The first, second, and third ionization energies of aluminum are 5781, 1817, and 2745 kJ mole⁻¹, respectively. Calculate the energy required for the fourth ionization.
a) 5781 kJ mole⁻¹
b) 1817 kJ mole⁻¹
c) 2745 kJ mole⁻¹
d) Data insufficient for calculation
