Assertion :The bond energies of $C l_{2}$ and $F_{2}$ are 243 and $155 k J m o l^{- 1}$ respectively. So, $X e F_{2}$ is much more stable compound than $X e C l_{2}$ . Reason: $Δ H$
$X e (g) ⟶ X e^{+} (g) + e^{-}; I$ (ionisation energy)
$X_{2} (g) ⟶ 2 X (g); D$ (Bond energy)
$2 X (g) + 2 e^{-} ⟶ 2 X^{-} (g); 2 E$ (electron affinity)
$X e^{+} (g) + 2 X^{-} (g) ⟶ X e X_{2} (g); - U$ (lattice energy).
......................................................................
$X e (g) + X_{2} (g) + X e X_{2} (g); Δ H_{f} = I + D + 2 F - U$
......................................................................
By taking values of $I, D, E a n d U$ , we find that $Δ H_{f}^{\circ} (X e F_{2})$ is more negative than that of $X e C l_{2}$ . Thus $X e F_{2}$ is much more stable.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Both Assertion and Reason are incorrect

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Solution

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
As we know, higher the atomic mass of noble gas and lower the size of halogen, more stable will be the compound as their $Δ H$ is higher.
So, order is $X e F_{2} > X e C l_{2} > X e I_{2}$ .
Also,
$X e (g) ⟶ X e^{+} (g) + e^{-}; I$ (ionisation energy)
$X_{2} (g) ⟶ 2 X (g); D$ (Bond energy)
$2 X (g) + 2 e^{-} ⟶ 2 X^{-} (g); 2 E$ (electron affinity)
$X e^{+} (g) + 2 X^{-} (g) ⟶ X e X_{2} (g); - U$ (lattice energy).
......................................................................
$X e (g) + X_{2} (g) + X e X_{2} (g); Δ H_{f} = I + D + 2 F - U$

Higher is the the mass of halogen, lower will be the $Δ H$ . Thus, $X e F_{2}$ is much more stable.

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