Question #dc82b

1 Answer
Jun 30, 2017

Answer:

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!