Match List-I with List-II.List-I
(Molecule) List-II
(Shape) (A) \(NH_3\) (I) Square pyramid (B) \(BrF_5\) (II) Tetrahedral (C) \(PCL_5\) (III) Trigonal pyramidal (D) \(CH_4\) (IV) Trigonal bipyramidal
Choose the correct answer from the options given below:
| List-I (Molecule) | List-II (Shape) |
|---|---|
| (A) \(NH_3\) | (I) Square pyramid |
| (B) \(BrF_5\) | (II) Tetrahedral |
| (C) \(PCL_5\) | (III) Trigonal pyramidal |
| (D) \(CH_4\) | (IV) Trigonal bipyramidal |
- A-IV, B-III, C-I, D-II
- A-II, B-IV, C-I, D-III
- A-III, B-I, C-IV, D-II
- A-III, B-IV, C-I, D-II
The Correct Option is C
Approach Solution - 1
To solve this question, we need to determine the correct shape for each molecule listed in List-I based on their molecular geometry, and then match them with the options in List-II. We will assess each molecule's electron pair and bond pair geometries:
- \(NH_3\) (Ammonia):
- Ammonia (\(NH_3\)) has a total of 5 valence electrons from Nitrogen and 1 valence electron from each of the 3 Hydrogens, making a total of 8 valence electrons.
- Nitrogen forms 3 sigma bonds with Hydrogen atoms and has one lone pair of electrons.
- The molecular geometry is Trigonal Pyramidal due to the lone pair that slightly distorts the angles. Thus, the shape for \(NH_3\) is III) Trigonal Pyramidal.
- \(BrF_5\) (Bromine Pentafluoride):
- In \(BrF_5\), the central atom Bromine has 7 valence electrons and bonds with 5 Fluorine atoms, using 5 electrons in bonding.
- There is 1 lone pair remaining on Bromine.
- The electron pair geometry is Octahedral, but due to the presence of a lone pair, its molecular geometry is a Square Pyramidal. Thus, the shape for \(BrF_5\) is I) Square Pyramid.
- \(PCl_5\) (Phosphorus Pentachloride):
- Phosphorus has 5 valence electrons and forms 5 bonds with Chlorine atoms with no lone pairs remaining.
- This yields a geometric structure that is Trigonal Bipyramidal with the Chlorine atoms.
- Thus, the shape for \(PCl_5\) is IV) Trigonal Bipyramidal.
- \(CH_4\) (Methane):
- The central atom Carbon forms 4 sigma bonds with Hydrogen atoms with no lone pairs.
- This results in a geometry that is Tetrahedral.
- Thus, the shape for \(CH_4\) is II) Tetrahedral.
Matching these descriptions with List-II, we get:
- A) \(NH_3\) - III) Trigonal Pyramidal
- B) \(BrF_5\) - I) Square Pyramid
- C) \(PCl_5\) - IV) Trigonal Bipyramidal
- D) \(CH_4\) - II) Tetrahedral
The correct answer is: A-III, B-I, C-IV, D-II.
Approach Solution -2
NH$_3$: Trigonalpyramidal ({sp}$^3$ hybridization with one lone pair on N).
BrF$_5$: Squarepyramidal ({sp}$^3${d}$^2$ hybridization with one lone pair on Br).
PCl$_5$: Trigonalbipyramidal ({sp}$^3${d} hybridization, no lone pairs).
CH$_4$: Tetrahedral ({sp}$^3$ hybridization, no lone pairs).
Matching: A-III, B-I, C-IV, D-II.
Final Answer: A-III, B-I, C-IV, D-II.
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