For the reaction sequence given below, the correct statement(s) is(are)

(In the options, X is any atom other than carbon and hydrogen, and it is different in P, Q and R)

The correct answer is: C−X bond enthalpy in P, Q and R follows the order R > P > Q.

To determine the correct statement, we must first identify the compounds P, Q, and R from the given reaction sequence:

Step 1: Formation of P
Cyclohexene reacts with $HBr$ via hydrohalogenation (electrophilic addition) to form P, which is bromocyclohexane.
Here, the substituent $X=Br$.

Step 2: Formation of Q
Bromocyclohexane (P) undergoes the Finkelstein reaction (typically using $NaI$ in acetone) to form Q, which is iodocyclohexane.
Here, the substituent $X=I$.

Step 3: Formation of R
Bromocyclohexane (P) undergoes the Swarts reaction (typically using a metal fluoride such as $AgF$) to form R, which is fluorocyclohexane.
Here, the substituent $X=F$.

Thus, we have:
P: Bromocyclohexane ($C-Br$ bond)
Q: Iodocyclohexane ($C-I$ bond)
R: Fluorocyclohexane ($C-F$ bond)

Now, let us evaluate the given options:

1. C−X Bond Length:
As we go down the halogen group ($F\to Cl\to Br\to I$), the atomic size of the halogen increases. Consequently, the overlap between the $2p$ orbital of carbon and the valence orbital of the halogen becomes poorer, leading to an increase in bond length.
The bond length order is:
$C-I>C-Br>C-F$
Which corresponds to the order: Q > P > R.
Thus, Option A is incorrect.

2. C−X Bond Enthalpy:
Bond enthalpy is inversely proportional to bond length. Shorter bonds have stronger orbital overlap and thus higher bond dissociation energy (bond enthalpy).
The bond enthalpy order is:
$C-F>C-Br>C-I$
Which corresponds to the order: R > P > Q.
Thus, Option B is correct.

3. Relative Reactivity toward S_N2 Reaction:
In S_N2 reactions, the rate of substitution is directly proportional to the leaving group ability of the halide. Since the weaker C−X bond is broken more easily, iodide is the best leaving group and fluoride is the poorest.
The reactivity order is:
$C-I>C-Br>C-F$
Which corresponds to the order: Q > P > R.
Thus, Option C is incorrect.

4. pKa of the Conjugate Acids of the Leaving Groups:
The leaving groups are $F^{-}$, $B{r}^{-}$, and $I^{-}$. Their conjugate acids are $HF$, $HBr$, and $HI$ respectively.
Acid strength follows the order: $HI>HBr>HF$.
Since $p{K}_a=-{\log }_{10}{K}_a$, a stronger acid has a lower $p{K}_a$.
The $p{K}_a$ order is:
$HF>HBr>HI$
Which corresponds to the order: R > P > Q.
Thus, Option D is incorrect.