Question

Which of the following match are incorrect?

(a) Goldschmidt aluminothermite process - ${Cr}_2{O}_3$
(b) Mac Arther cyanide process - $Fe$
(c) Mond process - $Ni$
(d) Van Arkel process - $Au$

• A. a, c
• B. c, d
• C. b, d
• D. a, b

Answer 1

The correct option is C b, d

Goldschmidt aluminothermite process is used for $C{r}_2{O}_3$ and Mond's process is used for Ni. Therefore a and c are correct.

Mac Arthur cyanide process is not used for Fe and Van Arkel process is not used for Au. Hence, b and d are incorrect.

Goldschmidt aluminothermite process: Certain Oxides of Mn, Cr do not reduce by carbon due to very high melting point. They are reduced by $Al$.

$C{r}_2{O}_3+2Al\to 2Cr+A{l}_2{O}_3$

$3M{n}_3{O}_4+8Al\to 9Mn+4A{l}_2{O}_3$

Cyanide process (Mac Arthur Forrest Process): This is a very good process for the removal of gold from tailings of amalgamation process or from even poorer ores. It is based on the fact that gold dissolves in dilute solution of sodium cyanide in presence of atmospheric oxygen with the formation of complex cyanide.

$A{g}_{2}S+4NaCN\to 2Na\left[Ag\right(CN{)}_2]+N{a}_{2}S$

Silver is obtained by reaction with Zn:

$2Na\left[Ag\right(CN{)}_2]+Zn\to N{a}_2[Zn\left(CN{)}_4\right]+2Ag$

Mond’s Method: It is specially used for $Ni$. The vapour phase refining of $Ni$ on heating in a stream of $CO$ forms Volatile $Ni(CO{)}_4$. These vapours when subjected to still higher temperature, decompose to give pure metal.

$Ni+4CO\stackrel{{50}^o}{\to }Ni(CO{)}_4\stackrel{{230}^o}{\to }Ni+4CO$

Van-Arkel method: This method is used for ultrapure metals like Ti, Zr etc. The metal is converted into its volatile stable compound such as $Ti{I}_4$, leaving behind impurities. The stable volatile compound thus formed is decomposed to get the pure metal.

$\frac{Ti}{\left(impure\right)}+2I_2\stackrel{500k}{\to }Ti{I}_4\stackrel{1675K}{\to }Ti+\frac{2I_2}{\left(pure\right)}$