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Question #44149

Difficulty: 8/10 | 🟢 Free

The Question

A container of volume $V = 8.21$ L at $300$ K contains an equimolar mixture of \ce{NF3} and \ce{N(SiH3)3} gases at a total pressure of $6$ atm.

Let $w$ be the total mass of the gas mixture in grams. Let $x$ be the total number of $sp^2$ hybridized atoms in one molecule of \ce{N(SiH3)3}. Let $y$ be the total number of stereochemically active lone pairs on the central Nitrogen atom in one molecule of \ce{NF3}. Let $z$ be the maximum number of coplanar atoms in the molecular skeleton of \ce{N(SiH3)3} (excluding hydrogen atoms).

Calculate the value of $w + x + y + z$ .

(Given: Atomic masses: N = 14, F = 19, Si = 28, H = 1. $R = 0.0821$ L atm K $^{-1}$ mol $^{-1}$ )

Options

A.
184
B.
182
C.
187
D.
95

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

Back Bonding Strength|Dipole Moment Vector Addition|VSEPR Lone Pair Positioning

Why This Question Is Tricky

Common Mistakes

This problem tests the critical exception of pπ-dπ back-bonding overriding standard VSEPR predictions.

Students often blindly apply VSEPR to N(SiH3)3, predicting an sp3 hybridized, trigonal pyramidal geometry similar to NF3 or NH3.

However, the delocalization of Nitrogen's lone pair into Silicon's vacant 3d orbitals planarizes the molecule (sp2), making the lone pair stereochemically inactive and increasing the number of coplanar skeletal atoms from 3 to 4.

Failing to account for this geometry shift leads directly to the trap answer 182 (where x=0 and z=3).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Back Bonding Strength
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free