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.

The elements giving minimum number of oxidation states are in extreme left and right in 3d series. This is because:

their elements have least number of unpaired electrons

The correct answer is option 1. their elements have least number of unpaired electrons.

Let us look into why elements at the extreme left (Scandium) and right (Zinc) of the 3d series show fewer oxidation states compared to other transition metals.

Transition metals are known for their ability to exhibit a variety of oxidation states. This variability arises from their ability to lose different numbers of electrons from their \(s\) and \(d\) orbitals.

Electronic Configurations and Oxidation States

a. Scandium (Sc)

Electronic Configuration: \([Ar] 3d^1 4s^2\)

Common Oxidation State: \(+3\)

Scandium typically loses all three of its outer electrons (both \(4s\) and \(3d\) electrons) to achieve the \(+3\) oxidation state. The single \(3d\) electron can be lost to form \(Sc^{3+}\), and this is the most stable oxidation state for Scandium.

Reason for Limited Oxidation States: Scandium has only one unpaired electron in the \(3d\) orbital, which limits its ability to exhibit a range of oxidation states.

b. Zinc (Zn)

Electronic Configuration: \([Ar] 3d^{10} 4s^2\)

Common Oxidation State: \(+2\)

Zinc typically loses the two \(4s\) electrons to form \(Zn^{2+}\). The \(3d\) electrons are not involved in bonding and remain paired.

Reason for Limited Oxidation States: Zinc has a completely filled \(3d\) orbital (\(3d^{10}\)), which is stable and less reactive. This configuration doesn’t favor the loss of additional electrons, leading to fewer oxidation states.

Factors Affecting Oxidation States

Number of Unpaired Electrons

The number of unpaired electrons determines the number of possible oxidation states:

Scandium: With only one unpaired electron, Scandium is limited to fewer oxidation states, typically \(+3\).

Zinc: With a completely filled \(3d\) orbital and no unpaired electrons, Zinc mainly shows the \(+2\) oxidation state.

Stability of Electron Configurations

Scandium: The loss of its \(3d\) electron and \(4s\) electrons to form a \(+3\) oxidation state is energetically favorable. However, having just one unpaired electron limits its ability to form multiple oxidation states.

Zinc: The \(3d^{10}\) configuration is stable. The removal of \(4s\) electrons does not disrupt this stable configuration, so Zinc predominantly forms the \(+2\) oxidation state and does not exhibit other oxidation states easily.

Electron Removal and Oxidation State Formation

Scandium: The removal of \(3d\) and \(4s\) electrons results in a stable ion, but the few unpaired electrons limit the range of oxidation states.

Zinc: Removal of \(4s\) electrons results in a stable \(Zn^{2+}\) ion, and the filled \(3d\) orbitals do not contribute to additional oxidation states.

Conclusion

The elements on the extreme left (Scandium) and right (Zinc) of the 3d series exhibit fewer oxidation states primarily due to the following reasons:

Scandium has only one unpaired electron, leading to a limited number of oxidation states. Zinc has a fully filled \(3d\) orbital, which stabilizes the \(+2\) oxidation state and prevents the formation of additional oxidation states.

Thus, the primary reason these elements show fewer oxidation states is that they have a limited number of unpaired electrons available for bonding. The correct answer is option 1: their elements have least number of unpaired electrons.