Question

The correct trend in the first ionization enthalpies of the elements in the 3rd period of periodic table is :

• A. $ Al
• B. $Si
• C. $S
• D. $Al

Hint:

Whenever you see a Nitrogen vs Oxygen or Phosphorus vs Sulfur comparison in Ionization Energy, the half-filled configuration always wins!

Answer 1

The Correct Option is A

The question requires us to determine the correct trend in the first ionization enthalpies of elements in the 3rd period of the periodic table.

The elements in the 3rd period of the periodic table are: Sodium (Na), Magnesium (Mg), Aluminum (Al), Silicon (Si), Phosphorus (P), Sulfur (S), Chlorine (Cl), and Argon (Ar). However, we are only concerned with Al, Si, P, S, and Cl as per the given options.

First ionization enthalpy is defined as the energy required to remove the outermost electron from a gaseous atom. Across a period, the ionization enthalpy generally increases as atomic size decreases and nuclear charge increases, making it harder to remove an electron.

Explanation for the given options:

• Option 1: Al < Si < S < P < Cl
• Option 2: Si < S < Al < P < Cl
• Option 3: S < Si < Al < P < Cl
• Option 4: Al < S < P < Si < Cl

The general trend in the 3rd period for ionization enthalpy is that it increases from left to right. However, there are small deviations due to subshell electronic configurations:

• Between Mg and Al: Al has a lower ionization enthalpy than Mg due to Al’s electron being removed from a new p-orbital, which is higher in energy than Mg's filled s-orbital.
• Between P and S: S has a slight decrease in ionization enthalpy compared to P because removing an electron from S relieves the electron-electron repulsion in the p-orbitals (pairing occurs in S).

Correct trend: Al < Si < S < P < Cl

The above option reflects these minor deviations and correctly orders all the elements based on their first ionization enthalpies.