CUSO4.5H2O,Find the Oxidation number of S?
The oxidation number of Sulfur (S) in CuSO4·5H2O can be calculated using the known oxidation numbers of the other elements and the overall charge of the compound.
The oxidation number of Copper (Cu) is +2.
The oxidation number of Oxygen (O) is usually -2, but in this case, each oxygen atom is bonded to a copper atom, making it a peroxide ion (O2-), so the oxidation number is -1.
The oxidation number of Hydrogen (H) is +1.
The overall charge of the compound is 0 since it is neutral.
Let the oxidation number of Sulfur (S) be x.
The compound can be written as Cu(SO4)·5H2O.
The sum of the oxidation numbers in a compound must equal the overall charge. Hence, we can set up the equation:
(+2) + x + 4(-2) + 5(+1) + 5(-2) = 0
Simplifying the equation, we get:
+2 + x - 8 + 5 - 10 = 0
Combining like terms, we get:
x - 11 = 0
Solving for x, we get:
x = +11
Therefore, the oxidation number of Sulfur (S) in CuSO4·5H2O is +11.