Question

Four successive members of the first row transition elements listed below with atomic numbers. Which one of them is expected to have the highest $E_{m{^{3+}}/m{^{2+}}}^{\circ}$value?

• A. $Cr(Z = 24)$
• B. $Mn (Z = 25)$
• C. $Fe (Z = 26)$
• D. $Co (Z = 27)$

Answer 1

The Correct Option is D

The question involves determining which of the given transition elements has the highest standard reduction potential for the conversion from the +3 to +2 oxidation state, denoted as $E_{m{^{3+}}/m{^{2+}}}^{\circ}$.

To answer this question, we must consider the electronic configurations and common oxidation states of the given elements, which are all adjacent transition metals: Chromium (Cr), Manganese (Mn), Iron (Fe), and Cobalt (Co).

1. Electronic Configurations:
   • $Cr \ (Z=24)$: [Ar] 3d⁵ 4s¹
   • $Mn \ (Z=25)$: [Ar] 3d⁵ 4s²
   • $Fe \ (Z=26)$: [Ar] 3d⁶ 4s²
   • $Co \ (Z=27)$: [Ar] 3d⁷ 4s²
2. The conversion from $M^{3+}$ to $M^{2+}$ involves a gain of one electron:
   • $M^{3+} + e^- \rightarrow M^{2+}$
3. In the series of first-row transition metals, the standard reduction potential $E^{\circ}$ reflects the stability of the given oxidation states. Stability in the lower oxidation state can increase this potential.
4. $Co^{3+}$ forms a stable low-spin complex, reducing easily to $Co^{2+}$, whereas the other elements typically favor other oxidation states. This makes the reduction potential for cobalt higher than for the others.
5. Thus, $Co \ (Z = 27)$ is likely to have the highest $E_{m{^{3+}}/m{^{2+}}}^{\circ}$ value among the given elements, due to the stability of its commonly observed oxidation states and its ability to form stable $d^6$ low-spin complexes.

Therefore, the element expected to have the highest $E_{m{^{3+}}/m{^{2+}}}^{\circ}$ value is $Co(Z = 27)$.