Respuesta :
Answer:
Option B
Explanation:
Energy required to remove outermost electron from an neutral atom is called ionization energy or first ionization energy ([tex]IE_1[/tex])
Energy required to remove an electron from M+ ion is called second ionization energy ([tex]IE_2[/tex]).
Energy required to remove an electron from M2+ ion is called third ionization energy [tex]IE_2[/tex]).
Therefore, among the given options:
[tex]Al+e^- \rightarrow Al^{+}(g),\ IE_1[/tex]
[tex]Al^++e^- \rightarrow Al^{2+}(g),\ IE_2[/tex]
[tex]Al^{2+}+e^- \rightarrow Al^{3+}(g),\ IE_3[/tex]
Therefore, the correct option is b
The third ionization of Al has been represented as [tex]\rm Al^2^+\;\+\;e^-\;\rightarrow\;Al^3^+[/tex]. Thus, option E is correct.
The ionization energy has been the energy required to remove the electron form the outermost orbital of a stable atom.
The first ionization energy has been the energy required to remove the first electron. The second ionization energy has been the removal of second electron from the atom, has been the second ionization energy, and so on.
The removal of an electron form the atom results in the ion with the positive charge. With every electron remove there has been a positive charge.
The ionization energy (IE) of Aluminum has been:
[tex]\rm Al\;\;+\;e^-\rightarrow\;Al^+\;IE_1\\Al^+\;\;+\;e^-\rightarrow\;Al^2^+\;IE_2\\Al^2^+\;+\;e^-\;\rightarrow\;Al^3^+\;IE_3[/tex]
The third ionization of Al has been resulted in formation of [tex]\rm Al^3^+[/tex]. Thus, the third ionization has been, [tex]\rm Al^2^+\;\+\;e^-\;\rightarrow\;Al^3^+[/tex]. Thus, option E is correct.
For more information about the ionization energy, refer to the link:
https://brainly.com/question/16243729
