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Question #44333

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following statements regarding the structural, electronic, and hybridization properties of Group 15 hydrides and fluorides at $298\text{ K}$ :

I. The $\ce{F-N-F}$ bond angle in $\ce{NF3}$ ( $102.5^\circ$ ) is smaller than the $\ce{H-N-H}$ bond angle in $\ce{NH3}$ ( $107.2^\circ$ ) in accordance with Bent's rule. II. The $\ce{F-P-F}$ bond angle in $\ce{PF3}$ is smaller than the $\ce{H-P-H}$ bond angle in $\ce{PH3}$ because the highly electronegative fluorine atoms demand more p-character in the $\ce{P-F}$ bonds. III. The hybridization of the central atom is restored in $\ce{PF3}$ compared to $\ce{PH3}$ because the highly electronegative fluorine atoms increase the effective nuclear charge ( $Z_{eff}$ ) on phosphorus, which significantly reduces the 3s-3p energy gap. IV. The fractional s-character of the lone pair in $\ce{PH3}$ is greater than $80\%$ , which is significantly higher than the s-character of the lone pair in $\ce{PF3}$ . V. The Lewis basicity of $\ce{PH3}$ is greater than that of $\ce{NH3}$ due to the lower electronegativity and larger polarizability of phosphorus.

Which of the following combinations of statements is CORRECT?

Options

A.
I, III, and IV
B.
I, II, and IV
C.
II, III, and V
D.
I, III, and V

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

Screening Effect in Inner Shells|Bent's Rule|Hybridization in Non-Standard Cases

Why This Question Is Tricky

Common Mistakes

The question tests the limits of Bent's Rule and Drago's Rule by introducing the 'Screening Effect' and effective nuclear charge ( $Z_{eff}$ ) as hidden variables.

Students often blindly apply Bent's Rule to predict that highly electronegative substituents always reduce bond angles (as seen in $\ce{NH3}$ vs $\ce{NF3}$ ).

However, comparing $\ce{PH3}$ and $\ce{PF3}$ reveals a chemical exception: $\ce{PH3}$ is unhybridized due to a large 3s-3p energy gap (Drago's Rule), while the electronegative fluorines in $\ce{PF3}$ increase $Z_{eff}$ on phosphorus, lowering the 3p orbital energy, reducing the gap, and restoring hybridization.

This makes the bond angle in $\ce{PF3}$ larger than in $\ce{PH3}$ , trapping students who ignore the orbital energy modulation caused by substituent electronegativity.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Classification of Elements and Periodicity in Properties - Class 11 Part 1
Concept	Screening Effect in Inner Shells
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free