in addition to previous Q:
for eg:
why is the oxidation state (#) of Ca in CaH2 "+2"? According to a p.table I have here +2 is the only oxidation # for Ca...but how is that figured?
One other eg:
the oxidation state (#) of the maganese atom in KMnO4.6H2O is "+7". How is this so & not: 2,3,or4?
If you assign charges to oxygen, it's (-2). There are 4 of these in the molecule. So 4*(-2) = -8. potassium is a +1 charge. So now you have (-8) +1 = (-7). Therefore, since there's only one Mn, that one atom is a +7 charge so that the -7 is balanced out to be neutral (since it's not an ion).
I hope this helps.. Just assign the charges for what you know, then calculate based upon the overall charge and the amount of that element in the molecule.
For the Ca ion, it is in group IIA (or 2 depending upon the system you are using). Alkali metals (group IA) and alkaline earth metals (IIA) are assigned oxidation states of +1 and {2 respectively. The other way you can do it is to know that hydrogen, in hydrides, (and this is a hydride) is
-1. Since there are two of them it makes the charge on H -2 for two H atoms; therefore, Ca must be +2 for the compound of CaH2 to be neutral.
To determine the oxidation state (#) of an element in a compound, you need to consider a few factors. Here's how you can figure out the oxidation state for each example provided:
For CaH2, start by assigning a charge of -1 to hydrogen since it is in a hydride compound. Since there are two hydrogen atoms, the total charge from hydrogen is -2.
Next, you know that the compound must be electrically neutral, so the sum of the charges for all the elements must add up to 0. Since the overall charge from hydrogen is -2, the oxidation state of calcium, denoted by "+x", must satisfy the equation: +x - 2 = 0. Solving this equation gives you x = +2. Therefore, the oxidation state of calcium in CaH2 is +2.
For KMnO4.6H2O, start by examining the charges of other elements involved. Oxygen is typically assigned an oxidation state of -2 in compounds, so having four oxygen atoms gives a total charge of -8 (-2 * 4 = -8).
Potassium is in group IA, so it has a standard oxidation state of +1.
To determine the oxidation state of manganese (Mn), you need to balance out the charges to make the compound neutral. Since the overall charge from oxygen is -8, and the charge from potassium is +1, the oxidation state of manganese, denoted by "+x", must satisfy the equation: +x - 8 + 1 = 0.
Simplifying the equation gives you x - 7 = 0, which means x = +7. Therefore, the oxidation state of manganese in KMnO4.6H2O is +7.
In summary, you can determine the oxidation state of an element in a compound by assigning charges to other elements, using the overall charge of the compound, and considering the standard oxidation states for elements in specific groups on the periodic table.