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Question #44181

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following thermodynamic and structural observations for Group 1 and Group 11 compounds: \textbf{I.} The magnitude of the experimental lattice energy (∣Uexp∣|U_{exp}|) of \ce{AgI} is significantly greater than its theoretical Born-Landé lattice energy (∣Uth∣|U_{th}|), whereas for \ce{AgF}, ∣Uexp∣≈∣Uth∣|U_{exp}| \approx |U_{th}|. \textbf{II.} The thermal decomposition of \ce{LiBF4(s)} to \ce{LiF(s)} and \ce{BF3(g)} occurs at a much lower temperature than that of \ce{CsBF4(s)}. \textbf{III.} The \ce{B-F} bond length in the \ce{BF4^-} ion is significantly longer than in the \ce{BF3} molecule.

Which of the following options correctly identifies the primary chemical principles responsible for these observations?

Options

  • A.

    I: Covalent character via polarization increases ∣Uexp∣|U_{exp}|; II: Large U(\ceLiF)−U(\ceLiBF4)U(\ce{LiF}) - U(\ce{LiBF4}) drives decomposition; III: Absence of pπ−pπp\pi-p\pi back-bonding in \ce{BF4^-}.

  • B.

    I: dπ−pπd\pi-p\pi back-bonding strengthens the lattice; II: Lower polarizing power of \ce{Cs^+} stabilizes the lattice kinetically; III: Steric repulsion between fluorines lengthens the bond.

  • C.

    I: Born-Landé overestimates electrostatic repulsion; II: High sublimation energy of Li metal drives decomposition; III: \ce{BF3} possesses dπ−pπd\pi-p\pi back-bonding.

  • D.

    I: \ce{AgI} has lower absolute lattice energy due to size; II: \ce{LiBF4} is kinetically labile due to small \ce{Li^+}; III: Hybridization change to sp3sp^3 inherently shortens the bond.

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Lattice Energy Factors|Back Bonding Strength

Why This Question Is Tricky

Common Mistakes

The question targets the 'Lattice Energy Trap' where students erroneously assume the Born-Landé equation perfectly describes all ionic solids.

It forces the recognition that Fajans' Rules (covalent character via polarization) create a measurable thermodynamic deviation (∣Uexp∣>∣Uth∣|U_{exp}| > |U_{th}|).

Furthermore, it tests the 'Hidden Variable' of lattice energy differences driving the thermal decomposition of polyatomic salts, rather than simple cation polarizing power.

Finally, it uses pπ−pπp\pi-p\pi back-bonding as a structural determinant, trapping students who misattribute bond length changes to steric hindrance or incorrect orbital overlaps (like dπ−pπd\pi-p\pi for Boron).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Bonding and Molecular Structure - Class 11 Part 1
ConceptLattice Energy Factors
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree