In Van Arkel method, if $ \mathrm{I}_{2} $ is introduced at $ 1800 \mathrm{~K} $ over impure zirconium metal, the product will be

Question

In Van Arkel method, if $ \mathrm{I}_{2} $ is introduced at $ 1800 \mathrm{~K} $ over impure zirconium metal, the product will be, iodide of the metal., pure metal., impurities react with iodine., none of these.

MARKS App · Accepted Answer

Zr impure + 2 I 2 ⟶ 1800 K Zrl 4 ; Zrl 4 ⟶ > 1800 K Zr pure + 2 I 2 and hence over 1800 ° C practically no reaction can take place between Zr and I 2 . Therefore, option 4 is correct.

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