Question

The \( 1^{\text{st}} \) ionization enthalpy for Mg is +737 kJ/mol. The most probable estimated value of the \( 2^{\text{nd}} \) ionization enthalpy of Mg is _______ kJ/mol.

• A. \( -906 \text{ kJ/mol} \)
• B. \( -856 \text{ kJ/mol} \)
• C. \( +1450 \text{ kJ/mol} \)
• D. \( +590 \text{ kJ/mol} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ionization enthalpy (IE) is the energy required to remove an electron from an isolated gaseous atom or ion. Successive ionization enthalpies always increase ($IE_1<IE_2<IE_3 \dots$) because removing an electron from a positively charged ion requires significantly more energy than removing it from a neutral atom. Additionally, ionization is strictly an endothermic process.
Step 2: Key Formula or Approach:
For any element: $IE_2>IE_1>0$.
Removing the second electron from a Group 2 metal usually takes roughly double the energy of the first.
Step 3: Detailed Explanation:
Given $IE_1 = +737 \text{ kJ/mol}$.
We must estimate $IE_2$.
1. Ionization energy represents an input of energy, hence it must always be a positive value. This immediately eliminates options (A) $-906 \text{ kJ/mol}$ and (B) $-856 \text{ kJ/mol}$.
2. The second ionization energy ($IE_2$) involves overcoming the stronger electrostatic pull of the newly formed $+1$ cation, ensuring that $IE_2$ is strictly greater than $IE_1$.
$IE_2>+737 \text{ kJ/mol}$.
This eliminates option (D) $+590 \text{ kJ/mol}$.
The only physically viable option remaining that is both positive and greater than $737 \text{ kJ/mol}$ is $+1450 \text{ kJ/mol}$.
Step 4: Final Answer:
The estimated value is +1450 kJ/mol.