1. H₂C₂O₄ (Oxalic Acid)
* H: Hydrogen usually has an oxidation number of +1. Since there are two hydrogen atoms, the total contribution from hydrogen is +2.
* O: Oxygen typically has an oxidation number of -2. With four oxygen atoms, the total contribution from oxygen is -8.
* C: Let the oxidation number of carbon be 'x'. There are two carbon atoms, so their total contribution is 2x.
To find 'x', we can use the fact that the sum of the oxidation numbers in a neutral compound is zero:
+2 + 2x - 8 = 0
2x = +6
x = +3
Therefore, the oxidation numbers are:
* H: +1
* C: +3
* O: -2
2. Mn₂O₇ (Manganese(VII) Oxide)
* O: Oxygen has an oxidation number of -2, and there are seven oxygen atoms, totaling -14.
* Mn: Let the oxidation number of manganese be 'y'. Since there are two manganese atoms, the total contribution is 2y.
The sum of the oxidation numbers is zero:
2y - 14 = 0
2y = +14
y = +7
Therefore, the oxidation numbers are:
* Mn: +7
* O: -2
3. Hg₂Cl₂ (Mercury(I) Chloride)
* Cl: Chlorine usually has an oxidation number of -1. There are two chlorine atoms, contributing a total of -2.
* Hg: Let the oxidation number of mercury be 'z'. Since there are two mercury atoms, the total contribution is 2z.
The sum of the oxidation numbers is zero:
2z - 2 = 0
2z = +2
z = +1
Therefore, the oxidation numbers are:
* Hg: +1
* Cl: -1
4. IF₅ (Iodine Pentafluoride)
* F: Fluorine always has an oxidation number of -1. There are five fluorine atoms, contributing a total of -5.
* I: Let the oxidation number of iodine be 'w'.
The sum of the oxidation numbers is zero:
w - 5 = 0
w = +5
Therefore, the oxidation numbers are:
* I: +5
* F: -1
