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Question #44317

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The Question

Elements $M$ and $N$ belong to Group 14 of the periodic table, located in the 5th and 6th periods, respectively. Their successive ionization energies ( $IE_1$ to $IE_4$ in $\text{kJ mol}^{-1}$ ) measured at $298\text{ K}$ are given below:

$M$ : $IE_1 = 709$ , $IE_2 = 1412$ , $IE_3 = 2943$ , $IE_4 = 3930$ $N$ : $IE_1 = 716$ , $IE_2 = 1450$ , $IE_3 = 3081$ , $IE_4 = 4083$

The reluctance of the $ns^2$ electron pair to participate in bonding in heavier elements is known as the Inert Pair Effect. Based on the provided thermodynamic data, calculate the exact additional energy barrier (in $\text{kJ mol}^{-1}$ ) specifically associated with the removal of this 'inert' $ns^2$ electron pair in the gaseous state for element $N$ as compared to element $M$ .

Options

A.
$291 \text{ kJ mol}^{-1}$
B.
$336 \text{ kJ mol}^{-1}$
C.
$45 \text{ kJ mol}^{-1}$
D.
$138 \text{ kJ mol}^{-1}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Inert Pair Effect|Successive Ionization Energies|Ionization Enthalpy (IE) Anomalies

Why This Question Is Tricky

Common Mistakes

The question tests the quantitative understanding of the Inert Pair Effect using successive ionization energies.

Students often fall into the trap of calculating the total ionization energy difference ( $IE_1$ through $IE_4$ , which is $336 \text{ kJ mol}^{-1}$ ) or the $p$ -electron difference ( $45 \text{ kJ mol}^{-1}$ ).

The 'Hidden Variable' is recognizing that the Inert Pair Effect specifically stabilizes the $ns^2$ electrons, meaning the energetic anomaly must be isolated strictly to $IE_3$ and $IE_4$ .

The poor shielding by $4f$ and $5d$ electrons causes a relativistic contraction, making the $6s$ electrons significantly harder to remove than the $5s$ electrons, which is the fundamental thermodynamic reason Pb(IV) is a strong oxidizing agent compared to Sn(IV).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Inert Pair Effect
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free