What is the oxidation number of mn in ca(mno4)2

+ 7.

Explanation:

We have general rules to assign the oxidation numbers:

Rule 1: The oxidation number of an element alone in its free (uncombined) state is zero.

For example, Al(s) or Zn(s).

This is also true for elements found in nature as diatomic (two-atom) elements  as: H₂, O₂, Cl₂, or I₂.

and also for sulfur, found as: S₈.

Rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion,

For example:  Na⁺ = +1, S⁻² = -2.

Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion.

Rule 4: The oxidation number of an alkali metal (group IA) in a compound is +1; the oxidation number of an alkaline earth metal (group IIA) in a compound is +2.

Rule 5: The oxidation number of oxygen in a compound is usually –2.

If, however, the oxygen is in a class of compounds called peroxides.

(for example, hydrogen peroxide), then the oxygen has an oxidation number of –1. If the oxygen is bonded to fluorine, the number is +1.

So, for our problem Ca(MnO₄)₂:

∴ oxidation no. of Ca + 2[(oxidation no. of Mn) + 4(oxidation no. of O) = 0.

(+2) + 2[(oxidation no. of Mn) + (- 8)] = 0.

2[(oxidation no. of Mn) + (- 8)] = - 2.

[(oxidation no. of Mn) + (- 8)] = -1.

∴ (oxidation no. of Mn) = - 1 + 8 = + 7.

the answer is either b or c