Redox Reactions — Chemistry STD 11 Science — Question

The mass (in $\mathrm{mg}$ ) of $\mathrm{S}$ obtained is. . . . . . .  [Use molar mass (in g mol ${ }^{-1}$ ) : $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{Br}=80$ ]

$(i)$ $\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {C{H_3} - {C^ \mathbf{-} }} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} 
\end{array}$

$(ii)$ $H _{2} C = CH - CH _{2}$

$(iii)$ $HC \equiv \stackrel{\ominus}{ C }$

$(iv)$ $\stackrel{\ominus}{ CH _{3}}$

$(v)$ $\stackrel{\ominus}{{ }_{ CN }}$

$(A)$ The decreasing order of atomic radii of group $13$ elements is $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$.

$(B)$ Down the group $13$ electronegativity decreases from top to bottom.

$(C)$ $\mathrm{Al}$ dissolves in dil. $\mathrm{HCl}$ and liberate $\mathrm{H}_2$ but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface.

$(D)$ All elements of group 13 exhibits highly stable +1 oxidation state.

$(E)$ Hybridisation of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion is $\mathrm{sp}^3 \mathrm{~d}^2$.

Choose the correct answer from the options given below: