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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

Match List-I (Molecules) with List-II (Shapes):

List-I: (a) PCl5PCl_5 (b) SF6SF_6 (c) BrF5BrF_5 (d) BF3BF_3

List-II: (i) Square pyramidal (ii) Trigonal planar (iii) Octahedral (iv) Trigonal bipyramidal

Choose the correct answer from the options given below:

A

(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

B

(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

C

(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

D

(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Step-by-Step Solution

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory and the principles of hybridisation described in the sources:

  1. PCl5PCl_5 (a): The central phosphorus atom undergoes sp3dsp^3d hybridisation. With five bonding pairs and no lone pairs, it adopts a trigonal bipyramidal geometry .
  2. SF6SF_6 (b): The central sulphur atom undergoes sp3d2sp^3d^2 hybridisation. With six bonding pairs, it forms a regular octahedral geometry .
  3. BrF5BrF_5 (c): Bromine has seven valence electrons. It forms five sigma bonds with fluorine and has one lone pair (sp3d2sp^3d^2 hybridisation). This 5 bond pair + 1 lone pair arrangement results in a square pyramidal shape .
  4. BF3BF_3 (d): Boron undergoes sp2sp^2 hybridisation, forming three bond pairs with no lone pairs. This results in a trigonal planar geometry .

Matching these correctly results in (a)-(iv), (b)-(iii), (c)-(i), and (d)-(ii), which corresponds to option 3.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Bonding and Molecular Structuremoleculeslistiishapeslistiisquare

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