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 \(801 { kJ/mol}\).