The ionization energies (IE) of the aluminum ion (A1P+) are 2744.7,577.6, 11577, and 1816.6 (kJ/mol).
Which is the fourth ionization energy of Al^3+?

The fourth ionization energy of Al³⁺ ion is 11577 kJ/mol.

Explanation:

The energy needed to remove an electron from the outermost shell of an neutral gaseous atom is termed as Ionization energy. In the case of Aluminum, it is given that 2744.5 , 577.6, 11577, 1816.6 kJ/mol. The IE required to remove an electron from the neutral atom is lesser when compared to the electron removal from an ion, that too if it is for 4 th electron removal from the ion with 3+ oxidation state. So the maximum amount given here is 11577 kJ/mol.

a electronegativity decreases beacuse the size of nucleus is bigger.