Which one of the following exhibits the large number of oxidation states?

The question asks which element exhibits the largest number of oxidation states among titanium (Ti, atomic number 22), vanadium (V, 23), chromium (Cr, 24), and manganese (Mn, 25). Oxidation states represent the hypothetical charges an atom would have if all bonds were ionic, and transition metals show variable oxidation states due to the involvement of both ns and (n-1)d electrons in bonding.

First, recall the electron configurations:

• Titanium (Ti, 22): [Ar] 4s² 3d²
• Vanadium (V, 23): [Ar] 4s² 3d³
• Chromium (Cr, 24): [Ar] 4s¹ 3d⁵ (exception for half-filled stability)
• Manganese (Mn, 25): [Ar] 4s² 3d⁵

Now, list the oxidation states each element can exhibit:

Titanium (Ti): The common oxidation states are +2 (losing 4s electrons), +3 (losing 4s and one 3d electron), and +4 (losing 4s and both 3d electrons). Less common states include 0 and +1, but typically, Ti shows 3 main oxidation states: +2, +3, +4.

Vanadium (V): The oxidation states include +2 (losing 4s electrons), +3 (losing 4s and one 3d electron), +4 (losing 4s and two 3d electrons), and +5 (losing 4s and all three 3d electrons). Less common states like +1 exist, but V primarily exhibits 4 oxidation states: +2, +3, +4, +5.

Chromium (Cr): Oxidation states range from +1 to +6. Specifically:

• +1: losing the 4s electron (rare)
• +2: losing the 4s electron
• +3: losing 4s and one 3d electron
• +4: losing 4s and two 3d electrons
• +5: losing 4s and three 3d electrons
• +6: losing 4s and all five 3d electrons (as in CrO₃)

Thus, chromium exhibits 6 oxidation states: +1, +2, +3, +4, +5, +6.

Manganese (Mn): Oxidation states range from +1 to +7. Specifically:

• +1: losing one 4s electron (in some complexes)
• +2: losing both 4s electrons (e.g., MnCl₂)
• +3: losing both 4s and one 3d electron (e.g., Mn₂O₃)
• +4: losing both 4s and two 3d electrons (e.g., MnO₂)
• +5: losing both 4s and three 3d electrons (e.g., K₃MnO₄)
• +6: losing both 4s and four 3d electrons (e.g., K₂MnO₄)
• +7: losing both 4s and all five 3d electrons (e.g., KMnO₄)

Manganese exhibits 7 oxidation states: +1, +2, +3, +4, +5, +6, +7. Additionally, manganese can show states like -3, -2, -1, and 0 in specific compounds, but the positive states from +1 to +7 are well-documented.

Comparing the number of oxidation states:

• Ti: 3 main states (+2, +3, +4)
• V: 4 main states (+2, +3, +4, +5)
• Cr: 6 states (+1 to +6)
• Mn: 7 states (+1 to +7)

Manganese exhibits the largest number of oxidation states (7) among the given options.

The reason for manganese having the highest number is its electron configuration ([Ar] 4s² 3d⁵). It can lose up to 7 electrons (both 4s and all five 3d electrons) to achieve the +7 state, and it also forms stable compounds in lower oxidation states due to the half-filled d-subshell stability.

Hence, the correct answer is Option D.