Question

The first ionization energy (in kJ/mol) of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ respectively

• A. 496, 737, 577, 786
• B. 496, 577, 737, 786
• C. 496, 577, 786, 737
• D. 786, 737, 577, 496

Answer 1

The correct option is A

496, 737, 577, 786

Explanation of the correct option:

The correct answer is option A:

1. The predicted sequence is $\mathrm{Na}<\mathrm{Mg}<\mathrm{Al}<\mathrm{Si}$.
2. Because of Magnesium's fully filled s-subshell and Aluminum's ${\mathrm{s}}^2{\mathrm{p}}^1$ structure, which makes it relatively simple for Aluminium to lose its outermost electron, the actual/experimental order turns out to be $\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$.
3. Thus, the first ionization energies of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ are 496, 737, 577, and 786 respectively.

Explanation of the incorrect options:

Option B:

1. The correct order of ionization energy is $\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$.
2. Thus, the respective first ionization energies of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ are not 496, 577, 737, and 786.

Option C:

1. The correct order of ionization energy is $\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$.
2. Thus, the respective first ionization energies of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ are not 496, 577, 786, and 737.

Option D:

1. The correct order of ionization energy is $\mathrm{Na}<\mathrm{Al}<\mathrm{Mg}<\mathrm{Si}$.
2. Thus, the respective first ionization energies of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ are not 786, 737, 577, and 496.

Hence, option A is correct.
The first ionization energy (in kJ/mol) of $\mathrm{Na},\mathrm{Mg},\mathrm{Al},$ and $\mathrm{Si}$ respectively 496, 737, 577, and 786.