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Question #44257

Difficulty: 8/10 | 🟢 Free

The Question

Consider the Group 6 transition metals: $M_1$ (3d series), $M_2$ (4d series), and $M_3$ (5d series). Which of the following options correctly identifies the trend in their first ionization enthalpies ( $IE_1$ ) and the true ground-state electronic configuration of $M_3$ ?

Options

A.
$IE_1(M_1) > IE_1(M_2) > IE_1(M_3)$ ; $M_3 = [\ce{Xe}] 4f^{14} 5d^5 6s^1$
B.
$IE_1(M_3) > IE_1(M_2) > IE_1(M_1)$ ; $M_3 = [\ce{Xe}] 4f^{14} 5d^4 6s^2$
C.
$IE_1(M_3) > IE_1(M_1) > IE_1(M_2)$ ; $M_3 = [\ce{Xe}] 4f^{14} 5d^5 6s^1$
D.
$IE_1(M_2) > IE_1(M_3) > IE_1(M_1)$ ; $M_3 = [\ce{Xe}] 4f^{14} 5d^4 6s^2$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Aufbau Principle Exceptions|Ionization Enthalpy (IE) Anomalies|Periodic Properties of Lanthanides

Why This Question Is Tricky

Common Mistakes

This question tests two interconnected exceptions.

First, students blindly apply the 'half-filled subshell stability' rule down the entire group, assuming Tungsten (W) is $5d^5 6s^1$ .

Second, they incorrectly apply the s-block periodic trend (IE decreases down a group) to the d-block.

The trap lies in failing to account for the Lanthanide contraction and relativistic effects, which stabilize the $6s$ orbital so much that the Aufbau exception is reversed for W, and its ionization enthalpy becomes the highest in the group.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Structure of Atom - Class 11 Part 1
Concept	Aufbau Principle Exceptions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free