1.

The correct orderof decreasing secondionization enthalpy of Ti(22),V(23), Cr(24)and Mn(25) is

Answer»

`Cr gt Mn gt V gt TI`
`V gt Mngt Cr gt Ti`
`Mn gt Cr gt TigtV`
`Ti gt V gtCr gt Mn`

Solution :Electronic configuration of the givenelements 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 removalof 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`.


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