Question

Question 6 0.15 pts Consider the following set of successive ionization energies: IE1 = 578kJ/mol IE2 = 1,820kJ/mol IE3 = 2,750kJ/mol IE4 = 3,600kJ/mol IE5 = 13,300kJ/mol To which second period element do these ionization values belong? Enter your answer as a chemical symbol (ex: He).

Answer 1

The given ionization energy values from the second period element corresponds to Aluminium (Al).

Ionization energy (IE) is defined as the amount of energy required to remove an electron from the outermost shell of an element, thus as one electron is removed the atom becomes positively charged and it becomes more difficult to remove more electron from this positively charged species. Thus, the value of sucessive first, second, third ionization energies,etc., increases gradually.

As we move left to right across a period ionization energy increases because atomic size decreases as a result the effective nuclear charge also increases. Hence it become difficult to remove an electron.

The first IE corresponds to 578 KJ/mol (too less compared to other values) from this we can infer that:

• The electron is not removed from the s-orbital (which is closest to the nucleus) as for doing so a lot of energy is required. Thus it is the p-orbitl of the second period whose electron is removed.
• As we move left to right the number of p-electrons increases, i.e., more towards in attaining stable half-filled or full filled configuration and then it requires a very high amount of energy to remove even the first electron.
• Thus, based on these observations the element corresponding to the given values is aluminium (Al) with electronic configuration 1s² 2s² 2p⁶ 3s² 3p¹

As it has only one electron in its p-orbital and by removing it we get 3s² orbital therefore it can be removed easily.