In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following $x$ , since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

$Mn^xI_2^-\\\\Mn ^xO_2^{-2}$

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

$Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4$

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

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