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: Understanding the Concept:
First ionization energy (IE\(_1\)) is the energy required to remove the most loosely bound electron from a neutral gaseous atom. Its value depends on factors like nuclear charge, atomic size, shielding effect, and the stability of the electronic configuration (e.g., half-filled and fully-filled orbitals).
Step 2: Detailed Explanation:
Let's analyze the ionization energy for each electronic configuration:
(D) ns\(^2\)np\(^6\): This is the configuration of a noble gas. Noble gases have a stable octet with fully filled p-orbitals. Removing an electron from such a stable configuration requires a very large amount of energy. Therefore, it will have the highest first ionization energy. So, (D) corresponds to the highest value, which is 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 extra stability results in a relatively high ionization energy. So, (C) will correspond to the second highest IE value, which is 1400 kJ mol\(^{-1}\) (II).
(A) ns\(^2\): This represents an alkaline earth metal. It has a fully filled s-orbital, which is relatively stable. Its IE will be higher than that of 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 is shielded by the inner ns\(^2\) electrons and is slightly further from the nucleus. This configuration will have the lowest ionization energy among the given options (excluding the noble gas). So, (B) corresponds to the lowest value, which is 800 kJ mol\(^{-1}\) (III).
Step 3: Final Answer:
Based on the analysis, the correct matching is:
(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 corresponds to option (4).