In van Arkel method, if I2 is introduced at 1800 K over impure zirconium metal, the product will be:
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a
iodide of the metal
b
pure metal
c
impurities react with iodine
d
none of these
answer is D.
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Detailed Solution
Zr (impure)+2I2→<1800 KZrI4;ZrI4→>1800KZr(pure)+2I2 and hence over 1800 K practically no reaction can take place between Zr and I2.Therefore, (4) option is correct.