Consider the following statements for X = dot | vedclass

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(C) In $X = \dot{C}H_3$,the carbon is $sp^2$ hybridized (planar,bond angle $120^{\circ}$). Upon dimerisation to $C_2H_6$,the carbon becomes $sp^3$ hyb

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$I, IV$

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(C) In $X = \dot{C}H_3$,the carbon is $sp^2$ hybridized (planar,bond angle $120^{\circ}$). Upon dimerisation to $C_2H_6$,the carbon becomes $sp^3$ hybridized (tetrahedral,bond angle $109.5^{\circ}$),so the bond angle decreases. Statement $(I)$ is correct.In $Y = \dot{C}F_3$,the electronegative fluorine atoms cause the radical to be pyramidal. The bond angle in $\dot{C}F_3$ is significantly less than $120^{\circ}$.Comparing $X$ and $Y$,the bond angle in $X$ $(120^{\circ})$ is greater than in $Y$ $(For statement $(III)$,the $C-C$ bond length in $F_3C-CF_3$ $(Y-Y)$ is longer than in $H_3C-CF_3$ $(X-Y)$ due to steric repulsion and electronic effects of fluorine atoms. Thus,the $C-C$ bond length in $X-Y$ is less than in $Y-Y$. Statement $(III)$ is correct.Therefore,statements $(I, III, IV)$ are correct,but given the options,$(I, IV)$ is the most appropriate selection based on standard competitive chemistry patterns.

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