Question

The first ionization enthalpies of Na, Mg and Si are respectively 496, 737 and 786 kJ mol⁻¹. What would be the first ionization enthalpy of Al in kJ mol⁻¹?

• A. 450
• B. 750
• C. 575
• D. 800

Hint:

Ionization enthalpies increase from left to right in a period and from bottom to top in a group due to the increasing nuclear charge and decreasing atomic radius.

Answer 1

The Correct Option is C

The ionization enthalpy increases across a period as the nuclear charge increases. Given the ionization enthalpies of elements Na, Mg, and Si, we expect that the ionization enthalpy of Al lies in between the values of Mg and Si. Thus, for Al, the ionization enthalpy is approximately the average of those of Mg and Si: \[ \text{Ionization Enthalpy of Al} = \frac{737 + 786}{2} = 761.5 \, \text{kJ mol}^{-1} \] But, the first ionization enthalpy of Al is slightly less than the average due to its slightly smaller size than Si. Therefore, the value of 575 kJ mol}^{-1} is the best estimate. Thus, the first ionization enthalpy of Al is 575 kJ mol}^{-1}.

Answer 2

Step 1: Understanding Ionization Enthalpy Trends
Ionization enthalpy generally increases across a period from left to right due to increasing nuclear charge.
However, there are exceptions caused by electron configurations and subshell stability.

Step 2: Given Data and Expected Trend
- Na (Group 1): 496 kJ/mol
- Mg (Group 2): 737 kJ/mol
- Si (Group 14): 786 kJ/mol
- Al (Group 13) lies between Mg and Si in the periodic table.

Step 3: Reason for Lower Ionization Enthalpy of Al than Expected
Al has electron configuration ending in 3p¹, which is higher in energy and easier to remove than 3s² electrons in Mg.
Thus, its ionization enthalpy is lower than expected by the general trend.

Step 4: Conclusion
The first ionization enthalpy of Al is approximately 575 kJ/mol, which fits the expected dip after Mg.
This is due to the ease of removing the 3p electron compared to the 3s electrons in Mg.