General Principles and Processes of Isolation of Metals JEE Main previous year questions with solutions

5 solved JEE Main questions on General Principles and Processes of Isolation of Metals, free to read — no sign-in needed. The full chapter has 127 questions; sign in to attempt the remaining 122 in the exam simulator.

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1. Q1JEE Main 2024 (29 Jan, Shift 1)
   Identify the incorrect pair from the following :
   1. A.Fluorspar- ${BF}_{3}$
   2. B.Cryolite- ${Na}_{3}{AlF}_{6}$
   3. C.Fluoroapatite- $3{Ca}_{3}{\left({PO}_{4}\right)}_{2}\cdot {CaF}_{2}$
   4. D.Carnallite- $KCl\cdot {MgCl}_{2}\cdot 6{H}_{2}O$
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   Answer: (A)

   Fluorine is present mainly as insoluble fluorides (fluorspar ${CaF}_{2}$, cryolite ${Na}_{3}{AlF}_{6}$ and fluoroapatite $3{Ca}_{3}({PO}_{4}{)}_{2}.{CaF}_{2}$ ). Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium, but is mainly sodium chloride solution (2.5% by mass). The deposits of dried up seas contain these compounds, e.g., sodium chloride and carnallite, $KCl.{MgCl}_{2}.6{H}_{2}O$.
2. Q2JEE Main 2022 (28 Jul, Shift 2)Entropy and Second law of thermodynamics
   Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$ Assertion $A$ : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state. Reason $R$ : The value of ${\Delta G}^{\Theta }$ becomes more on negative side as entropy is higher in liquid state than solid state. In the light of the above statements. Choose the most appropriate answer from the options given below
   1. A.Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$
   2. B.Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$
   3. C.$A$ is correct but $R$ is not correct
   4. D.$A$ is not correct but $R$ is correct
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   Answer: (A)

   The entropy is higher if the metal is in liquid state than when it is in solid state. The value of entropy change $(∆S)$ of the reduction process is more on positive side when the metal formed is in liquid state and the metal oxide being reduced is in solid state. Thus the value of ${∆}_{f}G$ becomes more on negative side and the reduction becomes easier. $\Delta G=\Delta H-T\Delta S$ $∵$ Entropy of liquid is more than solid $∴$ on melting the entropy increases and $\Delta G$ becomes more negative and hence it becomes easier to reduce metal.
3. Q3JEE Main 2021 (31 Aug, Shift 1)Reactions of Alcohols
   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Aluminium is extracted from bauxite by the electrolysis of molten mixture of ${Al}_{2}{O}_{3}$ with cryolite. Reason (R) : The oxidation state of $Al$ in cryolite is $+3.$ In the light of the above statements, choose the most appropriate answer from the options given below :
   1. A.Both (A) and (R) are correct and (R) is the correct explanation of (A).
   2. B.(A) is true but (R) is false.
   3. C.Both (A) and (R) are correct but (R) is not the correct explanation of (A).
   4. D.(A) is false but (R) is true.
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   Answer: (C)

   In electrolytic reduction of ${Al}_{2}{O}_{3}$ we use cryolite to increase conductivity and decrease melting point. Cryolyte is ${Na}_{3}{AIF}_{6},$ Oxidation number of $Al$ is $+3$
4. Q4JEE Main 2020 (06 Sep, Shift 2)Third law of thermodynamics
   For a reaction $4M\left(s\right)+{nO}_{2}\left(g\right)\rightarrow 2{M}_{2}{O}_{n}\left(s\right)$ The free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :
   1. A.the slope change from negative to positive
   2. B.the free energy change show a change from negative to positive value
   3. C.the slope changes from positive to negative
   4. D.the slope changes from positive to zero
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   Answer: (B)

   For oxide to be stable its $\Delta G$ of formation value should be negative. Gibbs free energy, also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant.
5. Q5JEE Main 2023 (15 Apr, Shift 1)
   Which one of the following is not an example of calcination?
   1. A.${CaCO}_{3}\rightarrow ^{∆}CaO+{CO}_{2}$
   2. B.${Fe}_{2}{O}_{3}.{xH}_{2}O\rightarrow ^{∆}{Fe}_{2}{O}_{3}+{xH}_{2}O$
   3. C.$2PbS+3{O}_{2}\rightarrow ^{∆}2PbO+2{SO}_{2}$
   4. D.${CaCO}_{3}.{MgCO}_{3}\rightarrow ^{∆}CaO+MgO+2{CO}_{2}$
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   Answer: (C)

   Calcination is a process in which ore is heated in the absence of air or air might be supplied in a limited quantity. Roasting involves the heating of ore lower than its melting point in the presence of air or oxygen. $2PbS+3{O}_{2}\rightarrow ^{∆}2PbO+2{SO}_{2}$ The above reaction is an example of roasting. Rest all the reactions are heated in the absence of oxygen.