Q24 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
The first ionisation enthalpy of Al in $\mathrm{kJ} \mathrm{mol}^{-1}$ is:
[Given the first ionisation enthalpy of $\mathrm{Na}, \mathrm{Mg}$ and Si in $\mathrm{kJ} \mathrm{mol}^{-1}$ are 497, 738 and 787 respectively]
- A857
- B769
- Correct answerC578
- D488
Solution
The first ionisation enthalpy values for the elements in the third period are as follows: Na (497), Mg (738), Al, and Si (787).
The general trend for ionisation enthalpy across a period is an increase due to the increase in effective nuclear charge. However, there are exceptions due to the stability of electronic configurations.
The electronic configuration of Mg is $[Ne] 3s^2$, which is a fully filled s-orbital, making it exceptionally stable. Consequently, the first ionisation enthalpy of Mg is higher than that of Al, which has the configuration $[Ne] 3s^2 3p^1$.
The electronic configuration of Al is $[Ne] 3s^2 3p^1$. Removing the $3p^1$ electron is easier than removing a $3s^2$ electron from Mg. Thus, the ionisation enthalpy of Al must be lower than that of Mg (738) but higher than that of Na (497).
Among the given options, 578 is the only value that lies between 497 and 738.
Answer: 578
The general trend for ionisation enthalpy across a period is an increase due to the increase in effective nuclear charge. However, there are exceptions due to the stability of electronic configurations.
The electronic configuration of Mg is $[Ne] 3s^2$, which is a fully filled s-orbital, making it exceptionally stable. Consequently, the first ionisation enthalpy of Mg is higher than that of Al, which has the configuration $[Ne] 3s^2 3p^1$.
The electronic configuration of Al is $[Ne] 3s^2 3p^1$. Removing the $3p^1$ electron is easier than removing a $3s^2$ electron from Mg. Thus, the ionisation enthalpy of Al must be lower than that of Mg (738) but higher than that of Na (497).
Among the given options, 578 is the only value that lies between 497 and 738.
Answer: 578
Hence, the correct answer is option: 'C'.