**JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.**

**Simulator**

**Previous Years JEE Advance Questions**

(A) 3 electrons in neutral medium

(B) 5 electrons in neutral medium

(C) 3 electrons in alkaline medium

(D) 5 electrons in acidic medium

**[JEE-2011]**

**Sol.**(A,C,D)

**[JEE- 2011]**

**Sol.**5

(A) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$

(B) $\mathrm{HNO}_{3}, \mathrm{NO}, \mathrm{N}_{2}, \mathrm{NH}_{4} \mathrm{Cl}$

(C) $\mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NO}, \mathrm{N}_{2}$

(D) $\mathrm{NO}, \mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2}$

**[JEE- 2012]**

**Sol.**(B)

(A) 0.48 M

(B) 0.96 M

(C) 0.24 M

(D) 0.024 M

**[JEE- 2012]**

**Sol.**(C)

Eq. of $\mathrm{CaOCl}_{2}=\mathrm{Eq}$, of $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

25 × M × 2 = 48 × 0.25

$\left[2 \mathrm{H}^{+}+\mathrm{CaOCl}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}^{2+}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{Cl}^{-}\right]$

(nf = 2)

M = 0.24

$\mathrm{I}^{-}+\mathrm{ClO}_{3}^{-}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Cl}^{-}+\mathrm{HSO}_{4}^{-}+\mathrm{I}_{2}$

The correct statement(s) in the balanced equation is / are :

(A) Stoichiometric coefficient of $\mathrm{HSO}_{4}^{-}$ is 6

(B) Iodide is oxidized

(C) Sulphur is reduced

(D) $\mathrm{H}_{2} \mathrm{O}$ is one of the products

**[JEE- 2014]**

**Sol.**(A,B,D)

Oxidation half reaction :

**[JEE – Adv. 2016]**

**Sol.**6

$\mathrm{MnCl}_{2}+\mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{8}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}+$ HCl (equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 g) was added in portions till the colour of the permanganate ion disappeard. The quantity of MnCl2 (in mg) present in the initial solution is

(Atomic weights in $\mathrm{g}$ mol $^{-1}: \mathrm{Mn}=55, \mathrm{Cl}=35.5$ )

**[JEE – Adv. 2018]**

**Sol.**126