For a homogeneous gaseous reaction $2{ \mathrm{A}}_{\left(\mathrm{g}\right)}+3{ \mathrm{B}}_{\left(\mathrm{g}\right)}\rightleftharpoons 4{\mathrm{C}}_{\left(\mathrm{g}\right)}+{\mathrm{D}}_{\left(\mathrm{g}\right)}{\mathrm{K}}_{\mathrm{eq}}=\frac{8}{(1.5{)}^3}.$ If in a $2$ litre rigid container starting with 4 moles of A and 6 moles of B equilibrium was established then identify the options which is / are correct.

$$\begin{gathered} \underset{4 - 2x}{\underset{4 \mathrm{moles}}{2{ \mathrm{A}}_{\left(\mathrm{g}\right)}}}+ \underset{6 -3x}{\underset{6 \mathrm{moles}}{3{ \mathrm{B}}_{\left(\mathrm{g}\right)}}} \rightleftharpoons \underset{4x}{\underset{\mathrm{V}=2\text{ litre }}{ 4{\mathrm{C}}_{\left(\mathrm{g}\right)}}}+\underset{x}{ \underset{ }{{\mathrm{D}}_{\left(\mathrm{g}\right)}}} \\ {\mathrm{K}}_{\mathrm{eq}}=\frac{8}{(1.5{)}^3} \\ {K}_{eq}=\frac{(4x{)}^{4}x}{(4-2x{)}^2(6-3x{)}^3}\times \frac{\frac{1}{V^4}\times \frac{1}{V}}{\frac{1}{V^2}\times \frac{1}{V^3}} \\ \frac{8}{(1.5{)}^3}=\frac{4^4{x}^5}{4(2-x{)}^{2}27(2-x{)}^3} \\ \Rightarrow \frac{8\times 4\times 27}{(1.5{)}^3(4{)}^4}={\left(\frac{x}{2-x}\right)}^5\Rightarrow {\left(\frac{x}{2-x}\right)}^5=1 \\ \frac{x}{2-x}=1\Rightarrow x=1 \\ [\mathrm{B}{]}_{\mathrm{eq}}=\frac{6-3\mathrm{x}}{\mathrm{V}}=\frac{6-3}{2}=\frac{3}{2}=1.5\mathrm{M} \\ [\mathrm{D}{]}_{\text{eq }}=\frac{\mathrm{x}}{\mathrm{V}}=\frac{1}{2}=0.5\mathrm{M} \\ [\mathrm{A}{]}_{\mathrm{eq}}=\frac{4-2\mathrm{x}}{\mathrm{V}}=\frac{2}{2}=1\mathrm{M} \\ [C{]}_{\mathrm{eq}}=\frac{4\mathrm{x}}{\mathrm{V}}=\frac{4\times 1}{2}=2\mathrm{M} \end{gathered}$$