1
JEE Advanced 2023 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
In the scheme given below, $\mathbf{X}$ and $\mathbf{Y}$, respectively, are

JEE Advanced 2023 Paper 1 Online Chemistry - p-Block Elements Question 2 English
A
$\mathrm{CrO}_4{ }^{2-}$ and $\mathrm{Br}_2$
B
$\mathrm{MnO}_4{ }^{2-}$ and $\mathrm{Cl}_2$
C
$\mathrm{MnO}_4{ }^{-}$and $\mathrm{Cl}_2$
D
$\mathrm{MnSO}_4$ and $\mathrm{HOCl}$
2
JEE Advanced 2023 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.

List - I List - II
(P) $\mathrm{P}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ (1) $\mathrm{P}(\mathrm{O})\left(\mathrm{OCH}_3\right) \mathrm{Cl}_2$
(Q) $\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ (2) $\mathrm{H}_3 \mathrm{PO}_3$
(R) $\mathrm{PCl}_5+\mathrm{CH}_3 \mathrm{COOH} \rightarrow$ (3) $\mathrm{PH}_3$
(S) $\mathrm{H}_3 \mathrm{PO}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{AgNO}_3 \rightarrow$ (4) $\mathrm{POCl}_3$
(5) $\mathrm{H}_3 \mathrm{PO}_4$
A
$\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 5$
B
$\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 2$
C
$\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 3$
D
$\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 5$
3
JEE Advanced 2022 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language
The reaction of $\mathrm{HClO}_{3}$ with $\mathrm{HCl}$ gives a paramagnetic gas, which upon reaction with $\mathrm{O}_{3}$ produces
A
$\mathrm{Cl}_{2} \mathrm{O}$
B
$\mathrm{ClO}_{2}$
C
$\mathrm{Cl}_{2} \mathrm{O}_{6}$
D
$\mathrm{Cl}_{2} \mathrm{O}_{7}$
4
JEE Advanced 2022 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language
The reaction of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{NaCl}$ in water produces a precipitate that dissolves upon the addition of $\mathrm{HCl}$ of appropriate concentration. The dissolution of the precipitate is due to the formation of
A
$\mathrm{PbCl}_{2}$
B
$\mathrm{PbCl}_{4}$
C
$\left[\mathrm{PbCl}_{4}\right]^{2-}$
D
$\left[\mathrm{PbCl}_{6}\right]^{2-}$
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