When sulfur loses one electron, it becomes a particularly stable, half-filled p subshell. The removal of this first electron requires less energy than the removal of an electron from phosphorus, which is initially a half-filled p subshell. This signifies that .

The question is incomplete, the complete question is;

When sulfur loses one electron, it becomes a particularly stable, half-filled p subshell. The removal of this first electron requires less energy than the removal of an electron from phosphorus, which is initially a half-filled p subshell. This signifies that the first ionization energy of sulfur is larger than the first ionization energy of phosphorus. the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of phosphorus is larger than the second ionization energy of sulfur.

the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus

Explanation:

Let us look back at the electronic configuration of each of the atoms;

Sulphur; [Ne] 3s² 3p⁴

Phosphorus; [Ne] 3s² 3p³

We can easily see that phosphorus has an exactly filled half filled 3p sublevel. This partially filled orbital has a great deal of stability associated with it.

On the other hand, sulphur can attain this stability that results from a half filled orbital by loosing one of its p electrons. The energy required for this process is much lower than the energy required to remove an electron from an already half filled 3p orbital of the phosphorus atom.

Hence, the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus.

the correct answer would be d. hope this !

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa