- ABoth the compounds form same product on treatment with alcoholic KOH.
- BBoth the compounds form same product on treatment with aq.NaOH.
- CBoth the compounds form same product on reduction.
- DBoth the compounds are optically active.
Explanation:
$\text{H}_3\text{C}-\text{CHCl}_2$ (ethylidene choride) and $\text{H}_2\text{C}-\text{CH}_2\\ \ \ \ \ \ \ | \ \ \ \ \ \ |\ \ \ \ \\ \ \ \ \ \ \text{Cl} \ \ \ \text{Cl}$ (ethylene dichloride) are isomers.
$\text{CH}_3\text{CHCl}_2\xrightarrow[\text{KOH}]{ \ \ \ \text{alc.} \ \ \ \ }\text{CH}\equiv\text{CH}+2\text{KCl}+2\text{H}_2\text{O}$
$\text{Cl}-\text{CH}_2-\text{CH}_2-\text{Cl}\xrightarrow{ \ \ \text{alc.KOH} \ \ }\text{CH}\equiv\text{CH}+2\text{KCl}+\text{H}_2\text{O}$
$\text{CH}_3\text{CHCl}_2\xrightarrow{ \ \ \text{CH}_3\text{OH} \ \ \ }\text{CH}_2=\text{CH}$
$\text{Cl}-\text{CH}_2-\text{CH}_2-\text{Cl}+\text{Zn}\xrightarrow{\text{CH}_3\text{OH} \ \ }\text{CH}_2=\text{CH}_2$
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$(i)$ $\left[ M ( NCS )_{6}\right]^{(-6+ n )}$
$(ii)$ $\left[ MF _{6}\right]^{(-6+ n )}$
$(iii)$ $\left[ M \left( NH _{3}\right)_{6}\right]^{ n ^{+}}$
Assertion $(A):$ $\mathrm{H}_2 \mathrm{Te}$ is more acidic than $\mathrm{H}_2 \mathrm{~S}$.
Reason $(R):$ The bond dissociation enthalpy of $\mathrm{H}_2 \mathrm{Te}$ is less than that of $\mathrm{H}_2 \mathrm{~S}$.
In the context of the above statements, choose the correct answer from the following options: