JEE Main 2024 — Chemical Bonding And Molecular Structure Question with Solution

To match the molecules in List I with their corresponding shapes in List II, we need to consider the electronic geometry and the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here’s the detailed analysis:

$$\mathrm{N}{\mathrm{H}}_3$$ (Ammonia): Ammonia has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons. This leads to a trigonal pyramidal shape.

$$\mathrm{Br}{\mathrm{F}}_5$$ (Bromine pentafluoride): Bromine pentafluoride has a central bromine atom bonded to five fluorine atoms and has one lone pair of electrons. This leads to a square pyramidal shape.

$$\mathrm{PC}{\mathrm{l}}_5$$ (Phosphorus pentachloride): Phosphorus pentachloride has a central phosphorus atom bonded to five chlorine atoms and no lone pairs. This results in a trigonal bipyramidal shape.

$$\mathrm{C}{\mathrm{H}}_4$$ (Methane): Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs, forming a tetrahedral structure.

Using the above analyses, we can match the molecules and shapes as follows:

• A ($$\mathrm{N}{\mathrm{H}}_3$$) - III (Trigonal pyramidal)
• B ($$\mathrm{Br}{\mathrm{F}}_5$$) - I (Square pyramid)
• C ($$\mathrm{PC}{\mathrm{l}}_5$$) - IV (Trigonal bipyramidal)
• D ($$\mathrm{C}{\mathrm{H}}_4$$) - II (Tetrahedral)

Therefore, the correct matching is option B:

Option B: A-III, B-I, C-IV, D-II