Final answer:
The order of increasing first ionization energy among Mg, O, S, Si is as follows: Mg (lowest), Si, S, and O (highest), based on their respective positions in the periodic table and the trends that ionization energy increases across a period and decreases down a group. The correct option is d.
Step-by-step explanation:
Order of Increasing First Ionization Energy
First ionization energy generally increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table. Therefore, when ranking the first ionization energy for the atoms Mg, O, S, Si, we consider their positions on the periodic table.
Magnesium (Mg) is in the second group, silicon (Si) is in the third group, sulfur (S) is in the sixteenth group, and oxygen (O) is in the sixteenth group but precedes sulfur.
Since ionization energy increases across a period and decreases down a group, we expect Mg to have the lowest ionization energy and O to have the highest among the four. The complete order from lowest to highest ionization energy is therefore Mg, Si, S, O.
Rationale for Ranking
Magnesium has the lowest first ionization energy because it is furthest to the left and has the fewest protons of the elements listed, providing less nuclear attraction to the valence electrons.
Silicon has a higher first ionization energy than Mg due to its increased nuclear charge and fewer electron shielding effects than higher group members.
Sulfur has a higher ionization energy than Si as it is in the same period but to the right, implying a greater nuclear charge and a smaller atomic radius, which increase ionization energy.
Oxygen has the highest first ionization energy of these atoms because it is the most rightward element in the period among the listed elements, indicating the highest effective nuclear charge acting on the electrons to be ionized. The correct option is d.