The correct order of decreasing second ionisation enthalpy of Ti(22), V (23), Cr (24) and $\mathrm{Mn}(25)$ is
The correct order of decreasing second ionisation enthalpy of Ti(22), V (23), Cr (24) and $\mathrm{Mn}(25)$ is
$\mathrm{Cr} > \mathrm{Mn} > \mathrm{V} > $ Ti
V $>$ Mn $>$ Cr $>$ Ti
Mn $>$ Cr $>$ Ti $>$ V
Ti $>$ V $>$ Cr $>$ Mn
Solution
Key Idea : The amount of energy required to remove an electron from a unipositive ion is called second IP which generally increases in a period from left to right.
The second ionisation energies of the first transition series increase almost regularly with increase in atomic number. However, the value for $\mathrm{Cr}$ is sufficiently higher than those of its neighbour, ie, (Mn). This is due to stable configuration of $\mathrm{Cr}^{+}$( $3 d^5$ exactly half filled).
Note : The halffilled and completely filled configurations are more stable.