Before You Begin
Question #44223
The Question
To investigate the thermodynamic stability of high oxidation state fluorides, a student models the formation of gaseous sulfur hexafluoride, , using a hypothetical gas-phase ionic Born-Haber cycle. The student assumes the formation of isolated and ions, which then condense to form the covalent molecule. Given the following thermodynamic data at 298 K: Standard enthalpy of formation of , Enthalpy of atomization of , Bond dissociation enthalpy of , Sum of the first six ionization enthalpies of , Electron gain enthalpy of , Calculate the hypothetical 'gas-phase condensation enthalpy' () in kJ/mol—the energy released when one mole of and six moles of combine to form one mole of .
Options
- A.
-27695 kJ/mol
- B.
-26486 kJ/mol
- C.
-25277 kJ/mol
- D.
-28172 kJ/mol
Check Your Answer
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 ) 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 () completely offsets the ionization energy cost (), proving that enthalpy of formation is a balance of these extremes, not just dictated by IE.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Thermodynamics - Class 11 Part 1 |
| Concept | Enthalpy Definition |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |