JEE MAIN - Chemistry (2024 - 8th April Morning Shift - No. 8)
Match List I with List II
LIST I (Molecule) |
LIST II (Shape) |
||
---|---|---|---|
A. | $$\mathrm{NH_3}$$ | I. | Square pyramid |
B. | $$\mathrm{BrF_5}$$ | II. | Tetrahedral |
C. | $$\mathrm{PCl_5}$$ | III. | Trigonal pyramidal |
D. | $$\mathrm{CH_4}$$ | IV. | Trigonal bipyramidal |
Choose the correct answer from the options given below:
Explanation
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
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