Question

Name the process:
\[Zr + 2{I_2}\xrightarrow{{{{600}^ \circ }c}}Zr{I_4}\xrightarrow{{{{1800}^ \circ }c}}Zr + 2{I_2}\]

Answer 1

Hint:In this question, we have to identify by which process Zirconium \[\left( {Zr} \right)\] is purified. Zirconium is a Group 4 element. This group lies in the d-block of the periodic table. Its atomic \[\left( {Zr} \right)\] number is 40 and its electronic configuration is \[[Kr]4{d^2}5{s^2}\].

Complete step by step answer:
Now, we will discuss all the given options in detail.
The first option is the Mond process. It is an example of vapor phase refining. This process is used for the refining of nickel. When impure nickel is heated in a current of \[CO\] at \[330\, - \,350\,K\] , it forms volatile nickel tetracarbonyl complex leaving the impurities which then heated to a higher temperature \[\left( {450 - 470\,K} \right)\] and undergoes decomposition to give a pure nickel. The above reaction is :
\[Ni + 4CO\xrightarrow{{330 - 350\,K}}Ni\left( {C{O_4}} \right)\xrightarrow{{450 - 470\,K}}Ni + 4CO\]
The second option is the Van Arkel process. It is also an example of vapor phase refining. This method is very useful for preparing ultra-pure (extremely pure) metals by removing the nitrogen and oxygen present in the form of impurity in certain metals such as titanium and zirconium. In this method, impure zirconium is heated in an evacuated vessel (a vessel which has vacuum conditions inside it) with iodine at \[{600^ \circ }C\], which forms a volatile compound zirconium tetraiodide \[\left( {Zr{I_4}} \right)\]. The volatile compound thus formed is separated. It is then decomposed by heating over tungsten filament at \[{1800^ \circ }C\] to give pure zirconium.
\[Zr + 2{I_2}\xrightarrow{{{{600}^ \circ }c}}Zr{I_4}\xrightarrow{{{{1800}^ \circ }c}}Zr + 2{I_2}\]
The third option is Hall’s process. This process is involved with the purification of Bauxite (an ore of aluminum). The method of purification depends upon the type of impurity present in the ore. If the ore contains more iron oxide, then it is called red Bauxite. In such a case, it is purified by Baeyer’s process or Hall’s process. In the Hall’ process, the ore is fused with sodium carbonate, when sodium meta-aluminate is formed.
\[A{l_2}{O_{3\,}} + N{a_2}C{O_3} \to 2NaAl{O_2} + C{O_2}\]
The sodium aluminate \[\left( {NaAl{O_2}} \right)\] is extracted with water due to which it dissolves and leaves behind oxide as residue. The water extracted containing sodium aluminate is heated to \[{60^ \circ }C\] and gas is passed through it to precipitate aluminum hydroxide \[\left( {Al{{\left( {OH} \right)}_3}} \right)\] which on ignition forms pure Bauxite \[\left( {A{l_2}{O_3}} \right)\]. The above reaction is:-
$2NaAl{O_2} + 3{H_2}O + C{O_2} \to 2Al{\left( {OH} \right)_3} + N{a_2}C{O_3}
2Al{\left( {OH} \right)_3} \to A{l_2}{O_3} + 3{H_2}O $
The fourth option is the Bosch process. It is an industrial process for manufacturing hydrogen by the catalytic reduction of steam with carbon monoxide. It is named after German chemist Carl Bosch.
The above reaction is:
\[C{O_2}\left( g \right)\, + \,2{H_2}(g) \to \,C\left( s \right)\, + \,2{H_2}O\,(g)\]
Hence, the correct option is 2.


Note:
Vapour phase refining is a method in which a crude metal is freed from impurities by first converting into a volatile compound by heating it with a specific reagent to a lower temperature and then decomposing the volatile compound to a higher temperature to give a pure metal.