Question

Question #dc82b

Answer 1

c. `Mn`

Explanation:

We need to find the oxidation state of `Mn` (manganese) for all 4 choices to determine which one is the lowest.

a. `MnF_2`

Fluorine's oxidation state can only be -1. However, there are two fluorine ions present, so their total charge equals -2.

Since compounds are always electrically neutral, the oxidation state of `Mn` must be +2 .

`+2 -2 = 0`

b. `MnH_3`

Since hydrogen has a higher electronegativity (2.2) than manganese (1.6), hydrogen's oxidation state is -1. However, there are 3 hydrogen ions present, so their total charge equals -3.

Since compounds are always electrically neutral, the oxidation state of `Mn` must be +3.

`+3 - 3 = 0`

c. `Mn`

The oxidation state is 0, since this an uncombined `Mn` atom.

d. `MnO_4^(-)`

This one is a bit trickier. This is a polyatomic ion, so it has an overall charge of 1-.

We do know that the oxidation state of oxygen is almost always -2. However, there are 4 oxygen ions here, so the total charge is -8. Using this information, we can set up an equation to solve for the oxidation state of `Mn`.

`Mn + (-8) = -1`
`Mn = +7`

The oxidation state of `Mn` is +7.

So the answer is c. `Mn`, because `Mn` has an oxidation state of 0.

Hope this hleps!