry -- Oxidation and Reduction

Calculate the oxidation number of the underlined elements:

Calculate the oxidation number of the underlined elements:

Let the oxidation number of each element to be found be x in each cases.

Here, ON of K, H = 1; ON of O = -2

a) KMnO4

It is a molecule hence, the ON of the molecule is zero.

1 + x + 4(-2) = 0

1 + x - 8 = 0

x - 7 = 0

x = 7

b) K2Cr2O7

It is a molecule hence, the ON of the molecule is zero.

2(1) + 2(x) + 7(-2) = 0

1 + x - 7 = 0

x - 6 = 0

x = 6

c) H2SO4

It is a molecule hence, the ON of the molecule is zero.

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

2 + x - 8 = 0

x - 6 = 0

x = 6

d) CO32-

It is a radical that has -2 charge so, the ON of the given radical is -2.

x + 3(-2) = -2

x - 6 = -2

x = 6 - 2

x = 4

e) Ni(CO)4

Carbon monoxide is a molecule so, its ON is zero.

x + 4 * (0) = 0

x + 0 = 0

x = 0

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