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