Question

\[Al + Mn{O_4}^ - \to Al\left( {OH} \right)_4^ - + Mn{O_2}\]
For this reaction the oxidation state of \[Al\] and \[Mn\] in the product are respectively:
A) \[ + 3\] and \[ + 4\]
B) \[ + 4\] and \[ + 4\]
C) \[ + 3\]and \[ + 7\]
D) \[ + 2\] and \[ + 4\]

Answer 1

Hint: We know that the sum of all the oxidation numbers of all atoms is equal to \[ - 1\] in aluminum hydroxide anion and the sum of all the oxidation numbers of all-atom is equal to zero in Manganese oxide.

Complete step by step answer:
Let us first focus on the product aluminum hydroxide:
We know that the oxidation state of hydroxide \[\left( {OH} \right)\] is \[ - 1\]
And the overall oxidation state of aluminum hydroxide is also \[ - 1\]
So we can find the oxidation state of aluminum by this following equation:
\[Al + 4\left( { - 1} \right) = - 1\]
By opening the bracket we get the following equation:
\[Al - 4 = - 1\]
By taking the numerical value we can find the oxidation state of aluminum:
\[Al = + 4 - 1\]
So the oxidation state of aluminum will be equal to:
\[Al = + 3\]
Now let us focus on the product Manganese oxide.
We know that the oxidation state of hydroxide \[\left( O \right)\] is \[ - 2\]
And the overall oxidation state of Manganese oxide is 0.
So we can find the oxidation state of Manganese by this following equation:
\[Mn + 2\left( { - 2} \right) = 0\]
By opening the bracket we get the following equation:
\[Mn - 4 = 0\]
By taking the numerical value we can find the oxidation state of Manganese:
\[Mn = 0 + 4\]
So the oxidation state of Manganese will be equal to:
\[Mn = + 4\]
So, from the reaction, the oxidation state of aluminum and Manganese is \[ + 3\] and \[ + 4\] respectively.

Therefore, we can conclude that the correct answer to this question is option A.

Note:We must know that the overall oxidation state of the anionic and cationic compound is equal to the charge associated with the molecule and for the neutral molecule the overall oxidation state is zero.