Calculate oxidation no. of underlined elements
1. (NH4)2SO4 (S is underlined)
2. CaH2 (Ca is underlined )
3.CH3COOH (Both C are underlined)
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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
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