Question

Which of the following statements is wrong?
(A) The first ionization potential of Al is less than the first ionization potential of Mg.
(B) The second ionization potential of Mg is greater than the second ionization potential of Na
(C) The first ionization potential of Na is less than the first ionization potential of Mg
(D) The third ionization potential of Mg is greater than the third ionization potential of Al.

Answer 1

Hint: Ionization potential is the energy required to remove one electron from the atom. The second ionization potential indicates the energy required to remove another electron after the first ionization.

Complete step by step solution:
We will check all the statements to find whether it is wrong or not but before that let’s see what is meant by ionization potential.
-Ionization potential is the energy required to remove one electron from the atom. As we remove the electron for the first time, it is called first ionization potential and then as the second electron is removed from the atom; the potential is called second ionization potential and so on.
(A) Al : $1s^{2}2s^{2}2p^{6}3s^{2}3p^1$
Mg :$1s^{2}2s^{2}2p^{6}3s^2$
Now, we will discuss the first ionization potential of Al and Mg. The valence orbital of Al has only an electron and so it will require less energy to remove it. In Mg, its valence orbital 3s is full and to remove one electron, it will require more energy. So, we can say that the first ionization potential of Al is less than the first ionization potential of Mg.
(B) Mg: $1s^{2}2s^{2}2p^{6}3s^2$ , $M{g}^{+}:1s^{2}2s^{2}2p^{6}3s^1$
Na:$1s^{2}2s^{2}2p^{6}3s^1$, $N{a}^{+}:1s^{2}2s^{2}2p^6$
Here, we will discuss the second ionization potential of Mg and Na. Upon losing the first electron, Mg and Na will become $M{g}^{+}$ and $N{a}^{+}$ respectively. Now as we will remove the second electron, the valence electron is in 3s orbital in the case of $M{g}^{+}$ and in case of $N{a}^{+}$, the valence orbital is full. So, it will require more energy to remove the second electron of Na. Thus, the second ionization potential of Mg is less than the second ionization potential of Na. So, statement (B) is wrong.
(C) The first ionization potential of Mg will be higher because the valence orbital of Mg which is 3s is full. While Na has only an electron in its valence orbital (3s). So, its potential will be less. Thus we can say that the first ionization potential of Na is less than the first ionization potential of Mg.
(D) Upon losing two electrons, electronic configuration of Mg and Al will be:
$$M{g}^{2+}:1s^{2}2s^{2}2p^6$$
$A{l}^{2+}:1s^{2}2s^{2}2p^{6}3s^1$
We can see that it will be easy to remove an electron from $A{l}^{2+}$ because there is only one electron and upon losing it, $A{l}^{2+}$ will attain noble gas configuration. It will be hard to remove an electron from $M{g}^{2+}$ because its valence orbital is full. So, we can say that the third ionization potential of Mg is greater than the third ionization potential of Al.

Therefore option (B) is the correct answer.

Note: Remember that second ionization potential shows the energy to remove one electron from the atom in which one electron is already removed. So, do not consider that second ionization potential is the sum of the energies required to remove two electrons.