Oxidation Number of P in Pb3O4
To determine the oxidation number of phosphorus (P) in lead tetroxide (Pb3O4), we start by analyzing the oxidation states of the other elements.
- Lead (Pb) typically has an oxidation state of +2.
- Oxygen (O) almost always has an oxidation state of -2.
Calculating the contribution from oxygen:
- There are 4 oxygen atoms: 4 × (-2) = -8.
Now, let’s set up the equation based on the overall charge of the compound (which is neutral):
- 3(Pb) + 4(O) + x(P) = 0
- 3(+2) + 4(-2) + x = 0
- 6 - 8 + x = 0
- x = 2
Thus, the oxidation number of P in Pb3O4 is +2.
Oxidation Number of C in C3O2
For carbon (C) in carbon dioxide (C3O2), we follow a similar approach:
- Oxygen (O) has an oxidation state of -2.
Calculating the contribution from oxygen:
- There are 2 oxygen atoms: 2 × (-2) = -4.
Setting up the equation:
- 3(C) + 2(O) + x(C) = 0
- 3x + 2(-2) = 0
- 3x - 4 = 0
- 3x = 4
- x = +4/3
Thus, the average oxidation number of C in C3O2 is +4/3.
Oxidation Number of Br in Br3O8
To find the oxidation number of bromine (Br) in bromine octoxide (Br3O8):
- Oxygen (O) has an oxidation state of -2.
Calculating the contribution from oxygen:
- There are 8 oxygen atoms: 8 × (-2) = -16.
Setting up the equation:
- 3(Br) + 8(O) + x(Br) = 0
- 3x + 8(-2) = 0
- 3x - 16 = 0
- 3x = 16
- x = +16/3
Thus, the average oxidation number of Br in Br3O8 is +16/3.