Determining oxidation numbers for remaining atoms in HIO3, H3PO4, Ba(OH)2, and BaPO4
I have no idea if I am doing this right I don't even know how to do this
For HIO3 I added the charges as +1 and 3(-2) which i got -5 then I did it the similar way for the others and the answers I got is for H3PO4 is -3
Then for Ba(OH)2 I got 0 then for BaPO4 I got 22.
Please show me how to do this correctly....
I'm explaining for HI03
First we assign oxidation numbers to Oxygen and Hydrogen then to any other atom.
So oxygen is -2*3 and H is +1.
So now let I's oxidation number be x.
+1+x+(3*-2)=0
(0 because the overall compound has 0 charge)
+1+x-6=0
x-5=0
x=5
To determine the oxidation numbers of atoms in a compound, it helps to understand some general rules:
1. The oxidation number of an element in its elemental form is always zero. For example, in H2, both hydrogen atoms have an oxidation number of 0.
2. In a neutral compound, the sum of the oxidation numbers of all the atoms is zero.
3. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
Now, let's determine the oxidation numbers for each element in the given compounds:
1. HIO3 (Iodic Acid)
We'll assign a variable for the oxidation number of iodine (I). Since oxygen is more electronegative than iodine, we assign oxygen an oxidation number of -2.
Let x be the oxidation number of iodine (I). Since there are three oxygen atoms, the equation becomes:
x + 3(-2) = 0
x - 6 = 0
x = +6
Therefore, the oxidation number of iodine (I) in HIO3 is +6.
2. H3PO4 (Phosphoric Acid)
Similar to the previous compound, we'll assign oxidation numbers to oxygen and hydrogen. We let x be the oxidation number of phosphorus (P).
The equation becomes:
3(1) + x + 4(-2) = 0
3 + x - 8 = 0
x - 5 = 0
x = +5
Therefore, the oxidation number of phosphorus (P) in H3PO4 is +5.
3. Ba(OH)2 (Barium Hydroxide)
In this compound, we'll determine the oxidation number for barium (Ba) and oxygen (O), while remembering that hydroxide is a polyatomic ion with a charge of -1.
For barium (Ba), the equation becomes:
Ba + 2(1) = 0
Ba + 2 = 0
Ba = -2
However, this result doesn't seem to make sense since barium is an alkaline earth metal and typically has an oxidation number of +2. It's important to note that in this case, we have an exception to Rule 2.
For oxygen (O), the equation becomes:
2(-2) = -2
Therefore, the oxidation number of barium (Ba) is +2 and the oxidation number of oxygen (O) is -2.
4. BaPO4 (Barium Phosphate)
In this compound, we'll determine the oxidation number for barium (Ba), phosphorus (P), and oxygen (O).
For barium (Ba), the equation becomes:
Ba + P + 4(-2) = 0
Ba + P - 8 = 0
Ba + P = 8
We can assign arbitrary variables for Ba and P. Let's assume Ba = +2 (based on Rule 2), so the equation becomes:
+2 + P = 8
P = +6
Therefore, the oxidation number of phosphorus (P) in BaPO4 is +6.
For oxygen (O), the equation becomes:
4(-2) = -8
Therefore, the oxidation number of oxygen (O) in BaPO4 is -2.
In summary, the oxidation numbers for the remaining atoms in the given compounds are:
- HIO3: I = +6
- H3PO4: P = +5
- Ba(OH)2: Ba = +2, O = -2
- BaPO4: Ba = +2, P = +6, O = -2