Question

List$-\text{I}$ contains chemical reaction used in metallurgy and List$-\text{II}$ contains the name of process.

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I)}\underset{\text{Impure}}{Ni}+4CO\stackrel{330K}{\to }Ni(CO{)}_4\stackrel{450K}{\to }\underset{\text{pure}}{Ni}+4CO& \text{(P) Leaching}\\ \text{(II)}4Au+4NaCN+O_2+2H_{2}O\underset{-4KOH}{\overset{}{\to }}& \text{(Q) Mond's reduction}\\ \underset{\text{Soluble}}{4Na\left[Au\right(CN{)}_2]}\stackrel{Zn}{\to }2Na\left[Zn\right(CN{)}_4]+4Au& \\ \text{(III)}TiC{l}_4+2Mg\to Ti+2MgC{l}_2& \text{(R) Auto reduction}\\ & \\ \text{(IV)}2C{u}_{2}S+{30}_2\to 2C{u}_{2}O+2S{O}_2& \text{(S) Kroll process}\\ C{u}_{2}S+2C{u}_{2}O\to 6Cu+S{O}_2& \end{array}$

Which of the following options has the correct combination considering List$\text{-I}$ and List$\text{-II}$?

• A. $\left(I\right)\to P,\left(II\right)\to Q,\left(III\right)\to S,\left(IV\right)\to R$
• B. $\left(I\right)\to Q,\left(II\right)\to P,\left(III\right)\to S,\left(IV\right)\to R$
• C. $\left(I\right)\to Q,\left(II\right)\to R,\left(III\right)\to S,\left(IV\right)\to P$
• D. $\left(I\right)\to P,\left(II\right)\to R,\left(III\right)\to S,\left(IV\right)\to Q$

Answer 1

The correct option is B $\left(I\right)\to Q,\left(II\right)\to P,\left(III\right)\to S,\left(IV\right)\to R$

(I) $Ni$ purification is done by Mond's process which includes vaporizing impure vapours with $CO$ gas.
So, $\left(I\right)\to Q$

(II) Leaching is often used if the ore is soluble in some suitable solvent.
The given process is Mac-Arthur forrest cyanide process used for metallurgy of Au and Ag using cyanide.
So, $\left(II\right)\to P$

(III) Metallurgical process for $Ti$ which involves $Mg$ is called Kroll's process.
So, $\left(III\right)\to S$

(iv) Metals like $Cu,Pb,Hg$ (less electro-positive) in form of sulphide are heated in air as to convert part of the ore into oxide which then react with remaining sulphide ore to give metal and $S{O}_2$.
No external agent is used in this process and hence, called auto reduction.
So, $\left(IV\right)\to R$.