In acidic solution Mn(VI) become unstable relative to Mn(VII) and Mn(IV)
3MnO42−+4H+→2MnO4−+MnO2+2H2O
So difference in oxidation state of product ions of Mn is = 3
JEE Main 2022 — Chemistry Physical Chemistry
Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is
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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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