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?


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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
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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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PeteLDWPeteLDW
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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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