# Question #dc82b

Jun 30, 2017

c. $M n$

#### Explanation:

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

a. $M n {F}_{2}$

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 $M n$ must be +2 .

$+ 2 - 2 = 0$

b. $M n {H}_{3}$

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 $M n$ must be +3.

$+ 3 - 3 = 0$

c. $M n$

The oxidation state is 0, since this an uncombined $M n$ atom.

d. $M n {O}_{4}^{-}$

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 $M n$.

$M n + \left(- 8\right) = - 1$
$M n = + 7$

The oxidation state of $M n$ is +7.

So the answer is c. $M n$, because $M n$ has an oxidation state of 0.

Hope this hleps!