Question

Given below are two statements: Statement I: The second ionization enthalpy of B, Al and Ga is in the order of B>Al>Ga.
Statement II: The correct order in terms of first ionization enthalpy is Si<Ge<Pb<Sn.
In light of the above statements, choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true.
• B. Both Statement I and Statement II are false.
• C. Statement I is true but Statement II is false.
• D. Statement I is false but Statement II is true.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ionization enthalpy generally decreases down a group due to increased atomic size.
However, deviations occur in the p-block due to poor shielding by d and f-orbitals (d-block contraction and lanthanide contraction), increasing the effective nuclear charge (\(Z_{eff}\)) on valence electrons.
Step 2: Key Formula or Approach:
Evaluate Group 13 second IE anomalies (B, Al, Ga).
Evaluate Group 14 first IE anomalies (Si, Ge, Sn, Pb).
Step 3: Detailed Explanation:
Evaluate Statement I:
We are looking at the Second Ionization Energy (IE\(_2\)) of Group 13 elements.
B (Group 13, Period 2): \(\text{B}^+\) is \(1s^2 2s^2\). Removing an electron from the small \(2s\) orbital requires very high energy.
Al (Group 13, Period 3): \(\text{Al}^+\) is \([\text{Ne}] 3s^2\).
Ga (Group 13, Period 4): \(\text{Ga}^+\) is \([\text{Ar}] 3d^{10} 4s^2\).
Because the \(3d\) electrons in Gallium shield the nucleus very poorly, the \(4s\) electrons experience a significantly higher effective nuclear charge compared to Aluminum. Thus, it is harder to remove the \(4s\) electron from \(\text{Ga}^+\) than the \(3s\) electron from \(\text{Al}^+\).
The actual order of IE\(_2\) is \(\text{B}>\text{Ga}>\text{Al}\).
The statement claims \(\text{B}>\text{Al}>\text{Ga}\), which is FALSE.
Evaluate Statement II:
We are looking at the First Ionization Energy (IE\(_1\)) of Group 14 elements.
The general trend is a decrease down the group: \(\text{C}>\text{Si}>\text{Ge}>\text{Sn}\).
However, for Lead (Pb), the presence of filled \(4f\) and \(5d\) orbitals causes severe lanthanide contraction. The very poor shielding heavily increases \(Z_{eff}\), making Pb's IE\(_1\) slightly higher than Sn's.
The actual order of IE\(_1\) is \(\text{C}>\text{Si}>\text{Ge}>\text{Pb}>\text{Sn}\).
The statement claims \(\text{Si}<\text{Ge}<\text{Pb}<\text{Sn}\), which asserts a completely reversed and incorrect trend.
Therefore, Statement II is FALSE.
Step 4: Final Answer:
Both Statement I and Statement II are false.