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Question #44170

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following 9 molecules: (I) \ce{N(SiH3)3} (II) \ce{N(CH3)3} (III) \ce{P(SiH3)3} (IV) \ce{N(CH3)2H} (V) \ce{N(GeH3)3} (VI) \ce{B(CH3)3} (VII) \ce{NH3} (VIII) \ce{NF3} (IX) \ce{N(Si(CH3)3)3}

Let $X$ be the number of molecules that possess a planar geometry around the central atom (i.e., the sum of the bond angles around the central atom is $\approx 360^\circ$ ). Let $Y$ be the number of molecules from the above list that act as strong Lewis bases and form stable adducts with \ce{BF3} at 298 K. Find the value of $10X + Y$ .

Options

A.
43
B.
53
C.
34
D.
44

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

Hybridization in Non-Standard Cases|Reaction Intermediates Stability|Resonance Stability Criteria

Why This Question Is Tricky

Common Mistakes

This question tests the limits and exceptions of VSEPR theory and Lewis acid-base chemistry.

The primary trap is assuming that lower electronegativity of Si compared to C makes \ce{N(SiH3)3} a stronger base than \ce{N(CH3)3}.

Students must recognize the 'Hidden Variable' of $p\pi-d\pi$ back-bonding, which delocalizes the lone pair, forces a planar geometry, and destroys nucleophilicity.

A secondary trap is over-extending this back-bonding to \ce{P(SiH3)3}, ignoring the poor $3p-3d$ orbital overlap which leaves it pyramidal.

Finally, it tests the inductive deactivation of the lone pair in \ce{NF3}.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Hybridization in Non-Standard Cases
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free