The correct order of decreasing second ionization enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is:

To determine the order of second ionization enthalpies, we must look at the electronic configurations of the unipositive ions ($M^+$) from which the second electron is removed:

• Ti (Z=22): Ground state $[Ar]3d^2 4s^2$. $Ti^+$ is $[Ar]3d^2 4s^1$. Electron removed from $4s$.
• V (Z=23): Ground state $[Ar]3d^3 4s^2$. $V^+$ is $[Ar]3d^3 4s^1$. Electron removed from $4s$.
• Cr (Z=24): Ground state $[Ar]3d^5 4s^1$ (exceptionally stable half-filled $d$-shell). $Cr^+$ is $[Ar]3d^5$. The second electron must be removed from the stable half-filled $3d$ orbital, requiring a very high amount of energy.
• Mn (Z=25): Ground state $[Ar]3d^5 4s^2$. $Mn^+$ is $[Ar]3d^5 4s^1$. Electron removed from $4s$.

Trends:

1. Cr vs others: The second ionization enthalpy of Cr is exceptionally high because it involves breaking the stable half-filled $d^5$ configuration ($1592 \text{ kJ mol}^{-1}$).
2. Mn, V, Ti: For these elements, the second electron is removed from the $4s$ orbital (or destabilizes the shell less than breaking a half-filled shell). The energy required generally follows the trend of increasing effective nuclear charge across the period. Thus, $Mn (1509) > V (1414) > Ti (1309)$ in kJ/mol .

Combining these, the correct order is $Cr > Mn > V > Ti$.