IE1IE2IE3IE4Al5781817274511577Si786157732324356a) Explain why IE2 > IE1 for Al.b) Explain why IE4 is enoormous for Al, & why it is much smaller for Si.

The amount of energy required to eject the most loosely bound electron of an isolated atom in the gaseous state to form a cation is known as ionization energy or ionization enthalpy. Given data: IE1 IE2 IE3 IE4 Al 578 1817 2745 11577 Si 786 1577 3232 4356 a) The electronic configuration of Al is as follows: 1s22s22p63s23p1 The removal first electron from Al (through first ionization, IE1) produces Al+ ion. Now the removal of another electron from this positively charged ion will require more energy. Also, for the removal of this electron, the fully filled 3s orbital will be disturbed. Thus, the second ionization enthalpy of Al will be greater than its first ionization enthalpy (IE2 > IE1) as it will require more energy. b) The electronic configuration of Al and Si are given below: Al→1s22s22p63s23p1Si→1s22s22p63s23p2 It is known that for the fourth ionization enthalpy (IE4), the fourth electron has to be removed from the atom. It can be seen that in case of Al, for the removal of fourth electron, the noble gas configuration has to be broken. This requires an enormous amount of energy. On the other hand, in case of Si, the electron from same shell (3s) has to be removed for fourth ionization. It requires very less energy as compared to the case of Al. Thus, fourth ionization enthalpy (IE4) is enormous for Al but it is smaller for Si.