Science : Chemical bonding: Problems on determination of Oxidation Number: Solved Example Problems with Answers, Solution

ON of neutral molecule is always zero

Let us take ON of H = +1 and ON of O = -2

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

(+2) + (-2) = 0 thus, ON of H is +1 and ON of O is -2

ON of Na = +1 and ON of Cl = -1

(+1) + (-1) = 0 thus, ON of Na is +1 and ON of Cl is -1

Let ON of S be (x) and we know ON of H = +1 and O = -2

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

(+2) + (x) + (-8) = 0

(x) = +6 therefore, ON of S is +6

Let ON of Cr be x and we know ON of K = +1 and O = -2

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

(+2) + (2x) + (-14) = 0

2x = +12

x = +6 therefore, ON of Cr in K2Cr2O7 is +6

Let ON of Fe be x and we know ON of S = +6 and O = -2

42. + 1 X (+6) + 4 X (-2) = 0

43. + (+6) + (-8) = 0

x = +2 therefore, ON of Fe in FeSO4 is +2

1. Find the oxidation number of Mn in KMnO4

2. Find the oxidation number of Cr in Na2Cr2O7

3. Find the oxidation number of Cu in CuSO4

4. Find the oxidation number of Fe in FeO

Tags : Solved Example Problems with Answers, Solution , 9th Science : Chemical bonding

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9th Science : Chemical bonding : Problems on determination of Oxidation Number | Solved Example Problems with Answers, Solution

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