Question

Which of the following correctly represents the third ionization of aluminum?

Answer 1

Answer:

Option B

Explanation:

Energy required to remove outermost electron from an neutral atom is called ionization energy or first ionization energy ($IE_1$)

Energy required to remove an electron from M+ ion is called second ionization energy ($IE_2$).

Energy required to remove an electron from M2+ ion is called third ionization energy $IE_2$).

Therefore, among the given options:

$Al+e^- \rightarrow Al^{+}(g),\ IE_1$

$Al^++e^- \rightarrow Al^{2+}(g),\ IE_2$

$Al^{2+}+e^- \rightarrow Al^{3+}(g),\ IE_3$

Therefore, the correct option is b