# Question 20e6a

Mar 6, 2017

Here's what I got.

#### Explanation:

Start with what you know for sure, which is that oxygen has an oxidation number of $\textcolor{b l u e}{- 2}$ in almost all the compounds it forms.

Remember that the sum of the oxidation numbers of each atom that is part of a compound must be equal to the net charge of said compound.

For water, you will have

"H"_ 2 stackrel(color(blue)(-2))("O")

Since you have $2$ hydrogen atoms, you will need an oxidation state of $\textcolor{b l u e}{+ 1}$ for each of them to get

$2 \times \left(+ 1\right) + 1 \times \left(- 2\right) = 0$

Therefore, you will have

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(-2))("O}}$

Do the same for sulfur dioxide. You will have

"S" stackrel(color(blue)(-2))("O")_2

Since you have $2$ oxygen atoms here, you will need an oxidation state of $\textcolor{b l u e}{+ 4}$ for the sulfur atom to get

$1 \times \left(+ 4\right) + 2 \times \left(- 2\right) = 0$

Therefore, you will have

${\stackrel{\textcolor{b l u e}{+ 4}}{\text{S")stackrel(color(blue)(-2))("O}}}_{2}$

Finally, move on to the permanganate ion, ${\text{MnO}}_{4}^{-}$. You will have

${\text{Mn" stackrel(color(blue)(-2))("O") _4}}^{-}$

This time, the ion sum of the oxidation numbers must be equal to $- 1$ since that is the net charge of the ion. You will need an oxidation state of $\textcolor{b l u e}{+ 7}$ for the manganese atom to get

$1 \times \left(+ 7\right) + 4 \times \left(- 2\right) = - 1$

Therefore, you will have

stackrel(color(blue)(+7))("Mn") stackrel(color(blue)(-2))("O") _4""^(-)#