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Question #44304

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The Question

Elements P,Q,R,P, Q, R, and SS belong to Group 17 of the periodic table. The magnitude of their standard electron gain enthalpies (∣ΔegH⊖∣|\Delta_{eg}H^\ominus|) follows the order Q>P>R>SQ > P > R > S, while their standard reduction potentials (EX2/X−∘E^\circ_{X_2/X^-}) follow the order P>Q>R>SP > Q > R > S.

A student incorrectly hypothesizes that the halogen with the highest magnitude of electron gain enthalpy will form the sodium salt with the highest limiting molar conductivity (Λm∘\Lambda_m^\circ) due to its strong ionic character. In reality, the highest Λm∘\Lambda_m^\circ belongs to the sodium salt of halogen XX.

Given the following data at 298 K298 \text{ K}: ∙\bullet Λm∘(\ceNaQ)=12.60 mS m2 mol−1\Lambda_m^\circ(\ce{NaQ}) = 12.60 \text{ mS m}^2 \text{ mol}^{-1} ∙\bullet Transport number of \ceNa+\ce{Na^+} in \ceNaQ\ce{NaQ} at infinite dilution is 0.4000.400. ∙\bullet Absolute ionic mobility of S−S^- at infinite dilution is 8.00×10−8 m2 V−1 s−18.00 \times 10^{-8} \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1}. ∙\bullet Absolute ionic mobility of P−P^- at infinite dilution is 5.60×10−8 m2 V−1 s−15.60 \times 10^{-8} \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1}. ∙\bullet Faraday constant F=96500 C mol−1F = 96500 \text{ C mol}^{-1}.

Identify halogen XX and calculate its Λm∘(\ceNaX)\Lambda_m^\circ(\ce{NaX}) in S cm2 mol−1\text{S cm}^2 \text{ mol}^{-1}.

Options

  • A.

    Halogen SS; 127.6 S cm2 mol−1127.6 \text{ S cm}^2 \text{ mol}^{-1}

  • B.

    Halogen QQ; 126.0 S cm2 mol−1126.0 \text{ S cm}^2 \text{ mol}^{-1}

  • C.

    Halogen PP; 104.4 S cm2 mol−1104.4 \text{ S cm}^2 \text{ mol}^{-1}

  • D.

    Halogen SS; 12.76 S cm2 mol−112.76 \text{ S cm}^2 \text{ mol}^{-1}

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Correct Answer

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Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Electron Gain Enthalpy (EGE)|Molar Conductivity Trend|Concentration Units Dependency

Why This Question Is Tricky

Common Mistakes

This problem tests the critical distinction between gas-phase thermodynamic properties (Electron Gain Enthalpy) and aqueous transport properties (Molar Conductivity).

A common trap is assuming that the halogen with the highest EGE (Chlorine) or highest electronegativity (Fluorine) will exhibit the highest conductivity due to 'greater ionic character'.

In reality, aqueous conductivity is governed by Stokes' Law and the Born solvation model; the smallest bare ion (F-) has the highest charge density, leading to the largest hydrated radius and lowest mobility.

Thus, the largest bare ion (I-) has the highest conductivity.

A secondary mathematical trap involves the unit conversion from mS·m²/mol to S·cm²/mol, penalizing unit-blind dimensional analysis.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterClassification of Elements and Periodicity in Properties - Class 11 Part 1
ConceptElectron Gain Enthalpy (EGE)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree