QJEEM24Ch03fd: JEE Main 2024 Chemistry - Class 11 JEEM24Ch03fd Solution: Match List I with List II .tg…

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:

$\mathrm{NH_3}$ (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.

$\mathrm{BrF_5}$ (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.

$\mathrm{PCl_5}$ (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.

$\mathrm{CH_4}$ (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 ( $\mathrm{NH_3}$ ) - III (Trigonal pyramidal)
• B ( $\mathrm{BrF_5}$ ) - I (Square pyramid)
• C ( $\mathrm{PCl_5}$ ) - IV (Trigonal bipyramidal)
• D ( $\mathrm{CH_4}$ ) - II (Tetrahedral)

Therefore, the correct matching is option B:

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