What is the oxidation state of manganese in MnO4?
The oxidation state of an element in a compound refers to the number of electrons that the atom has gained or lost in order to form the compound. It is a fundamental concept in chemistry that helps us understand the electronic structure and reactivity of elements. In the case of MnO4, which is known as permanganate, the oxidation state of manganese is crucial in determining its chemical properties and reactions.
Manganese is a transition metal that can exhibit various oxidation states, ranging from +2 to +7. In the permanganate ion (MnO4-), manganese is in its highest oxidation state, which is +7. This is because oxygen typically has an oxidation state of -2, and there are four oxygen atoms in the permanganate ion. To balance the overall charge of the ion, which is -1, the oxidation state of manganese must be +7.
The oxidation state of manganese in MnO4 can be determined by using the following equation:
x + 4(-2) = -1
Solving for x, we find that the oxidation state of manganese in MnO4 is +7. This high oxidation state indicates that manganese has lost seven electrons, making it highly reactive and capable of undergoing reduction reactions.
The permanganate ion is a powerful oxidizing agent and is widely used in various applications, such as water purification, dyeing, and analytical chemistry. Its strong oxidizing properties are due to the high oxidation state of manganese, which allows it to readily accept electrons from other substances.
In summary, the oxidation state of manganese in MnO4 is +7, which is its highest oxidation state. This high oxidation state makes permanganate a potent oxidizing agent and has significant implications for its chemical properties and applications.
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