Question

Match List-I with List-II 

Choose the correct answer from the options given below: 
 

• 1. (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
• 2. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
• 3. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
• 4. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Hint:

Remember the special stability associated with half-filled (np\(^3\), nd\(^5\)) and fully-filled (ns\(^2\), np\(^6\), nd\(^{10}\)) subshells. This leads to higher ionization energies than expected from the general trend across a period. Noble gases always have the highest IE in their period.

Answer 1

The Correct Option is D

Step 1: Introduction to Ionization Energy:
First ionization energy (IE\(_1\)) is the energy needed to remove the least tightly held electron from a neutral, gaseous atom. This value depends on factors like the positive charge of the nucleus, the atom's size, shielding effects, and the stability of the electron arrangement (e.g., half-filled and completely filled orbitals).
Step 2: Analysis of Ionization Energies:
Let's examine the ionization energy for each electronic configuration:
(D) ns\(^2\)np\(^6\): This represents a noble gas configuration. Noble gases have a stable octet due to their completely filled p-orbitals. Removing an electron from such a stable arrangement requires a significant amount of energy, making it the highest first ionization energy. Thus, (D) corresponds to the highest value: 2100 kJ mol\(^{-1}\) (I).
(C) ns\(^2\)np\(^3\): This configuration has a half-filled p-subshell. Half-filled orbitals are more stable than partially filled orbitals due to symmetrical electron distribution and maximum exchange energy. This stability leads to a relatively high ionization energy. So, (C) corresponds to the second highest IE: 1400 kJ mol\(^{-1}\) (II).
(A) ns\(^2\): This represents an alkaline earth metal. It has a full s-orbital, which is relatively stable. Its IE will be higher than the ns\(^2\)np\(^1\) configuration because the electron in the np orbital is further from the nucleus and better shielded. So, (A) corresponds to 900 kJ mol\(^{-1}\) (IV).
(B) ns\(^2\)np\(^1\): This represents an element from Group 13. The single electron in the p-orbital is relatively easy to remove because it's shielded by the inner ns\(^2\) electrons and further from the nucleus. This configuration will have the lowest ionization energy among the options (excluding the noble gas). So, (B) corresponds to the lowest value: 800 kJ mol\(^{-1}\) (III).
Step 3: Matching:
Based on the analysis:
(A) ns\(^2\) \(\rightarrow\) (IV) 900
(B) ns\(^2\)np\(^1\) \(\rightarrow\) (III) 800
(C) ns\(^2\)np\(^3\) \(\rightarrow\) (II) 1400
(D) ns\(^2\)np\(^6\) \(\rightarrow\) (I) 2100
This matches option (4).