p Block Elements (Group 15, 16, 17 & 18) JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on p Block Elements (Group 15, 16, 17 & 18), free to read — no sign-in needed. The full chapter has 55 questions; sign in to attempt the remaining 50 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   During sodium nitroprusside test of sulphide ion in an aqueous solution, one of the ligands coordinated to the metal ion is converted to
   1. A.$\mathrm{NOS}^{-}$
   2. B.$\mathrm{SCN}^{-}$
   3. C.$\mathrm{SNO}^{-}$
   4. D.$\mathrm{NCS}^{-}$
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   Answer: (A)

   $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NO})\right]+\mathrm{Na}_2 \mathrm{~S} \rightarrow \mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5(\mathrm{NOS})\right]$ Sodium nitroprusside purple solution
2. Q2JEE Advanced Adv 2025 (Paper 1)
   The heating of $\mathrm{NH}_4 \mathrm{NO}_2$ at $60-70^{\circ} \mathrm{C}$ and $\mathrm{NH}_4 \mathrm{NO}_3$ at $200-250^{\circ} \mathrm{C}$ is associated with the formation of nitrogen containing compounds $\mathbf{X}$ and $\mathbf{Y}$, respectively. $\mathbf{X}$ and $\mathbf{Y}$, respectively, are
   1. A.$\mathrm{N}_2$ and $\mathrm{N}_2 \mathrm{O}$
   2. B.$\mathrm{NH}_3$ and $\mathrm{NO}_2$
   3. C.NO and $\mathrm{N}_2 \mathrm{O}$
   4. D.$\mathrm{N}_2$ and $\mathrm{NH}_3$
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   Answer: (A)

   $\begin{aligned} & \mathrm{NH}_4 \mathrm{NO}_2 \xrightarrow[60-70^{\circ} \mathrm{C}]{\Delta} \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow[200-250^{\circ} \mathrm{C}]{\Delta} \mathrm{N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O}\end{aligned}$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are :
   1. A.$\left[\mathrm{PF}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}$and $\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}$
   2. B.$\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PCl}_4 \mathrm{~F}_2\right]^{-}$and $\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}$
   3. C.$\mathrm{PF}_3$ and $\mathrm{PCl}_3$
   4. D.$\mathrm{PF}_5$ and $\mathrm{PCl}_3$
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   Answer: (B)

   $\mathrm{PCl}_5$ when fluorinated in a polar organic solvent, ionic isomers are formed. $\left[\mathrm{PCl}_4^{+}\right]\left[\mathrm{PCl}_4 \mathrm{~F}_2\right]^{-}$(colorless crystals) $\left[\mathrm{PCl}_4^{+}\right]\left[\mathrm{PF}_6\right]^{-}$(white crystals)
4. Q4JEE Advanced Adv 2022 (Paper 2)
   The reaction of $Pb{\left({NO}_{3}\right)}_{2}$ and $NaCl$ in water produces a precipitate that dissolves upon the addition of $HCl$ of appropriate concentration. The dissolution of the precipitate is due to the formation of
   1. A.${PbCl}_{2}$
   2. B.${PbCl}_{4}$
   3. C.${\left[{PbCl}_{4}\right]}^{2-}$
   4. D.${\left[{PbCl}_{6}\right]}^{2-}$
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   Answer: (C)

   ${Pb}^{2+}$ on reaction with ${Cl}^{-}$, produces white precipitate of ${PbCl}_{2}$ ${Pb}^{2+}+2{Cl}^{-}⟶{PbCl}_{2}↓$ This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate $\left(II\right)$ ion ${PbCl}_{2}↓+2{Cl}^{-}⟶{\left[{PbCl}_{4}\right]}^{2-}$ On addition of potassium chromate (K2CrO4 ) solution to lead chloride a yellow precipitate of lead chromate is obtained.
5. Q5JEE Advanced Adv 2022 (Paper 2)
   The reaction of ${HClO}_{3}$ with $HCl$ gives a paramagnetic gas, which upon reaction with ${O}_{3}$ produces
   1. A.${Cl}_{2}O$
   2. B.${ClO}_{2}$
   3. C.${Cl}_{2}{O}_{6}$
   4. D.${Cl}_{2}{O}_{7}$
   Show answer & solution

   Answer: (C)

   ${HClO}_{3}$ reacts with $HCl$ according to the following equation, $2{HClO}_{3}+2HCl⟶2{ClO}_{2}+{Cl}_{2}+2{H}_{2}O$ ${ClO}_{2}$ molecule is paramagnetic, as it contains odd number of electrons. $2{ClO}_{2}+2{O}_{3}⟶{Cl}_{2}{O}_{6}+2{O}_{2}$ ClO2 is used as a bleaching agent for paper pulp and textiles and in water treatment