Find the oxidation number of: A. chlorine in ClO4- ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) B. manganese in MnO4- ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) C. cobalt in Co ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) Find the oxidation number of: A. iron in Fe(OH)2 ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) B. nickel in Ni ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) C. manganese in MnO4-
Find the oxidation number of:
| A. | chlorine in ClO4- | ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) |
|---|---|---|
| B. | manganese in MnO4- | ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) |
| C. | cobalt in Co | ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) |
Find the oxidation number of:
| A. | iron in Fe(OH)2 | ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) |
|---|---|---|
| B. | nickel in Ni | ___(-7-6-5-4-3-2-10+1+2+3+4+5+6+7) |
| C. | manganese in MnO4- |
The sum of oxidation number of all the atoms present in any anion = charge on the anion
A) chlorine in ClO4-
Since O will be having -2 oxidation number as it has 6 valence electron and needs 2 electron to complete octet ( O can not expand its octet due to non availability of vacant d orbitals )
Assuming the oxidation number of Cl is y
Hence y + (-2) X 4 = Oxidation number of Cl + oxidation number of O X 4 = charge on ion = -1
Hence y -8 = -1
=> y = oxidation number of Cl = +7
B) manganese in MnO4-
Since O will be having -2 oxidation number as it has 6 valence electron and needs 2 electron to complete octet ( O can not expand its octet due to non availability of vacant d orbitals )
Assuming the oxidation number of Mn is y
Hence y + (-2) X 4 = Oxidation number of Mn + oxidation number of O X 4 = charge on ion = -1
Hence y -8 = -1
=> y = oxidation number of Mn = +7
C) cobalt in Co
Assuming the oxidation number of Mn is y
Hence y = Oxidation number of Mn = charge on ion = 0
Hence y = 0
=> y = oxidation number of Co = 0
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