In: Chemistry
Identify the element of Period 2 that has the following successive ionization energies, in kJ/mol.
IE1 = 1314 IE2 = 3389 IE3 = 5298 IE4 = 7471 IE5 = 10992 IE6 = 13329 IE7 = 71345 IE8 = 84087
Li |
Ne |
B |
O |
Element corresponds to the given IE trend is: Oxygen (O)
Oxygen have 6 valance electron; After removal of 6 electrons (after 6th Ionization), it attains a stable nearest Nobel gas configuraation of He.
So; a sudden jump in 7th Ionization energy seen from IE6 to IE7.
IE7 will be much greater (Since because of He nobel gas configuration).