What is the oxidation number of Cr2(SO4)3
zero.
Cr = +3
S is +6
O is -2
To determine the oxidation number of an element in a compound, you need to consider its electronegativity, valence electrons, and the overall charge of the compound.
In the case of Cr2(SO4)3, we can break down the compound into its components:
1. "Cr" represents the element chromium.
2. "SO4" represents the polyatomic ion sulfate, which consists of one sulfur atom (S) and four oxygen atoms (O).
To find the oxidation number of chromium (Cr), we can follow these steps:
1. Identify the oxidation number of sulfate (SO4):
- Oxygen (O) typically has an oxidation number of -2.
- The overall charge of sulfate (SO4) is -2. Therefore, the sum of the oxidation numbers must equal -2. Considering there are four oxygen atoms, the total oxidation number for oxygen is (4 x -2) = -8.
- Therefore, the oxidation number of sulfur (S) in sulfate (SO4) is +6, as the sum of -8 and +6 equals -2.
2. For Cr2(SO4)3, the overall charge of the compound is 0 since it is not specified.
- The total oxidation number of sulfate (SO4) is (3 x -2) = -6.
- Since there are two chromium (Cr) atoms, the combined oxidation numbers of the chromium atoms must equal +6 for the sum to be -6 and achieve a compound with a charge of 0.
- Therefore, the oxidation number of each chromium (Cr) atom in Cr2(SO4)3 is +3.
In summary, the oxidation number of each chromium (Cr) atom in Cr2(SO4)3 is +3.