Respuesta :
The first ionization energies from highest to lowest for the given elements are as such P>S>Si>Al.
Answer: D
Explanation
Ionization energy, the energy required to remove the valence shell electron from the atom and thus converting the atoms to ions.
The ionization energy should be more than the binding energy as then only, the outermost electron can be removed by this energy.
If the energy is required to remove electrons from the outermost shell, then it will be termed as first ionization energy.
If the energy is required to remove the electrons from the second outermost shell, then that will be termed as second ionization energy.
As here, the first ionization energy should be compared that means the n value is 1, as the energy is required to remove electrons from the outermost shell.
So the n value will be one, so the ionization energy will be fully dependent on the atomic number of the element.
So here we know that ,the atomic number of aluminium (Al) is 13, phosphorous (P) is 15, silicon (Si) is 14 and sulfur (S) is 16.
So according to atomic number, the ionization energy will be increasing in the order of P>S>Si>Al.
The ionization energy of P is more than S while theoretically, the S ionization energy should be more than P.
This is because of the electron shielding effect occurring in S which reduces the binding force acting on the outermost shell of S by the nucleus. Thus , the increasing order of 1st ionization energy is P>S>Si>Al.
