The oxidation number of sulphur in Cuso4,5H2ois what
The oxidation number of sulfur in CuSO4 · 5H2O is +6.
To find the oxidation number of sulfur in CuSO4,5H2O, we can follow the steps below:
1. Identify and assign oxidation numbers to other elements in the compound.
- The oxidation number of copper (Cu) is +2 since it is a common oxidation state for copper.
- The oxidation number of oxygen (O) is -2, except in peroxides where it is -1. Since there are four oxygen atoms in CuSO4, the total oxidation number contributed by oxygen is -8.
- The oxidation number of hydrogen (H) is +1 in most compounds.
2. Determine the total oxidation number of the compound.
- The compound has a neutral charge, so the sum of the oxidation numbers must be zero.
- From the oxidation numbers assigned to copper and oxygen, we have: (+2) + (-8) = -6.
- Since there are no other elements in the compound, the total oxidation number must come from sulfur.
3. Calculate the oxidation number for sulfur.
- Let's assume the oxidation number of sulfur is x.
- The compound CuSO4 contains one copper atom and one sulfur atom. Therefore, the total oxidation number for copper and sulfur must be zero.
- So, we can write the equation: (+2) + x + (-8) = 0.
- Simplifying the equation: x - 6 = 0.
- Adding 6 to both sides, we get: x = 6.
Therefore, the oxidation number of sulfur (S) in CuSO4,5H2O is +6.