The fourth ionization energy of aluminum is significantly higher than the first three ionization energies. This jump can be attributed to the electronic structure of the aluminum atom and the energy required to remove an electron from a more stable electronic configuration.
When the first three electrons are removed from aluminum, the resulting Al^{3+} ion has a noble gas configuration, similar to that of neon. This stable configuration makes the Al^{3+} ion much less likely to lose another electron. As a result, removing a fourth electron to form Al^{4+} requires significantly more energy because the electron is being removed from a more stable and lower energy state.
In addition, the fourth ionization involves removing an electron from the outermost shell that is increasingly closer to the nucleus, where the effective nuclear charge experienced by the remaining electrons is higher. This means that the remaining electrons are held more tightly by the positive charge of the nucleus, requiring even greater energy to overcome this attraction.
In summary, the high fourth ionization energy of aluminum is due to the stability of the Al^{3+} ion and the increased effective nuclear charge felt by the outer electrons during the ionization process.