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Question #44223

Difficulty: 8/10 | 🟢 Free

The Question

To investigate the thermodynamic stability of high oxidation state fluorides, a student models the formation of gaseous sulfur hexafluoride, \ceSF6(g)\ce{SF6(g)}, using a hypothetical gas-phase ionic Born-Haber cycle. The student assumes the formation of isolated \ceS6+(g)\ce{S^{6+}(g)} and \ceF−(g)\ce{F^{-}(g)} ions, which then condense to form the covalent \ceSF6(g)\ce{SF6(g)} molecule. Given the following thermodynamic data at 298 K: Standard enthalpy of formation of \ceSF6(g)\ce{SF6(g)}, ΔfH∘=−1209 kJ/mol\Delta_f H^\circ = -1209 \text{ kJ/mol} Enthalpy of atomization of \ceS(s)\ce{S(s)}, ΔatomH∘=277 kJ/mol\Delta_{atom} H^\circ = 277 \text{ kJ/mol} Bond dissociation enthalpy of \ceF2(g)\ce{F2(g)}, ΔdissH∘=159 kJ/mol\Delta_{diss} H^\circ = 159 \text{ kJ/mol} Sum of the first six ionization enthalpies of \ceS(g)\ce{S(g)}, ∑i=16IEi=27700 kJ/mol\sum_{i=1}^6 IE_i = 27700 \text{ kJ/mol} Electron gain enthalpy of \ceF(g)\ce{F(g)}, ΔegH∘=−328 kJ/mol\Delta_{eg} H^\circ = -328 \text{ kJ/mol} Calculate the hypothetical 'gas-phase condensation enthalpy' (ΔcondH∘\Delta_{cond} H^\circ) in kJ/mol—the energy released when one mole of \ceS6+(g)\ce{S^{6+}(g)} and six moles of \ceF−(g)\ce{F^{-}(g)} combine to form one mole of \ceSF6(g)\ce{SF6(g)}.

Options

  • A.

    -27695 kJ/mol

  • B.

    -26486 kJ/mol

  • C.

    -25277 kJ/mol

  • D.

    -28172 kJ/mol

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Enthalpy Definition|Variable Oxidation State Stability|Stability of High OS Fluorides/Oxides

Why This Question Is Tricky

Common Mistakes

The question targets the misconception that high oxidation states (like +6 in \ceSF6\ce{SF6}) are thermodynamically prohibited due to the astronomically high sum of successive ionization enthalpies.

Students often fail to realize that the formation of multiple strong bonds with highly electronegative ligands (like F or O) releases an immense amount of energy.

By forcing the student to calculate the hypothetical 'condensation enthalpy' in an ionic Born-Haber cycle, the problem quantitatively reveals the 'Spoiler': the bonding/lattice energy (−27695 kJ/mol-27695 \text{ kJ/mol}) completely offsets the ionization energy cost (27700 kJ/mol27700 \text{ kJ/mol}), proving that enthalpy of formation is a balance of these extremes, not just dictated by IE.

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Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptEnthalpy Definition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree