Oxidation Numbers
AlH(SO4)3 and the overall charge is -2
BaCr2O7
http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm
Here is a better page.
http://www.chemteam.info/Redox/Redox-Rules.html
To determine the oxidation numbers of the elements in a compound, we need to apply some rules and guidelines.
AlH(SO4)3 is a compound consisting of aluminum, hydrogen, and sulfate ions. In order to determine the oxidation numbers of the elements, we will assign oxidation numbers based on a set of rules:
1. The oxidation number of an atom in its elemental form is always zero. This means that the oxidation number of hydrogen (H) in AlH(SO4)3 is zero.
2. The sum of the oxidation numbers in a compound must be equal to the overall charge of the compound. In this case, the overall charge of AlH(SO4)3 is -2.
3. For polyatomic ions, the sum of the oxidation numbers is equal to the charge of the ion. The sulfate ion (SO4) has a charge of -2, so the sum of the oxidation numbers of sulfur and oxygen in SO4 will be -2.
Now, let's determine the oxidation number for each element in AlH(SO4)3:
Since hydrogen (H) has an oxidation number of zero, we can omit it from our calculations.
Let's assign the oxidation number of sulfur (S) as x. Since there are four oxygen atoms in SO4, each with an oxidation number of -2 (the sum of the oxidation numbers must equal the charge, which is -2), the equation will be:
x + (4 * -2) = -2
x - 8 = -2
x = +6
Therefore, the oxidation number for sulfur in SO4 is +6.
Now let's assign the oxidation number for aluminum (Al) as y. Since we know that the sum of the oxidation numbers must be equal to -2, and we have already assigned the oxidation number for sulfur as +6, the equation will be:
y + (3 * -2) + 6 = -2
y - 6 + 6 = -2
y = -2
Therefore, the oxidation number for aluminum in AlH(SO4)3 is -2.
In the compound BaCr2O7, we apply the same principles to determine the oxidation numbers. The overall charge is not given, so we assume that the compound is neutral (i.e., the sum of the oxidation numbers equals zero).
Let's assign the oxidation number for barium (Ba) as x. Since barium is an alkali earth metal, it has a fixed oxidation number of +2 in most compounds.
Now let's assign the oxidation number for chromium (Cr) as y. Since there are two chromium atoms in the compound, the equation will be:
2y + (7 * -2) = 0
2y - 14 = 0
2y = 14
y = +7
Therefore, the oxidation number for chromium in BaCr2O7 is +7.
In summary:
- The oxidation number of sulfur in AlH(SO4)3 is +6.
- The oxidation number of aluminum in AlH(SO4)3 is -2.
- The oxidation number of barium in BaCr2O7 is +2.
- The oxidation number of chromium in BaCr2O7 is +7.