Question

If four different elements A, B, C and D having outer electronic configuration as,
\(\text{A} = 3s^2 3p^4, \text{B} = 3s^2 3p^5, \text{C} = 4s^2 4p^4, \text{D} = 4s^2 4p^5\)
identify the element with lowest ionization enthalpy \((\Delta_i \text{H}_1)\)

• A. A
• B. B
• C. C
• D. D

Hint:

Down group → IE decreases due to increased atomic size.

Answer 1

The Correct Option is C

Step 1: Identify elements.
• A: \(3s^2 3p^4\) → Sulfur (Period 3, Group 16)
• B: \(3s^2 3p^5\) → Chlorine (Period 3, Group 17)
• C: \(4s^2 4p^4\) → Selenium (Period 4, Group 16)
• D: \(4s^2 4p^5\) → Bromine (Period 4, Group 17)

Step 2: Apply trends.
• Ionization enthalpy decreases down the group
• Increases across the period Thus: \[ \text{S} > \text{Se}, \quad \text{Cl} > \text{Br} \]

Step 3: Compare all. \[ \text{Lowest IE} = \text{Selenium (C)} \] Conclusion: \[ \text{Answer = (C)} \]