The oxidation number of sulphur in CUSO4.5H2O IS
QUATION
To determine the oxidation number of sulphur (S) in CUSO4.5H2O, we need to use some rules and concepts related to oxidation numbers.
1. Rule: The sum of oxidation numbers in a compound is always equal to zero.
Copper (Cu) and oxygen (O) have fixed oxidation numbers in most compounds. Copper has an oxidation number of +2, and oxygen has an oxidation number of -2.
Given that, let's now calculate the oxidation number of sulphur in CUSO4.5H2O:
The compound CUSO4 contains one copper atom, one sulphur atom, and four oxygen atoms. The compound CUSO4.5H2O additionally contains five water molecules (H2O).
Let's consider the oxidation number of copper (Cu) as +2 and oxygen (O) as -2.
We can set up an equation to find the oxidation number of sulphur:
(+2) + (oxidation number of sulphur) + 4(-2) + 5(oxidation number of hydrogen in H2O) = 0
Simplifying the equation:
+2 + (oxidation number of sulphur) - 8 + 10(oxidation number of hydrogen in H2O) = 0
The oxidation number of hydrogen in water (H2O) is +1 since hydrogen usually has an oxidation number of +1 in compounds.
2 + (oxidation number of sulphur) - 8 + 10(1) = 0
Simplifying further:
(oxidation number of sulphur) + 2 = 8
(oxidation number of sulphur) = 6
Therefore, the oxidation number of sulphur (S) in CUSO4.5H2O is +6.