What will be the product formed when the crude metal of Zirconium is heated in an evacuated vessel with iodine?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(ZrI_4\)
In the van Arkel method crude zirconium is heated with iodine in an evacuated vessel. Zirconium (Z = 40, [Kr] 4d2 5s2) uses all four valence electrons and forms zirconium(IV) iodide, which, being covalent, is volatile and distils away from the impurities (oxygen and nitrogen):
\( \ce{Zr + 2I2 -> ZrI4} \)
The vapour is decomposed on a tungsten filament heated electrically to about 1800 K, and pure zirconium deposits on the filament:
\( \ce{ZrI4} \xrightarrow{1800\,\mathrm{K}} \ce{Zr + 2I2} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Vapour Phase Refining – …
