- ABleaching powder
- BWhite vitriol
- ✓Mohr's salt
- DMicrocosmic salt
$2KMn{O_4} + 8{H_2}S{O_4} + 10FeS{O_4}\, \to $
${K_2}S{O_4} + 2MnS{O_4} + 5F{e_2}{(S{O_4})_3} + 8{H_2}O$
Mohr’s salt decolourises $KMn{O_4}$ by reducing $M{n^{ + 7}}$ ions to $M{n^{ + 2}}$ ions.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{|l|l|} \hline Column\,\,I & Column\,\,II \\ \hline (A)\,XeF_6 & (i)\,\,distortedoctahedral \\ \hline (B)\,XeO_3 & (ii)\,\,square\,\,planar \\ \hline (C)\,XeOF_4 & (iii)\,\,pyramidal \\ \hline (D)\,XeF_4 & (iv)\,\,square\,\,pyramidal \\ \hline \end{array}$
[Atomic numbers of $Cr =24$ and $Mn =25$ ]
$(A)$ $Cr ^{2+}$ is a reducing agent
$(B)$ $Mn ^{3+}$ is an oxidizing agent
$(C)$ Both $Cr ^{2+}$ and $Mn ^{3+}$ exhibit $d^4$ electronic configuration
$(D)$ When $Cr ^{2+}$ is used as a reducing agent, the chromium ion attains $d^5$ electronic configuration
(Take $pH_2 = 1\,atm$ ), $T = 298\,K$.
