Determine the oxidation state of each element in Sn(MnO4)2

Sure, let's determine the oxidation state of each element in Sn(MnO4)2.

1. The oxidation state of Sn: In most stable compounds, the oxidation state of tin (Sn) is +2 or +4. In Sn(MnO4)2, the oxidation state of Sn is +2.

2. The oxidation state of Mn: In MnO4-, the oxidation state of manganese (Mn) is +7. Since there are two MnO4- ions in the compound, the total oxidation state contributed by Mn is +14.

3. The oxidation state of O: In most compounds, the oxidation state of oxygen (O) is -2. Since there are eight oxygen atoms in the compound (4 from each MnO4-), the total oxidation state contributed by O is -16.

In a neutral compound, the sum of the oxidation states of all elements is zero. Therefore, in Sn(MnO4)2, the sum of the oxidation states of Sn, Mn, and O is +2 +14 -16 = 0, which confirms our determinations.