


∴X→Cu(NO3)2
JEE Main 2021 — Chemistry Physical Chemistry
An inorganic Compound ′X′ on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also Compound ′X′ gives precipitate ′Y′, when its solution in dilute HCl is treated with H2S gas. The precipitate ′Y′ on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution, Compound ′X′ is:
Held on 20 Jul 2021 · Verified 6 Jul 2026.
Co(NO3)2
Pb(NO2)2
Cu(NO3)2
Pb(NO3)2
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80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
What is the energy (in J atom$^{-1}$) required for the following process? $Li^{2+}(g) \rightarrow Li^{3+}(g) + e^-$ (Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18}$ J atom$^{-1}$)
Consider the following reduction processes : $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{s}), \mathrm{E}^{0}=-1.66 \mathrm{~V}$ $\mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^{0}=+0.77 \mathrm{~V}$ $\mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^{0}=+1.81 \mathrm{~V}$ $\mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{s}), \mathrm{E}^{\circ}=-0.74 \mathrm{~V}$ The tendency to act as reducing agent decreases in the order :
X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$to $\mathrm{I}_{2}$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$. The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$.
Identify the correct statements from the following: A. Heisenberg uncertainty principle is applicable to electrons. B. The size of $2p_x$ orbital is less than the size of $3p_x$ orbital. C. The energy of $2s$ orbital of H atom is equal to the energy of $2s$ orbital of Li. D. The electronic configuration of Cr is $[\text{Ar}] 3d^5 4s^1$ Choose the correct answer from the options given below:
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