- A$SiO_4^{4-} > PCl_3 > NCl_3 > SbH_3 > H_2Te$
- B$SiO_4^{4-} > NCl_3 > PCl_3 > SbH_3 > H_2Te$
- ✓$SiO_4^{4-} > H_2Te > SbH_3 > PCl_3 > NCl_3$
- D$NCl_3 > PCl_3 > Si_4^{4-} > SbH_3 > H_2Te$
$Cl_2 (g) + 2H_2O (l) \to HCl(aq) + HOCl$
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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C - C{H_2} - C{H_2}OH}
\end{array}$
$CH_2OH-CHOH-CH_2OH$ $\xrightarrow{{KHS{O_4}/\Delta }}(X)\mathop {\xrightarrow{{{{({C_2}{H_5}O)}_3}Al}}}\limits_\Delta (Y)$
$(Y)$ will be:
$(i)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}OH}}$ $ (CH_3)_2CH - CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH - CH_2Br \xrightarrow{{{C_2}{H_5}O^-}} $ $(CH_3)_2CH - CH_2OC_2H_5 + Br^-$
The mechanisms of reactions $(i)$ and $(ii)$ are respectively