What is the oxidation number for copper in CuSO4? This is a common question in chemistry, especially when dealing with inorganic compounds. The oxidation number, also known as the oxidation state, is a measure of the degree of oxidation of an atom in a chemical compound. It is an important concept in understanding the chemical behavior and reactivity of elements in various compounds.
The compound CuSO4, which is copper(II) sulfate, consists of copper, sulfur, and oxygen atoms. In this compound, the oxidation number of copper can be determined by using the rules of oxidation numbers. According to the rules, the oxidation number of oxygen is usually -2, and the overall charge of the compound is neutral.
To find the oxidation number of copper in CuSO4, we can start by assigning the oxidation number of oxygen as -2. Since there are four oxygen atoms in the compound, the total oxidation number contributed by oxygen is -8. The sulfur atom in CuSO4 is usually assigned an oxidation number of +6, which is consistent with its position in the periodic table.
Now, let’s denote the oxidation number of copper as x. The sum of the oxidation numbers of all atoms in a neutral compound must equal zero. Therefore, we can set up the following equation:
x + (+6) + (-8) = 0
Simplifying the equation, we get:
x – 2 = 0
Adding 2 to both sides of the equation, we find:
x = +2
Hence, the oxidation number for copper in CuSO4 is +2. This indicates that copper has lost two electrons and is in a +2 oxidation state in the compound. Understanding the oxidation number of copper in CuSO4 can help us predict its reactivity and chemical behavior in various reactions.
