Amongest Be, B, Mg and Al, the second ionization potential is maximum for

Options

A. B
B. Be
C. Mg
D. A1

Answer

A

Explanation

The E.Cs of the ions obtained after removal of first electron from these elements are `: Be^(+) (1s^(2) 2s^(1))`
`B^(+) (1s^(2) 2s^(2) ) ,Mg^(+) (1s^(2) 2s^(2) 3s^(1))` and `A1^(+) (1s^(2) 2p^(2) 2p^(6) 3s^(2))`
Since `Mg^(+)` and `A1^(+)` belong to 3rd period while `Be^(+)` and `B^(+)` belong to 2nd period , therefore due to bigger size the `Delta_(i)H_(2) `of `Mg^(+)` and `A1^(+)` are lower than those of `Be^(+)` and `B^(+)`
Since `B^(+)` the next electron has to be removed from a more stable tally filled 2s- orbital while in `Be^(+)` it has to be lost from the less stable hafl- filled 2s- orbital adn furthermore loss of an electron from `Be^(+)` gives more stable `Be^(2+)` ion with inet gas configuration therefore `Delta_(i)H_(2)` of B is higher than that of Be , i.e., option ( a) is correct.