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

Options

A. `Mn gt Cr gt Ti gt V`
B. `Cr gt Mn gt V gt Ti`
C. `Tigt V gt Crgt Mn`
D. `V gt Mngt Cr gt Ti`

Answer

B

Explanation

The given elements `Ti, V, Cr`, and `Mn` are transition elements of the fourth period. In general, the ionization enthalphy increases from left to right across the period due to the increase of effective nuclear change. Thus, the expected order of `Delta_(i)H_(2)`should be `Mn gt Crgt Vgt Ti`, but the actural order is `Cr(1592kJ mol^(-1))gt Mn (1509) gt V(1414)gt Ti (1309)`. the unusually high value for `Cr` is on account of the removel of electron form extra stable `d^(5)` configuration:
`Cr(3d^(5)4s^(1)) overset(Delta_(i)H_(1))rarr Cr^(+)(3d^(5)) overset(Delta_(i)H_(2))rarrCr^(++)(3d^(4))`