Question

Match the following: 

• A. $A-I, B-IV, C-III, D-II$
• B. $A-II, B-IV, C-III, D-I$
• C. $A-II, B-III, C-IV, D-I$
• D. $A-I, B-III, C-IV, D-II$

Hint:

Understanding the periodic trends and electronic configurations helps accurately predict and match properties like ionization energies.

Answer 1

The Correct Option is C

Step 1: Assign each element its correct ionization enthalpy based on periodic trends.
First ionization enthalpies typically increase across a period from left to right, and within a group, it decreases from top to bottom:
${Be (Beryllium)}$ typically has a lower ionization energy compared to elements in p-block like Boron.
${O (Oxygen)}$ should have a lower ionization energy than Nitrogen due to the stability of the half-filled p orbital in nitrogen.
${N (Nitrogen)}$ often has higher ionization energy due to its electronic configuration.
${B (Boron)}$ being in Group 13, has ionization energy less than Be but higher than many p-block elements.
Step 2: Match the given enthalpies to the elements based on the option (3) which is the correct answer.
$A$ is matched with II $899 { kJ/mol}$.
$B$ is matched with III $1314 { kJ/mol}$.
$C$ is matched with IV $1402 { kJ/mol}$.
$D$ is matched with I $801 { kJ/mol}$.