Hydrogen JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Hydrogen, free to read — no sign-in needed.

1. Q1JEE Advanced Adv 2022 (Paper 1)
   LIST-$I$ contains compounds and LIST-$II$ contains reactions LIST-$I$ LIST-$II$ $\left(I\right)$ ${H}_{2}{O}_{2}$ $\left(P\right)$ $Mg{\left({HCO}_{3}\right)}_{2}+Ca{\left(OH\right)}_{2}\rightarrow$ $\left(II\right)$ $Mg{\left(OH\right)}_{2}$ $\left(Q\right)$ ${BaO}_{2}+{H}_{2}{SO}_{4}\rightarrow$ $\left(III\right)$ ${BaCl}_{2}$ $\left(R\right)$ $Ca\left({OH}_{2}\right)2+{MgCl}_{2}\rightarrow$ $\left(IV\right)$ ${CaCO}_{3}$ $\left(S\right)$ ${BaO}_{2}+HCl\rightarrow$ $\left(T\right)$ $Ca{\left({HCO}_{3}\right)}_{2}+Ca{\left(OH\right)}_{2}\rightarrow$ Match each compound in LIST-$I$ with its formation reaction(s) in LIST-$II$, and choose the correct option
   1. A.$I\rightarrow Q;II\rightarrow P;III\rightarrow S;IV\rightarrow R$
   2. B.$I\rightarrow T:II\rightarrow P\cdot III\rightarrow Q:IV\rightarrow R$
   3. C.$I\rightarrow T;II\rightarrow R;III\rightarrow Q;IV\rightarrow P$
   4. D.$I\rightarrow Q;II\rightarrow R;III\rightarrow S;IV\rightarrow P$
   Show answer & solution

   Answer: (D)

   In this method calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off$\left(P\right)Mg{\left({HCO}_{3}\right)}_{2}+2Ca{\left(OH\right)}_{2}⟶2{CaCO}_{3}↓+Mg{\left(OH\right)}_{2}↓+2{H}_{2}O$ $\left(T\right)Ca{\left({HCO}_{3}\right)}_{2}+Ca{\left(OH\right)}_{2}⟶2{CaCO}_{3}↓+2{H}_{2}O$ $\left(R\right)Ca{\left(OH\right)}_{2}+{MgCl}_{2}⟶Mg{\left(OH\right)}_{2}+{CaCl}_{2}$ Acidifying barium peroxide and removing excess water by evaporation under reduced pressure gives hydrogen peroxide $\left(Q\right){BaO}_{2}+{H}_{2}{SO}_{4}⟶{BaSO}_{4}+{H}_{2}{O}_{2}$ $\left(S\right){BaO}_{2}+2HCl⟶{BaCl}_{2}+{H}_{2}{O}_{2}$ $I\rightarrow Q$ $II\rightarrow R$ $III\rightarrow S$ $IV\rightarrow P$
2. Q2JEE Advanced Adv 2017 (Paper 2)
   Which of the following combinations will produce ${\text{H}}_{2}$ gas?
   1. A.Zn metal and $\text{N}\text{a}\text{O}\text{H}\left(\text{a}\text{q}\right)$
   2. B.Au metal and $\text{N}\text{a}\text{C}\text{N}\left(\text{a}\text{q}\right)$ in the presence of air
   3. C.Cu metal and conc. $\text{H}\text{N}{\text{O}}_{3}$
   4. D.Fe metal and conc. $\text{H}\text{N}{\text{O}}_{3}$
   Show answer & solution

   Answer: (A)

   (i) $\text{Z}\text{n}+2\text{N}\text{a}\text{O}\text{H}\rightarrow \text{N}{\text{a}}_{2}\text{Z}\text{n}{\text{O}}_{2}+{\text{H}}_{2}$ (ii) $4\text{A}\text{u}+8\text{N}\text{a}\text{C}\text{N}+{\text{O}}_{2}+2{\text{H}}_{2}\text{O}\rightarrow 4\text{N}\text{a}\left[\text{A}\text{u}{\left(\text{C}\text{N}\right)}_{2}\right]+4\text{N}\text{a}\text{O}\text{H}$ (iii) $\begin{matrix}\text{C}⁡\text{u}⁡ & + & 4\text{H}⁡\text{N}⁡{\text{O}⁡}_{3} & \rightarrow & \text{C}⁡\text{u}⁡{\left(\text{N}⁡{\text{O}⁡}_{3}\right)}_{2} & + & 2\text{N}⁡{\text{O}⁡}_{2} & +2{\text{H}⁡}_{2}\text{O}⁡ & \\ & & \left(\text{conc.}\right) & & & & & & \end{matrix}$ (iv) Formation of passive layer of $\text{F}{\text{e}}_{2}{\text{O}}_{3}$ on the surface of Fe and $\text{N}{\text{O}}_{2}$ gas is evolved. $\text{F}\text{e}+4\text{H}\text{N}{\text{O}}_{3}\rightarrow \text{F}\text{e}{(\text{N}{\text{O}}_{3})}_{3}+\text{N}{\text{O}}_{2}+\dfrac{1}{2}{\text{O}}_{2}$
3. Q3JEE Advanced Adv 2014 (Paper 2)
   Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a
   1. A.reducing agent, oxidising agent
   2. B.reducing agent, reducing agent
   3. C.oxidising agent, oxidising agent
   4. D.oxidising agent, reducing agent
   Show answer & solution

   Answer: (A)

   $Kl{O}_{4}+{H}_{2}{O}_{2}\rightarrow Kl{O}_{3}+{H}_{2}O+{O}_{2}$ In this reaction${H}_{2}{O}_{2}$ act as reducing agent $N{H}_{2}OH+3{H}_{2}{O}_{2}\rightarrow HN{O}_{3}+4{H}_{2}O$ In this reaction ${H}_{2}{O}_{2}$ act as oxidising agent.
4. Q4JEE Advanced Adv 2010 (Paper 1)
   The reagent(s) used for softening the temporary hardness of water is(are)
   1. A.$\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
   2. B.$\mathrm{Ca}(\mathrm{OH})_2$
   3. C.$\mathrm{Na}_2 \mathrm{CO}_3$
   4. D.$\mathrm{NaOCl}$
   Show answer & solution

   Answer: B,C,D

   Softening the temporary hardness of water means conversion of bicarbonate into calcium carbonate (precipitation) $$\begin{gathered} \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \\ 2 \mathrm{CaCO}_3 \downarrow+2 \mathrm{H}_2 \mathrm{O} \\ \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow \\ \mathrm{CaCO}_3 \downarrow+2 \mathrm{NaHCO}_3 \end{gathered}$$ Even $\mathrm{NaOCl}$ can be used, as in aqueous medium $\overline{\mathrm{O}} \mathrm{Cl}$ gets hydrolysed and makes solution alkaline enough to cause precipitation of $\mathrm{CaCO}_3$. $$\begin{aligned} \mathrm{NaOCl}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{HOCl}+\mathrm{NaOH} \\ \mathrm{OH}^{-}+\mathrm{HCO}_3^{-} & \longrightarrow \mathrm{CO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \end{aligned}$$ Representative elements (Hydrogen and its compounds) Conceptual II