Start Learning Free

Before You Begin

Question #44870

Difficulty: 8/10 | 🟢 Free

The Question

Equimolar 0.01 m0.01\text{ m} aqueous solutions of \ceZrCl4\ce{ZrCl4} and \ceHfCl4\ce{HfCl4} are prepared at 298 K298\text{ K}. Let ΔTb(\ceZr)\Delta T_{b(\ce{Zr})} and ΔTb(\ceHf)\Delta T_{b(\ce{Hf})} be their respective boiling point elevations. Let Kh(\ceZr)K_{h(\ce{Zr})} and Kh(\ceHf)K_{h(\ce{Hf})} be the equilibrium constants for their primary hydrolysis step: \ceM4+(aq)+H2O(l)⇌M(OH)3+(aq)+H+(aq)\ce{M^{4+}(aq) + H2O(l) \rightleftharpoons M(OH)^{3+}(aq) + H+(aq)}. Assuming dilute ideal behavior, which of the following sets of relations correctly compares these systems?

Options

  • A.

    ΔTb(\ceZr)<ΔTb(\ceHf),  Kh(\ceZr)<Kh(\ceHf)\Delta T_{b(\ce{Zr})} < \Delta T_{b(\ce{Hf})} ,\; K_{h(\ce{Zr})} < K_{h(\ce{Hf})}

  • B.

    ΔTb(\ceZr)=ΔTb(\ceHf),  Kh(\ceZr)=Kh(\ceHf)\Delta T_{b(\ce{Zr})} = \Delta T_{b(\ce{Hf})} ,\; K_{h(\ce{Zr})} = K_{h(\ce{Hf})}

  • C.

    ΔTb(\ceZr)>ΔTb(\ceHf),  Kh(\ceZr)>Kh(\ceHf)\Delta T_{b(\ce{Zr})} > \Delta T_{b(\ce{Hf})} ,\; K_{h(\ce{Zr})} > K_{h(\ce{Hf})}

  • D.

    ΔTb(\ceZr)<ΔTb(\ceHf),  Kh(\ceZr)>Kh(\ceHf)\Delta T_{b(\ce{Zr})} < \Delta T_{b(\ce{Hf})} ,\; K_{h(\ce{Zr})} > K_{h(\ce{Hf})}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

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

Concepts Behind This Question

Post-Lanthanoid Radii Equality|Van’t Hoff Factor (i) Application|Boiling Point Elevation (Solvent vs Solution)

Why This Question Is Tricky

Common Mistakes

Tests the 'Chemical Exception' to the textbook rule that 'Lanthanoid contraction makes Zr and Hf chemically identical.' Students often assume equimolar solutions of \ceZrCl4\ce{ZrCl4} and \ceHfCl4\ce{HfCl4} will have identical Van't Hoff factors (ii) and colligative properties.

The trap is failing to account for the higher effective nuclear charge (ZeffZ_{eff}) of Hf(IV), which increases its Lewis acidity and drives more extensive hydrolysis.

This thermodynamic difference alters the particle count in solution, unpredictably changing the boiling point elevation.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterD and F Block Elements - Class 12
ConceptPost-Lanthanoid Radii Equality
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree