Question

which of the following statements is false concerning the first ionization energy of sulfur? all the ionization energies listed below refer to first ionization energies. select one: the ie of s is higher than ie of p the ie of s is higher than ie of se the ie of s is higher than ie of mg the ie of s is lower than ie of cl the ie of s is lower than ie of s+

Explanation:

• For option 1:
• Phosphorus (\(P\)) has a higher first - ionization energy than sulfur (\(S\)). Phosphorus has a half - filled \(3p\) subshell (\(3p^{3}\)), which is a relatively stable electron configuration. Removing an electron from sulfur (\(3p^{4}\)) is easier because after losing one electron, it attains a half - filled \(3p\) subshell.
• For option 2:
• As we move down a group (sulfur (\(S\)) and selenium (\(Se\)) are in the same group), the first - ionization energy decreases. This is because the atomic radius increases, and the outermost electrons are less tightly held by the nucleus. So, \(IE(S)>IE(Se)\).
• For option 3:
• Magnesium (\(Mg\)) has a \(3s^{2}\) electron configuration. Sulfur is to the right of magnesium in the periodic table. As we move from left to right across a period (from \(Mg\) to \(S\)), the nuclear charge increases, and the atomic radius decreases. So, \(IE(S)>IE(Mg)\).
• For option 4:
• Chlorine (\(Cl\)) is to the right of sulfur in the same period. As we move from left to right across a period, the nuclear charge increases, and the atomic radius decreases. So, \(IE(Cl)>IE(S)\).
• For option 5:
• The first - ionization energy of \(S\) is the energy required to remove an electron from \(S\) (\(S

ightarrow S^{+}+e^{-}\)), and the first - ionization energy of \(S^{+}\) is the energy required to remove an electron from \(S^{+}\) (\(S^{+}
ightarrow S^{2 +}+e^{-}\)). Since \(S^{+}\) has a higher effective nuclear charge (because it has lost one electron) acting on the remaining electrons, \(IE(S)<IE(S^{+})\).

Answer:

The IE of S is higher than IE of P