MCQ
Bleaching action of $S{O_2}$ is due to
- ✓Reduction
- BOxidation
- CHydrolysis
- DIts acidic nature
✓
Answer
Correct option: A.
Reduction
a
(a) $2H_2O$ + $SO_2$ $\to$ $H_2SO_4$ + $2[H]$ (nascent hydrogen)
(a) $2H_2O$ + $SO_2$ $\to$ $H_2SO_4$ + $2[H]$ (nascent hydrogen)
Coloured flower $ + \,2[H] \to $Colourless flower
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Copper cannot replace….. from solution
Co-ordination number of $Fe$ in the complexes ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }},$ ${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }}$ and ${\left[ {FeC{l_4}} \right]^ - }$ would be respectively
→The heats of combustion of carbon and carbon monoxide are $-393.5$ and $-283.5\, kJ\, mol^{-1},$ respectively. The heat offormation (in $kJ$) of carbon monoxide per mole is :
→The stability of carbanions in the following.
→$(i)\,\,RC \equiv \mathop C\limits^ \ominus $
$(ii)\,\,[IMAGE]$
$(iii)\,\,{R_2}C = \mathop C\limits^ \ominus H$
$(iv)\,\,{R_3}C - \mathop C\limits^ \ominus {H_2}$
An element has successive ionization enthalpies as $940\, (first), 2080, 3090, 4140, 703 0, 7870,16000$ and $19500\, kJ \,mol^{-1}.$ To which group of the periodic table does this element belong?
→Which of the following will show optical isomerism ?
$\begin{matrix}
Ph-CH-C{{H}_{2}}-C{{H}_{2}}\xrightarrow[\Delta ]{Zn-Cu} \\
\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\,\,\,\,\,Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$ Product
→Ph-CH-C{{H}_{2}}-C{{H}_{2}}\xrightarrow[\Delta ]{Zn-Cu} \\
\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\,\,\,\,\,Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Br\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$ Product
Product of the above reaction is
Identify the position where $E.A.S.$ can take place
→Consider the following changes
→$M(s) \to M(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\, ........(1)$
$M(s) \to M^{2+} (g) + 2e^-\,\,\,\,\,\,\,\,.......(2)$
$M(g) \to M^+(g) + e^-\,\,\,\,\,\,\,\,\,\,\,.........(3)$
$M^+ (g) \to M^{2+} (g) + e^-\,\,\,\,\,\,\,\,\,.........(4)$
$M(g) \to M^{2+} (g) +2e^-\,\,\,\,\,\,\,\,\,\,\,..........(5)$
The second ionization energy of $M$ could be calculated from the energy values assoclated with
For the given reactions
→$Sn ^{2+}+2 e ^{-} \rightarrow Sn$
$Sn ^{4+}+4 e ^{-} \rightarrow Sn$
The electrode potentials are; $E _{ Sn ^{2+} / Sn }^{\circ}=-0.140\, V$ and $E _{ Sn ^{4+} / Sn }^{\circ}=0.010\, V$. The magnitude of standard electrode potential for $Sn ^{4+} / Sn ^{2+}$ i.e. $E _{ Sn ^{4+} / Sn ^{2+}}^{\circ}$ is $.....\times 10^{-2}\, V$. (Nearest integer)