The correct of decreasing second ionisation enthalpy of `Ti(22),V(23),Cr(24)` and `Mn(25)` is

Options

A. `Cr gt Mn gt V gt Ti`
B. `V gt Mngt Cr gt Ti`
C. `Mn gt Cr gt TigtV`
D. `Ti gt V gtCr gt Mn`

Explanation

Electronic configuration of the given elements are `Ti=[Ar]^(18)3d^(2) 4s^(2) ,V= [Ar]3d^(3) 4s^(2) `,
`Cr= [Ar]^(18) 3d^(5) 4s^(1) ,Mn= [ Ar]^(18) 3d^(5) 4s^(2)`.
Their effective nuclear charges increase from Ti to Mn, therefore, their 1st ionization enthalpies increase in the same order, i.e., `Mn gt Cr gt V gt Ti`. However,after the removal of 1st electron,Cr acquiresstable`3d^(5)` configuration. Hence,it shows exceptional behaviour and has very high 2nd ionization enthalpy. For the remaining elements, the trend remains the same.Thus, 2nd ionization enthalpies will be in the order `:`
`Cr gt Mn gt V gt Ti`.