What is the oxidation state of S in Rb2SO4?
is the charge = 0?
+6. Here is a site that tells you how to figure oxidation state. (Broken Link Removed)
To determine the oxidation state of S (sulfur) in Rb2SO4 (rubidium sulfate), we need to consider the oxidation states of the other elements in the compound.
Let's start by assigning a variable for the oxidation state of S. Let's say x represents the oxidation state of S.
Next, we know that Rb (Rubidium) has an oxidation state of +1. Since there are two Rb atoms in Rb2SO4, the total oxidation state contribution of Rb is +2.
The sulfate ion (SO4^2-) has an overall charge of -2. Since oxygen usually has an oxidation state of -2, the total oxidation state contribution of oxygen is -8. Since there are four oxygen atoms, the total oxidation state contribution of oxygen is -8 * 4 = -32.
The overall charge of Rb2SO4 is zero (neutral compound), so the sum of the oxidation states of all the atoms must be zero. Therefore, we can set up the following equation:
(+2) + x + (-32) = 0
Simplifying the equation, we have:
x - 30 = 0
Now, solving for x:
x = 30
Therefore, the oxidation state of S in Rb2SO4 is +30.