d and f Block Elements JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on d and f Block Elements, free to read — no sign-in needed. The full chapter has 19 questions; sign in to attempt the remaining 14 in the exam simulator.

1. Q1JEE Advanced Adv 2024 (Paper 2)
   Paragraph: When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $\mathbf{P}$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $\mathbf{P}$. Addition of zinc chloride to $\mathbf{P}$ in a slightly acidic medium results in a sparingly soluble complex Q. Question: The number of zinc ions present in the molecular formula of $\mathbf{Q}$ is
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   Answer: 2.5

   $\mathrm{Zn}^{2+}+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$ OR $\mathrm{Zn}^{2+}+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow \mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
2. Q2JEE Advanced Adv 2024 (Paper 2)
   Paragraph: When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex $\mathrm{P}$ is formed. In a strong acidic medium, the equilibrium shifts completely towards $\mathrm{P}$. Addition of zinc chloride to $\mathrm{P}$ in a slightly acidic medium results in a sparingly soluble complex Q. Question: The number of moles of potassium iodide required to produce two moles of $\mathbf{P}$ is
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   Answer: 2

   $\mathrm{KI}+\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\frac{1}{2} \mathrm{I}_2$ Moles of $\mathrm{KI}$ required $=2$
3. Q3JEE Advanced Adv 2023 (Paper 1)
   In the scheme given below, $X$ and $Y$, respectively, are $\text{Metal halide}\rightarrow ^{aq.NaOH}\text{White precipitate}\left(P\right)+\text{Filtrate}\left(Q\right)$ $P⟶^{{PbO}_{2}(excess)}{}^{aq.{H}_{2}{SO}_{4}}_{heat}X\left(\text{a coloured species in solution}\right)$ $Q⟶^{MnO(OH{)}_{2},Conc.{H}_{2}{SO}_{4}}_{warm}Y\left(\text{gives blue-coloration with}KI-\text{starch paper}\right)$
   1. A.${CrO}_{4}^{2-}$ and ${Br}_{2}$
   2. B.${MnO}_{4}^{2-}and{Cl}_{2}$
   3. C.${MnO}_{4}^{-}and{Cl}_{2}$
   4. D.${MnSO}_{4}andHOCl$
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   Answer: (C)

   The metal halide can be manganese dichloride. It gives white precipitate of manganese hydroxide. ${MnCl}_{2}\rightarrow ^{aq.NaOH}Mn{\left(OH\right)}_{2}{}_{\left(P\right)}+NaCl_{\left(Q\right)}$ The white precipitate manganese hydroxide on reaction with lead dioxide and concentrated sulphuric acid gives purple coloured solution permanganate. $Mn{\left(OH\right)}_{2}\rightarrow _{Conc.{H}_{2}{SO}_{4},heat}^{{PbO}_{2}\left(excess\right)}{MnO}_{4}^{-}\left(purplesolution\right)$ The sodium chloride gives ${Cl}_{2}$ on reaction with $MnO{\left(OH\right)}_{2},conc.{H}_{2}{SO}_{4}$. The chlorine formed can oxidise iodide to iodine which gives blue colouration with starch paper.
4. Q4JEE Advanced Adv 2019 (Paper 2)
   Consider the following reactions (unbalanced) $Zn+Hotconc.{H}_{2}S{O}_{4}\rightarrow G+R+X$ $Zn+conc.NaOH\rightarrow T+Q$ $G+{H}_{2}S+{NH}_{4}OH\rightarrow Z\left(aprecipitate\right)+X+Y$ Choose the correct option(s).
   1. A.The oxidation state of $Zn$ in $T$ is $+1$
   2. B.Bond order of $Q$ is $1$ in its ground state
   3. C.$Z$ is dirty white in colour
   4. D.$R$ is a $V$ -shaped molecule
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   Answer: B,C,D

   $\text{Zn}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{Hot\&Conc.}\right)}\rightarrow {\text{ZnSO}}_{\text{4}}{}_{\left(\text{N}\right)}+{\text{SO}}_{\text{2}}{}_{®}+{\text{H}}_{\text{2}}\text{O}_{\left(\text{X}\right)}$ $\text{Zn}+\text{2NaOH}_{\left(\text{Conc.}\right)}\rightarrow {\text{NaZnO}}_{\text{2}}{}_{\left(\text{T}\right)}+{\text{H}}_{\text{2}}{}_{\left(\text{Q}\right)}$ ${\text{ZnSO}}_{4}+{\text{H}}_{\text{2}}\text{S}+{\text{NH}}_{4}\text{OH}\rightarrow \text{ZnS}_{\left(\text{S}\right)}+{\text{H}}_{\text{2}}\text{O}_{\left(\text{X}\right)}+{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{Y}\right)}$
5. Q5JEE Advanced Adv 2017 (Paper 2)
   The option(s) with only amphoteric oxides is(are)
   1. A.$C{r}_{2}{O}_{3},CrO,SnO,PbO$
   2. B.$NO,{B}_{2}{O}_{3},PbO,Sn{O}_{2}$
   3. C.$C{r}_{2}{O}_{3},BeO,SnO,Sn{O}_{2}$
   4. D.$ZnO,A{l}_{2}{O}_{3},PbO,Pb{O}_{2}$
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   Answer: C,D

   (iii) $C{r}_{2}{O}_{3},BeO,SnO,Sn{O}_{2}$ All are amphoteric oxides as they react with both acids as well as bases (iv) $ZnO,A{l}_{2}{O}_{3},PbO,Pb{O}_{2}$ All are amphoteric oxides as they react with both acids as well as bases