Read the passage given and answer the question.

The transition elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at extreme ends stems from either too few electrons to lose or share (Sc, Ti) on too many d-electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus early in the early series scandium (II) is virtually unknown and titanium (IV) is more stable than Ti(III) or Ti(II). At the other end, the only oxidation state of zinc is +2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum as s- and d-electrons upto  manganese \((Ti^{IV}O_2, V^VO_2^+, Cr^{VI}O_4^{2}, Mn^{VII}O_4 )\) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are \(Fe^{II, III}\), \(Co^{II, III}\), \(Cu^{I, II}\), \(Zn^{II}\).

The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., \(V^{II}\), \(V^{III}\), \(V^{IV}\), \(V^V\). This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit or two.

An interesting feature in the variability of oxidation states of the d-block elements is noticedamong the groups (groups 4 through 10). Although in the p-block, the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, \(Mo^{(VI)}\) and \(W^{(VI)}\) are found to be more stable than \(Cr^{(VI)}\). The \(Cr^{(VI)}\) in the form of dichromate in acidic medium is a strong agent, whereas \(MoO_3\) and \(WO_3\) are not.

Low oxidation states are found when a complex compound has ligands capable of \(\pi \)-acceptor character in addition to the \(\sigma \)-bonding. For example, in \(Ni(CO)_4\) and \(Fe(CO)_5\), the oxidation state of nickel and iron is zero.

Highest oxidation state shown by 'Mn' is:

+7

The correct answer is option 3. +7.

Oxidation states, or oxidation numbers, indicate the degree of oxidation of an atom in a compound. For manganese (Mn), these states range from \(-3\) to \(+7\). Here's how we determine the highest oxidation state:

Common Oxidation States of Manganese

a. Oxidation State of \(+2\)

Compounds: \(MnO\), \(MnCl_2\), etc.

Example: Manganese(II) oxide (\(MnO\)) shows manganese with an oxidation state of \(+2\).

b. Oxidation State of \(+3\)

Compounds: \(Mn_2O_3\)

Example: Manganese(III) oxide (\(Mn_2O_3\)) shows manganese with an oxidation state of \(+3\).

c. Oxidation State of \(+4\)

Compounds: \(MnO_2\)

Example: Manganese(IV) oxide (\(MnO_2\)) shows manganese with an oxidation state of \(+4\).

d. Oxidation State of \(+6\)

Compounds: \(MnO_3\), \(K_2MnO_4\) (where manganese is in the \(+6\) oxidation state)

Example: Manganese(VI) oxide (\(MnO_3\)) shows manganese with an oxidation state of \(+6\). Manganese also appears in the \(+6\) state in potassium manganate (\(K_2MnO_4\)).

e. Oxidation State of \(+7\)

Compounds: \(KMnO_4\), \(MnO_4^{2-}\) (in permanganates)

Example: Potassium permanganate (\(KMnO_4\)) is a classic example where manganese exhibits its highest oxidation state of \(+7\). In this compound:

Potassium (K) is \(+1\).

Oxygen (O) is \(-2\).

The total charge on permanganate (\(MnO_4^-\)) is \(-1\).

To find the oxidation state of Mn in \(KMnO_4\):

\(\text{Oxidation state of } Mn + 4 \times (-2) = -1\)

\(\text{Oxidation state of } Mn - 8 = -1\)

\(\text{Oxidation state of } Mn = +7\)

Significance of the +7 Oxidation State

Manganese can achieve a maximum oxidation state of \(+7\) due to its ability to lose all seven of its valence electrons. This is characteristic of the transition metals in the periodic table, which can exhibit multiple oxidation states due to the involvement of both \(s\) and \(d\) orbitals in bonding.

Conclusion

The highest oxidation state that manganese can exhibit is \(+7\). This state is typically found in compounds like potassium permanganate (\(KMnO_4\)), where manganese is fully oxidized.

Thus, the correct answer is option 3: +7.