Question

Calculate oxidation no. of underlined elements

1. (NH4)2SO4 (S is underlined)

2. CaH2 (Ca is underlined )

3.CH3COOH (Both C are underlined)

Answer 1

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Answer 2

Answer:

1. Let X be the oxidation no. of S

Therefore,

2 x O.N of NH4 + X + 4 x O.N of O = 0

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

X = +6

Hence, the oxidation number of S in (NH4)2SO4 is +6

2. Let X be the oxidation no. of Ca

Therefore,

X + 2 x O.N of H = 0

X + 2 x (-1) = 0 (oxidation number of H is (-1) in this compound)

X = +2

Hence, the oxidation number of C in CaH2 is +2

3. The molecule contains two carbon atoms.

Let the oxidation no. of C1= X

Therefore,

X +3 x O.N of H = 0

X + 3 x (+1) = 0

X = -3

Hence the oxidation number of C1 = -3

Now, ‘

Let the O.N OF C2 = Y

Therefore,

Y + 2 x O.N of O + O.N of H = 0

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

Y = +3

Hence, the oxidation number of C2 is +3