I have actually tried to recognize what the oxidation state for Mn in $ceMnO4^-$ is. So much i have interpreted that O have the oxidation state of -2 so because there is 4 O the number turns to -8. What happens to the minus at O? and what is the oxidation state for Mn?



The amount of the oxidation states of all of the atoms demands to equal the total charge top top the ion.

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For example, for $ceNH4+$, if each hydrogen atom has actually an oxidation state of +1, and the in its entirety charge is +1, then we deserve to solve for the oxidation state that nitrogen:

$$eginaligned 4(+1) + OS_ceN &= +1\OS_ceN &= +1 -4(+1) \OS_ceN &= -3endaligned$$

Similarly, we deserve to do $ceSO4^2-$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve because that the oxidation state that sulfur:

$$eginaligned 4(-2) + OS_ceS &= -2\OS_ceS &= -2 -4(-2) \OS_ceS &= +6endaligned$$

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answered jan 6 "19 in ~ 12:28

Ben NorrisBen Norris
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Seeing together Oxygen has an oxidation state that -2 and you have 4 Oxygen atoms, the complete charge added by Oxygen = -8. Because the total charge the this compound = -1, climate in this situation Mn = +7+7 - 8 = -1

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answered jan 6 "19 in ~ 13:32

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As you claimed Oxygen has actually the oxidation state of -2 therefore if we add up every the oxygens they have a -8 oxidation state.

As the oxidation state of the ion is same to the charge, the Mn needs to "balance" the -8 to get it up to -1. Because of this the oxidation state that Mn in Mno4- is +7.

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answered jan 6 "19 in ~ 13:33

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