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) or too many d-electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus, early in the 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 of 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 noticed among 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 oxidising 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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → +7
Up to manganese, the highest oxidation state of a 3d element equals the sum of its 4s and 3d electrons. Manganese is [Ar] 3d54s2, so its highest state is 5 + 2 = +7. It is shown in the permanganate ion and in the oxide:
\( \ce{KMnO4} \): x + 1 + 4(−2) = 0 ⇒ x = +7; \( \ce{Mn2O7} \): 2x + 7(−2) = 0 ⇒ x = +7
NCERT (2026-27 edition):
Oxidation States – “… The 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.”