- ✓$2.5$
- B$4$
- C$2$
- D$3$
At equilibrium, $1$ mole of $CO$ is present
Hence, $2-1=1$ moles of $CO$ has reacted.
$1$ mole of $CO$ will react with $1$ mole of $Cl_2$ to from $1$ mole of $COCl_2$.
$3-1=2$ moles of $Cl_2$ remains at equilibrium
The equlibrium constant
${K_c} = \frac{{[COC{l_2}]}}{{[CO][C{l_2}]}} = \frac{{\frac{{1\,mol}}{{5\,L}}}}{{\frac{{1\,mol}}{{5\,L}} \times \frac{{2\,mol}}{{5\,L}}}} = 2.5$
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$(A)$ 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 $\mathrm{sp}^3 \mathrm{~d}^2$.
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