Question

Find the oxidation number of Mn in $\left[ MnO_{ 4 } \right] ^{ - }$ ion?

Answer 1

Hint: Try to find the most common oxidation state of other elements then we can find the oxidation number of the required element by balancing the total charge on the ion with calculated charge.

Complete step by step solution:
 We first check the charge of another element in ion. Oxygen is one of the most common elements in our day to day life. We know that oxygen exhibits oxidation states from -2 to +2. But +2 and +1 are exhibited only when it forms a bond with fluorine. 0 is exhibited when it forms a bond with itself. -1 is exhibited in per oxo compounds which are quite unstable. But a given ion does not contain per oxo oxygen. Therefore in the given ion charge of oxygen atom is -2.
Let charge of Mn be x. According to the law of conservation of charge, the sum of charges of each atom must be equal to the total charge on an ion or compound.
Sum of charge of each atom is $x\quad +\quad 4\quad \times \quad \left( -2 \right)$. Since there are 4 oxygen atoms we multiplied charge by 4. Now equating the charge on ion and calculated charge.
\[x\quad -\quad 8\quad =\quad -1\]
Sending -8 to R.H.S of the equation results in a change of sign.
\[x\quad =\quad -1\quad +\quad 8\quad =\quad +7\]
Therefore, the oxidation state of Mn is +7.

Note: We must take the general oxidation state of other elements, not other oxidation states which element rarely exhibits. We should also take the number of atoms correctly while calculating the oxidation number.