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Question #44282
The Question
Zirconium (\ce{Zr}, ) and Hafnium (\ce{Hf}, ) both crystallize in a hexagonal close-packed (hcp) lattice. Extrapolating the group trend without considering the lanthanoid contraction, the expected atomic radius of \ce{Hf} would be . However, due to the poor shielding of the electrons, the actual atomic radius of \ce{Hf} is , which is slightly smaller than that of \ce{Zr} (). Given the density of \ce{Zr} is , calculate the difference between the actual density of \ce{Hf} and its expected density (if the lanthanoid contraction did not exist). Assume the hcp ratio is identical in all cases.\n\n[Given: , , ]
Options
- A.
- B.
- C.
- D.
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
Effective Nuclear Charge (Zeff)|d-orbital contraction|Post-Lanthanoid Radii Equality
Why This Question Is Tricky
Common Mistakes
This question tests the quantitative macroscopic consequences of the Lanthanoid Contraction.
The 'Hidden Variable' is the slight overcompensation of the contraction, making \ce{Hf} () slightly smaller than \ce{Zr} ().
Students often memorize that their radii are 'identical' and will erroneously use a 1:1 volume ratio, leading to the trap answer .
The problem demands rigorous application of the relationship rather than relying on generalized periodic trends.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Classification of Elements and Periodicity in Properties - Class 11 Part 1 |
| Concept | Effective Nuclear Charge (Zeff) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |