Oxidation numbers of \(\mathrm{Mn}\) in its compounds \(\mathrm{MnCl}_{2}\) \(\mathrm{Mn}(\mathrm{OH})_{3}, \mathrm{MnO}_{2}\) and \(\mathrm{KMnO}_{4}\) respectively are (a) \(+2,+4,+7,+3\) (b) \(+2,+3,+4,+7\) (c) \(+7,+3,+2,+4\) (d) \(+7,+4,+3,+2\)

Understanding the oxidation state of manganese (Mn) in manganese(II) chloride, MnCl2, is simplified by recognizing that chlorine (Cl) consistently exhibits an oxidation number of -1. Given the presence of two Cl atoms, we have a cumulative negative charge of -2. To balance this within a neutral molecule, Mn must compensate with a positive charge. The equation for the oxidation states is thus \( x + 2(-1) = 0 \), where \( x \) represents Mn's oxidation state. Solving this readily gives us an oxidation state of +2 for Mn in MnCl2.
This result confirms that MnCl2 is composed of Mn ions with a +2 charge paired with Cl ions each holding a -1 charge. In other words, each chlorine relinquishes an electron, which is then accepted by the manganese ion, leading to its +2 oxidation state.

When considering the manganese(III) hydroxide, Mn(OH)3, each hydroxide ion (OH) has an assigned oxidation number of -1. Since Mn(OH)3 contains three OH groups, their collective oxidation charge is -3. To figure out Mn's oxidation number, we rely on the fact that the compound is electrically neutral. By establishing the equation \( x + 3(-1) = 0 \), where \( x \) signifies the oxidation state of Mn, we find that solving yields an oxidation state of +3 for Mn.

This tells us that the Mn in Mn(OH)3 has essentially 'taken' three electrons in total from the hydroxide ions and, therefore, needs to exhibit a +3 oxidation state to maintain neutrality in the compound.

The oxidation state of Mn in manganese(IV) oxide, or MnO2, is deduced by considering the usual oxidation number for oxygen (O), which is -2. With two oxygen atoms in the formula, they collectively bring a -4 charge to the compound. To balance this and achieve a neutral molecule, Mn must present a compensating positive value. The calculation follows the formula \( x + 2(-2) = 0 \). Through simple algebra, we find that Mn holds an oxidation state of +4 in MnO2.

This analysis clarifies that Mn in the compound is in a +4 oxidation state, offsetting the charge from the two oxygen atoms each at -2, resulting in a stable, neutral molecule.

In the case of potassium permanganate, KMnO4, potassium (K) contributes a +1 oxidation state, while each oxygen is -2. As there are four oxygen atoms, their total oxidation charge is -8. The neutral nature of the compound dictates that the sum of all oxidation numbers must total zero. This leads us to the equation \( +1 + x + 4(-2) = 0 \), where \( x \) is Mn's oxidation state. Solving this equation shows that Mn must have an oxidation state of +7 in KMnO4.

The result allows us to understand that in KMnO4, the Mn atom has a high oxidation state of +7, which is balanced by one K atom at +1 and four O atoms in total contributing -8, thus explaining the strong oxidizing nature of potassium permanganate.