
- A$4-$isopropyl $1-6-$methyl octane
- ✓$3- $methyl$-5-(1’-$methylethyl$)$ octane
- C$3-$methyl$-5-$isopropyl octane
- D$6-$methyl$-4-(1’$methylethyl$) $octane


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$\mathrm{X}_2(g) \rightleftharpoons 2 \mathrm{X}(g)$
The standard reaction Gibbs energy, $\Delta_r G^{\circ}$, of this reaction is positive. At the stiur of the reaction, there is one mole of $X_2$ and no $X$. As the reaction proceeds, the number of roles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equitibrium }}$ is the number of moles of $\mathrm{X}$ formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given : $R=0.083 \mathrm{~L} \mathrm{bar}^{-1} \mathrm{~mol}^{-1}$ )
($1$) The equilibrium constant $K_P$ for this reaction at $298 \mathrm{~K}$, in terms of $\beta_{\text {equilibs, um }}$, is
($A$) $\frac{8 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$ ($B$) $\frac{8 \beta_{\text {equititrium }}^2}{4-\beta_{\text {equilibrium }}^2}$ ($C$) $\frac{4 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$ ($D$) $\frac{4 \ell_{\text {equitibrium }}^2}{4-\beta_{\text {equilibrium }}^2}$
($2$) The $INCORRECT$ statement among the following, for this reaction, is
($A$) Decrease in the total pressure will result in formation of more moles of gaseous $\mathrm{X}$
($B$) At the start of the reaction, dissociation of gaseous $\mathrm{X}_2$ takes place spontaneously
($C$) $\beta_{\text {equilibrium }}=0.7$
($D$) $\quad K_c<1$
Given the answer question ($1$) and ($2$)
(Given, $\mathrm{R}=8.3\; \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$$, \ln \left(\frac{2}{3}\right)=0.4$ $\left.e^{-3}=4.0\right)$

Assertion $(A):$ The ionic radii of $O ^{2-}$ and $Mg ^{2+}$ are same.
Reason $( R )$ : Both $O ^{2-}$ and $Mg ^{2+}$ are isoelectronic species.
In the light of the above statements, choose the correct answer from the options given below
