Answer
- First ionization: $$Al(g)\to Al^+(g)+e^-$$
- Second ionization: $$Al^+(g)\to Al^{2+}+e^-$$
- Third ionization: $$Al^{2+}\to Al^{3+}+e^-$$
The first ionization requires the least amount of energy.
Work Step by Step
1) Ionization energy is the minimum energy required to remove an electron from the ground state of the isolated gaseous atom or ion. The atom that has electrons removed is said to be ionized.
Therefore, the equations that show the process that describe:
- The first ionization energy of aluminum atom:
The first ionization energy is the minimum energy required to remove the first electron from a neutral atom. Therefore, the equation is, $$Al(g)\to Al^+(g)+e^-$$
- The second ionization energy of aluminum atom:
The second ionization energy is the minimum energy required to remove the second electron from the ion after the first one has been removed. The equation, thus, is, $$Al^+(g)\to Al^{2+}+e^-$$
- The third ionization energy of aluminum atom:
The third ionization energy is the minimum energy required to remove the third electron from the ion after the second one has been removed. Thus the equation is $$Al^{2+}\to Al^{3+}+e^-$$
2) The first ionization energy is the least since the first ionization removes an electron from a neutral atom. All the other ionizations remove electrons from cations, which means the core of the aluminum atom has stronger effective nuclear charge upon the electrons. Therefore, to move an electron out of there requires higher energy than to remove the first one.
Thus, all the other ionization energies are higher than the first ionization energy.
