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:

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

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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