Question

Oxidation number of boron in na b(no3)4

Answer 1

We have the molecular formula:

`NaB(NO_3)_4,`

The first number of oxidation that we need to find is oxygen, which is -2 (you can see it in the periodic table) and multiply that number with the number of oxygen (12 oxygens because 3 x 4 = 12):

`\text{Oxidation of oxygen= -2}\cdot12=-24.`

It is giving us a really low number because is negative, so the other elements must be positive, for example, the unique oxidation number of Na is +1 and the unique number of boron is +3, and based on that we do an algebraic sum to find the oxidation number of N and the result of this sum must be zero:

`-24+3+1=-20.`

N have negative and positive oxidation numbers (2, 3, 4 and 5). In the formula, we have 4 nitrogens and that number is multiplying by anyone of the oxidation numbers. In our case is 5, because 5 x 4 = 20.

The sum of all the number oxidations must be zero:

`Na^(+1)B^(+3)(N^(+5)O^(-2)_3)_4.`

The super indexes indicate the oxidation number of each element.

So the conclusion is the following:

`\text{Oxidation number:}`
`Na=\text{ +1, B=+3, N=+5 and O=-16.}`