The element with which of the following outer electron configuration may exhibit the largest number of oxidation states in its compounds :

The question asks which element, based on its outer electron configuration, may exhibit the largest number of oxidation states in its compounds. The options are:

A. $$ 3d^5 4s^2 $$

B. $$ 3d^8 4s^2 $$

C. $$ 3d^7 4s^2 $$

D. $$ 3d^6 4s^2 $$

First, we identify the elements corresponding to each configuration. The atomic number is determined by the total electrons in the configuration. For option A, $$ 3d^5 4s^2 $$ has 5 electrons in the 3d orbital and 2 in the 4s orbital, totaling 7 valence electrons. This corresponds to manganese (Mn), atomic number 25. For option B, $$ 3d^8 4s^2 $$ has 8 + 2 = 10 valence electrons, corresponding to nickel (Ni), atomic number 28. For option C, $$ 3d^7 4s^2 $$ has 7 + 2 = 9 valence electrons, corresponding to cobalt (Co), atomic number 27. For option D, $$ 3d^6 4s^2 $$ has 6 + 2 = 8 valence electrons, corresponding to iron (Fe), atomic number 26.

Oxidation states depend on the number of electrons an element can lose or share. Transition metals can exhibit multiple oxidation states due to electrons in both the 4s and 3d orbitals. The 4s electrons are lost first, followed by the 3d electrons. The stability of half-filled or fully filled d orbitals also influences the range of oxidation states.

We now examine the common oxidation states for each element:

- Manganese (Mn, $$ 3d^5 4s^2 $$): The half-filled $$ d^5 $$ configuration (after losing the two 4s electrons) is stable, allowing Mn to lose electrons from +2 to +7. Common oxidation states include +2 (e.g., MnCl₂), +3 (e.g., Mn₂O₃), +4 (e.g., MnO₂), +5 (e.g., K₃MnO₄), +6 (e.g., K₂MnO₄), and +7 (e.g., KMnO₄). Thus, Mn exhibits at least six oxidation states.

- Nickel (Ni, $$ 3d^8 4s^2 $$): Common oxidation states are +2 (most stable, e.g., NiO), and less commonly +1, +3, and +4. For example, Ni(IV) exists in NiO₂. However, Ni typically shows fewer oxidation states, around three to four.

- Cobalt (Co, $$ 3d^7 4s^2 $$): Common oxidation states are +2 (e.g., CoCl₂) and +3 (e.g., Co₂O₃). Less common states include +1 (e.g., [Co(CO)₄]⁻), +4 (e.g., CoO₂), and +5. Co exhibits up to five oxidation states, but +4 and higher are rare.

- Iron (Fe, $$ 3d^6 4s^2 $$): Common oxidation states are +2 (e.g., FeCl₂) and +3 (e.g., FeCl₃). Less common states include +4 (e.g., FeO₂), +6 (e.g., K₂FeO₄), and occasionally +1 or +5. Fe shows up to four common oxidation states, though higher states are unstable.

Comparing the number of oxidation states:

- Mn (option A) exhibits six oxidation states (+2 to +7).

- Ni (option B) exhibits about three to four oxidation states.

- Co (option C) exhibits up to five oxidation states, but fewer are common.

- Fe (option D) exhibits about four oxidation states.

Manganese shows the widest range and the largest number of oxidation states due to its stable half-filled d⁵ configuration, allowing it to lose electrons progressively from +2 to +7. Therefore, the element with configuration $$ 3d^5 4s^2 $$ exhibits the largest number of oxidation states.

Hence, the correct answer is Option A.