What is the oxidation state of Manganese in KMnO4 (s) and sulfur in H2S?
Rule 1 is that the oxidation states must add to zero for a compound or to the charge on an ion.
Rule 2 is representative elements have their "normal" charge; i.e., group I has +1, group II has +3, group III has +3, etc.
Rule 3 is that H is +1 (except in unusual cases of hydrides) and O has -2 (except in unusual cases of peroxides or superoxides). How does that work for KMnO4.
K is +1, O is -2 so 4*-2 = -8. So to make KMnO4 zero Mn must be +7.
Check. +1(for K) + (+7)(for Mn) + (-8) (for 4 oxygen) = 0
Now you do that for H2S. Remember H is +1 each.
To determine the oxidation state of an element in a compound, you need to know the typical oxidation states of the other elements involved and apply some rules.
Let's start with KMnO4 (potassium permanganate). We know that oxygen typically has an oxidation state of -2 in compounds, and potassium has an oxidation state of +1 in most cases.
Next, we must determine the oxidation state of manganese (Mn). Since the compound is neutral, the sum of the oxidation states of all the elements must equal zero. In KMnO4, there are four oxygen atoms with a total oxidation state of -2 each, which equals -8 (-2 x 4 = -8).
To balance the equation, the oxidation state of Mn must be such that: +1 (from potassium) + x (from Mn) - 8 (from oxygen) = 0. Solving for x, we obtain the oxidation state of manganese in KMnO4 as +7.
Now let's consider H2S (hydrogen sulfide). Hydrogen typically has an oxidation state of +1 in compounds, and oxygen is generally -2. However, in this case, we are dealing with sulfur (S).
For H2S to be neutral, the total oxidation state of hydrogen must balance the oxidation state of sulfur. Since there are two hydrogen atoms, their total oxidation state is +2 (+1 x 2 = +2).
To balance this, the oxidation state of sulfur (S) must be such that: +2 (from hydrogen) + x (from S) = 0. Solving for x, we find that the oxidation state of sulfur in H2S is -2.
Therefore, the oxidation state of manganese in KMnO4 is +7, and the oxidation state of sulfur in H2S is -2.