- ✓$I^-$
- B$S(s)$
- C$NO_3^ - \left( {aq} \right)$
- D$Cr_2O_7^{2 - }$
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$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
Here $20\, mL$ of $0.1\, M\, KMnO_4$ is equivalent to
$\mathrm{Li}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{Na}_{2} \mathrm{O}_{2}, \mathrm{KO}_{2}, \mathrm{MgO}$ and $\mathrm{K}_{2} \mathrm{O}$

$(1)$ An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
$(2)$ For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number
$(3)$ According to wave mechanics, the ground state angular momentum is equal to $\frac {h}{2\pi }$
$(4)$ The plot of $\Psi \,\,Vs\,\,r$ for various azimuthal quantum numbers, shows peak shifting towards higher $r$ value