Q.The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds. (Note: more than one of the given options may be correct.)

The exhibition of multiple oxidation states is primarily governed by the availability of electrons in different subshells that can be lost or shared, leading to varying stable electronic configurations.

The key idea is: Variable Valency in Transition and P-block Elements.

Here's the essential reasoning:

1. Elements exhibit more than one oxidation state when they can lose a variable number of electrons from their valence shell or from both valence and penultimate shells.
2. $s$-block elements, like those with $3s^1$ configuration, typically lose all their $s$ electrons to achieve a stable noble gas configuration, exhibiting only one oxidation state (+1). …

The ability of an element to exhibit multiple oxidation states is directly linked to the availability and energy levels of its valence electrons. Atoms achieve stability by forming bonds, which often involves losing, gaining, or sharing electrons. When an atom has valence electrons in different subshells (e.g., $s$ and $p$, or $s$ and $d$) that are relatively close in energy, it can lose or share a varying number of these electrons under different chemical environments. This leads to the exhibition of more than one oxidation state.

Here's a breakdown of how this applies to the given electronic configurations:

1. Understanding Variable Oxidation States:

   • s-block elements (Group 1 and 2): These elements typically have only $ns^1$ or $ns^2$ valence electrons. They tend to lose all their valence electrons to achieve a stable noble gas configuration, resulting in a single, characteristic oxidation state (+1 for Group 1, +2 for Group 2).
   • p-block elements: These elements have $ns^2 np^{1-6}$ valence electrons. They can exhibit multiple oxidation states because they can lose only the $p$-electrons, or both $s$ and $p$-electrons. For heavier p-block elements, the $ns^2$ electrons might be reluctant to participate in bonding (inert pair effect), leading to oxidation states that are two units less than the group oxidation state. They can also gain electrons to achieve a noble gas configuration, leading to negative oxidation states.
   • d-block elements (Transition Metals): These elements are well-known for exhibiting a wide range of oxidation states. This is because the $(n-1)d$ orbitals and $ns$ orbitals are very close in energy. Consequently, electrons from both the $ns$ subshell and the $(n-1)d$ subshell can participate in bonding, leading to a variety of possible oxidation states.

2. Analysing Each Option:

   (i) $3s^1$

   • This configuration corresponds to an element in Group 1, Period 3 (Sodium, Na).
   • It has only one valence electron in the $3s$ orbital.
   • To achieve a stable noble gas configuration (like Neon), it readily loses this single electron.
   • Therefore, it will exhibit only one oxidation state, which is +1.
   • This option does not exhibit more than one oxidation state.

   (ii) $3d^1 4s^2$

   • This configuration corresponds to a d-block element (Scandium, Sc).
   • The valence electrons are in the $4s$ and $3d$ subshells. The $4s$ and $3d$ orbitals are very close in energy.
   • In principle the $4s$ electrons could be lost separately, but in practice scandium's chemistry is dominated by the loss of all three valence electrons together: in its compounds scandium exhibits only the +3 oxidation state.
   • So this configuration does not give more than one oxidation state.

   (iii) $3d^2 4s^2$ …