Question #c2dc6

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Question #c2dc6

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Answer 1 · 2017-01-19T22:58:15

Oxidation numbers are defined as the theoretical charge for the pure ion.

Well, since oxygen is on the second to last column of the periodic table, it commonly has an oxidation number of $- 2$ $color(blue)(-2)$ . Therefore, four of them have a total oxidation number of $- 8$ $-8$ ( $4 \times - 2 = - 8$ $4xx-2 = -8$ ).

The oxidation numbers must all add to give the total charge of ${MnO}_{4}^{-}$ $"MnO"_4^(-)$ (which is also not ${MnO}_{4}$ $"MnO"_4$ ). Given that the total charge of the polyatomic ion is $- 1$ $-1$ , we now need to use the total oxidation number of the $O$ $"O"$ atoms within the polyatomic ion.

The oxidation number of manganese is then:

$- 1 - (- 8) = + 7$ $-1 - (-8) = color(blue)(+7)$

Answer 2 · 2017-01-19T23:04:03

The oxidation number of $Mn$ $"Mn"$ in ${MnO}_{4}^{-}$ $"MnO"_4^"-"$ is +7.

Explanation:

${MnO}_{4}$ $"MnO"_4$ does not exist.

I think you may be referring to the permanganate ion, ${MnO}_{4}^{-}$ $"MnO"_4^"-"$ .

You must follow several rules to determine oxidation numbers.

The important rules for this question are:

The oxidation number of oxygen in a compound is usually –2.
The sum of all oxidation numbers in an ion is equal to its charge.

Per Rule 1, the oxidation number of $O$ $"O"$ is -2.

Write the oxidation number above the $O$ $"O"$ in the formula:

${[Mn {-2 O}_{4}]}_{4}^{-}$ $["Mn"stackrelcolor(blue)("-2")("O")_4]^"-"$ .

The four $O$ $"O"$ atoms together have a total oxidation number of -8.

Write this number below the $O$ $"O"$ atom:

${[Mn {-2 O}_{4}]}_{4}^{-}$ $["Mn"stackrelcolor(blue)("-2")("O")_4]^"-"$
$-8 m m m m m m$ $stackrelcolor(blue)("-8")(color(white)(mmmmmm))$

Per Rule 2, the sum of all the oxidation numbers must be -1.

The oxidation number of $Mn$ $"Mn"$ must be +7, because +7 - 8 = -1.

Put a +7 below the $Mn$ $"Mn"$ .

${[Mn {-2 O}_{4}]}_{4}^{-}$ $["Mn"stackrelcolor(blue)("-2")("O")_4]^"-"$
$+7 m m l l -8 l$ $stackrelcolor(blue)("+7")(color(white)(mmll))stackrelcolor(blue)("-8")(color(white)(l))$

There is only one $Mn$ $"Mn"$ atom, so its oxidation number must be +7.

Write the oxidation number above the $Mn$ $"Mn"$ .

${[+7 Mn {-2 O}_{4}]}_{4}^{-}$ $[stackrelcolor(blue)("+7")("Mn")stackrelcolor(blue)("-2")("O")_4]^"-"$
$+7 m m l l -8 l$ $stackrelcolor(blue)("+7")(color(white)(mmll))stackrelcolor(blue)("-8")(color(white)(l))$

You now have a formula with the oxidation number of each atom above its symbol and the total oxidation numbers of those atoms below their symbols.

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Question #c2dc6

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