Math list `I` with list `Ii` and select the correct answer with the code given below
`{:(" ListI",ListII),(("Successie ionization energies"//KJ mol^(-1)),"(Elements)"),(" "IE_(1)" "IE_(2)" "IE_(3),),((i)2080" "3963" "6130,(a)H),((ii)520" "7297" "11810,(b)Li),((iii)900" "1758" "14810,(c)Be),((iv)800" "2428" "3660,(d)B),(,(e)Ne):}`

Answer

D

Explanation

`H (1s^(1))` cannot have `IE_(2)`and `IE_(3).Li(1s^(2)2s^(1))` will have the lowest `IE_(1)` but will have very high `IE_(2).Be (1s^(2)2s^(2))` will have low `IE_(1)`(relatively higher than `Li`) but will be characterized by very high `IE_(3)`. The `IE_(1)` of `B(1s^(2)2s^(2)2p^(1))` will be less than the `IE_(1)` of `Be.Ne (1s^(2)2s^(2)2p^(6))` will have the highest `IE_(1)`. Thus, option `(4)` is correct.