The correct order of decreasing second ionisation enthalpy of $\text{Ti}(22),\text{V}(23),\text{Cr}(24)$ and $\text{Mn}(25)$ is:

Second ionization enthalpy involves removing an electron from $M^{+}$.Configurations of $M^{+}$:${\text{Ti}}^{+}:[Ar]3d^{2}4s^1$${\text{V}}^{+}:[Ar]3d^{3}4s^1$${\text{Cr}}^{+}:[Ar]3d^5$ (Half-filled, very stable)${\text{Mn}}^{+}:[Ar]3d^{5}4s^1$${\text{Cr}}^{+}$ has a half-filled $d$ -subshell, making it very difficult to remove the second electron. ${\text{Mn}}^{+}$ is next due to effective nuclear charge. Order: $\text{Cr}>\text{Mn}>\text{V}>\text{Ti}$.