d and f Block Elements: Diagram-Based Questions (7)

$Sc$: not stable as $M^{2+}$ (forms $Sc^{3+}$). For others: $Ti$ −1.63, $V$ −1.18, $Cr$ −0.91, $Mn$ −1.18, $Fe$ −0.44, $Co$ −0.28, $Ni$ −0.25, $Cu$ +0.34, $Zn$ −0.76 V.

Most are negative — they’re more reactive than $H_2$, so $M$ is oxidized to $M^{2+}$ in standard conditions.

The reduction potential depends on three energy steps: sublimation, ionization (1st + 2nd), and hydration. For $Cu$:

• Sublimation enthalpy: high (Cu has filled $3d^{10}$, strong metallic bonding)
• Ionization: relatively high (filled subshell stable)
• Hydration enthalpy of $Cu^{2+}$: less negative than expected

The high sum of sublimation + ionization is not offset by hydration. So forming $Cu^{2+}$ from $Cu(s)$ is energy-uphill, and the equilibrium favors $Cu(s)$ — positive $E^\circ$.

$Mn^{2+}$ has $3d^5$ — a half-filled subshell, which is exceptionally stable (exchange energy maximum).

So oxidation of $Mn$ to $Mn^{2+}$ is favored more than the simple trend predicts. The half-filled stability lowers the energy of $Mn^{2+}$, making the reduction potential more negative.

The general trend (becoming less negative across the period) reflects increasing nuclear charge → tighter binding of valence electrons → less easy to oxidize. $Mn$ and $Cu$ deviate due to half-filled and filled subshell stability of $Mn^{2+}$ ($3d^5$) and $Cu^{0}$ ($3d^{10}4s^1$).