d and f Block Elements - Oxidation states and Ionization enthalpy

Why is the $E^{\ominus}$ value for the $Mn^{3+}/Mn^{2+}$ couple much more positive than that for $Cr^{3+}/Cr^{2+}$?

$Mn^{2+}$ has a stable $[Ar] 3d^5$ configuration (half-filled), while $Mn^{3+}$ has $[Ar] 3d^4$. $Cr^{3+}$ has a stable half-filled $t_{2g}$ level ($d^3$).

The high positive value for $Mn^{3+}/Mn^{2+}$ indicates that $Mn^{2+}$ is very stable relative to $Mn^{3+}$. This is due to the extra stability of the exactly half-filled $d^5$ subshell in $Mn^{2+}$. Conversely, $Cr^{3+}$ is more stable than $Cr^{2+}$ because $Cr^{3+}$ has a half-filled $t_{2g}$ set of orbitals in an octahedral field.

Account for the fact that $Zn$ has the lowest enthalpy of atomization ($126 \text{ kJ mol}^{-1}$) in the $3d$ series.

$Zn$ has the electronic configuration $[Ar] 3d^{10} 4s^2$.

In $Zn$, all electrons are paired in the $d$-orbitals. There are no unpaired electrons available for metallic bonding (delocalization). This results in weak metallic bonds and consequently a low enthalpy of atomization compared to other transition metals like $V$ or $Cr$ which have many unpaired electrons.

Compare the $IE_1$ and $IE_2$ of $Cr$ ($Z=24$) and $Mn$ ($Z=25$).

$IE_1(Cr) < IE_1(Mn)$ but $IE_2(Cr) > IE_2(Mn)$.

For $IE_1$, $Mn$ is higher because of higher nuclear charge. For $IE_2$, removing an electron from $Cr^+$ ($3d^5$) requires breaking a stable half-filled configuration, whereas removing an electron from $Mn^+$ ($3d^5 4s^1$) involves removing the remaining $4s$ electron, which is easier.