Oxidation Numbers:
My question to you is how did Al get a +3 Oxid.#, and S got 6... What is the mathematical explanation for that. I understand the rules that O is almost always -2 for its' oxid #.
Would O be 10 bc you'd multple 4*3 and minus 2?
What compound are you thinking of?
Aluminum sulfate!
Al is in group IIIA and is +3, I think, almost always. But you can tell by looking at the formula of Al2(SO4)3. The tells you oxidation number of Al is +3 and the 2 tell you the sulfate ion is -2. How to do S.
Al is 2*+3 each = +6.
O is 12*-2 = -24
Remember Al2(SO4)3 is zero overall.
So 2*Al + 12*O + 3S = 0
6 + (-24) + 3S = 0
3S = 18
S = +6
Voila
Another way.
You know SO4 polyatomic ion is -2
4*O is -8 so S must be +6 to leave a -2 charge on the ion.
To determine the oxidation numbers of elements in a compound, you need to apply a set of rules. Let's go through the rules and apply them to aluminum (Al) and sulfur (S).
Rule 1: The oxidation number of an element in its elemental state is always 0. This means that when an element exists by itself as a pure element, it has an oxidation number of 0.
Rule 2: In most compounds, oxygen has an oxidation number of -2. Therefore, in most cases, you can assign oxygen an oxidation number of -2, unless stated otherwise.
Rule 3: The sum of oxidation numbers in a neutral compound is always 0. For example, the sum of the oxidation numbers of the elements in Al2O3 (aluminum oxide) must equal 0 since it is a neutral compound.
Now, let's apply these rules to Al and S in the compounds:
1. Aluminum (Al):
In aluminum oxide (Al2O3), we have two aluminum atoms (Al). According to Rule 3, the sum of the oxidation numbers must be 0. Let's assume the oxidation number of aluminum is x.
So, 2x + 3(-2) = 0 (since there are three oxygen atoms, each with an oxidation number of -2)
Solving this equation, we get 2x - 6 = 0
2x = 6
x = 3
Therefore, the oxidation number of aluminum (Al) in aluminum oxide (Al2O3) is +3.
2. Sulfur (S):
In sulfur hexafluoride (SF6), we have one sulfur atom (S) surrounded by six fluorine atoms (F). According to Rule 3, the sum of the oxidation numbers must be 0. Let's assume the oxidation number of sulfur is y.
So, y + 6(-2) = 0 (since there are six fluorine atoms, each with an oxidation number of -1)
Solving this equation, we get y - 12 = 0
y = 12
Therefore, the oxidation number of sulfur (S) in sulfur hexafluoride (SF6) is +6.
Regarding your question about oxygen (O), it is almost always -2 in most compounds, unless stated otherwise. In this case, each oxygen atom in Al2O3 contributes -2 to the overall oxidation number of the compound. Since there are four oxygen atoms, the total contribution is -8. However, it is important to remember that the oxidation number of oxygen is not mathematically calculated by multiplying the number of oxygen atoms by -2. Instead, you need to consider the specific rules and its valence electrons.
I hope this clarifies the concept of oxidation numbers and how they are determined!