whats the oxidation number for all the elements in the following compound

BaNaPO4

http://www.chemteam.info/Redox/Redox-Rules.html

To determine the oxidation number for each element in the compound BaNaPO4, we need to consider the known oxidation numbers of certain elements and make some assumptions.

Let's start by determining the oxidation numbers of the more common elements:

1. Barium (Ba): As an alkaline earth metal, barium typically has an oxidation number of +2.

2. Sodium (Na): Sodium is also an alkali metal, so it usually has an oxidation number of +1.

3. Oxygen (O): Oxygen almost always has an oxidation number of -2, with few exceptions.

Now, let's consider the oxidation number for phosphorus:

Assuming that oxygen has an oxidation number of -2, we can set up an equation to calculate phosphorus's oxidation number:

(oxidation number of Ba) + (oxidation number of Na) + (oxidation number of P) + 4 * (oxidation number of O) = 0

Substituting the known values:

(+2) + (+1) + (oxidation number of P) + 4 * (-2) = 0

Simplifying the equation:

2 + 1 + (oxidation number of P) - 8 = 0

3 + (oxidation number of P) = 8

(oxidation number of P) = 8 - 3 = +5

So, the oxidation number for phosphorus (P) in BaNaPO4 is +5.

In summary, here are the oxidation numbers for each element in the compound BaNaPO4:
- Barium (Ba): +2
- Sodium (Na): +1
- Phosphorus (P): +5
- Oxygen (O): -2