What is the oxidation number of S in CUSO4. 5H2O
To find the oxidation number of S (sulfur) in the compound CUSO4.5H2O, we need to determine the oxidation numbers of the other elements and apply the principle of charge balance.
1. Start by considering the oxidation number of elements that always have a consistent oxidation state:
- Hydrogen (H) usually has an oxidation number of +1. In this case, there are five H atoms, so the total contribution from H is +5.
- Oxygen (O) usually has an oxidation number of -2. In this case, there are four O atoms in the sulfate (SO4) ion and ten O atoms in the water (H2O) molecules, so the total contribution from O is -24.
2. Apply the principle of charge balance:
- Copper (Cu) has a known oxidation number of +2. The sulfate ion (SO4) has a charge of -2. Considering that the overall compound is neutral, the total oxidation number of Cu and S should balance out the total negative charge.
- With this information, we can calculate the oxidation number of S:
- Since Cu has a charge of +2, and there is only one Cu atom, its contribution is +2.
- Adding the charges together, we have: +2 + (-24) + (+5) = -17.
3. Determine the oxidation number of S:
- The sum of the oxidation numbers of all elements in a compound should equal zero because the compound is neutral. From step 2, we found that the sum is -17, so we need to find the oxidation number of S that balances this out.
- Since there is only one S atom, its oxidation number is + 17. Therefore, the oxidation number of S in CUSO4.5H2O is +17.
Note: It is important to remember that oxidation numbers are not necessarily related to the actual charge on the atom but are rather a formalism used to keep track of electron transfer in chemical reactions.