Match List-I with the List-IIList-I
(Compound /
Species) List-II
(Shape / Geometry) (A) \(SF_4\) (I) Tetrahedral (B) \(BrF_3\) (II) Pyramidal (C) \(BrO_{3}^{-}\) (III) See saw (D) \(NH^{+}_{4}\) (IV) Bent T-shape
Choose the correct answer from the options given below:
| List-I (Compound / Species) | List-II (Shape / Geometry) |
|---|---|
| (A) \(SF_4\) | (I) Tetrahedral |
| (B) \(BrF_3\) | (II) Pyramidal |
| (C) \(BrO_{3}^{-}\) | (III) See saw |
| (D) \(NH^{+}_{4}\) | (IV) Bent T-shape |
- A-II, B-III, C-I, D-IV
- A-III, B-IV, C-II, D-I
- A-II, B-IV, C-III, D-I
- A-III, B-II, C-IV, D-I
The Correct Option is B
Approach Solution - 1
(A) SF$_4$: The sulfur atom in SF$_4$ undergoes sp$^3$d hybridization, resulting in a see-saw geometry.
(B) BrF$_3$: Bromine in BrF$_3$ exhibits sp$^3$d hybridization with two lone pairs, resulting in a bent T-shape geometry.
(C) BrO$_3^-$: The bromine atom in BrO$_3^-$ is sp$^3$ hybridized, resulting in a pyramidal geometry.
(D) NH$_4^+$: The nitrogen atom in NH$_4^+$ undergoes sp$^3$ hybridization, forming a tetrahedral geometry.
Approach Solution -2
To determine the correct match between List-I (Compound / Species) and List-II (Shape / Geometry), we need to understand the molecular geometry of each compound based on the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here is a step-by-step analysis:
- Compound (A): \(SF_4\)
- Valence Electrons Count: Sulfur has 6 valence electrons and each fluorine atom contributes 1 electron. Total = \(6 + (4 \times 1) = 10\) electrons.
- Geometry: The molecule has 4 bonded pairs and 1 lone pair, leading to a see-saw shape based on VSEPR theory.
- Therefore, \(SF_4\) corresponds to \(III\) (See-saw). - Compound (B): \(BrF_3\)
- Valence Electrons Count: Bromine has 7 valence electrons and each fluorine atom contributes 1 electron. Total = \(7 + (3 \times 1) = 10\) electrons.
- Geometry: The molecule has 3 bonded pairs and 2 lone pairs, leading to a bent T-shaped structure.
- Therefore, \(BrF_3\) corresponds to \(IV\) (Bent T-shape). - Compound (C): \(BrO_3^{-}\)
- Valence Electrons Count: Bromine has 7 valence electrons, 3 oxygen atoms each bring 6 electrons, plus 1 extra electron for the negative charge. Total = \(7 + (3 \times 6) + 1 = 26\) electrons.
- Resonance: The structure is better visualized, considering the resonance forms, giving bromine a formal charge.
- Geometry: Assume 3 bonded groups and no lone pair for the central bromine, leading to a pyramidal shape.
- Therefore, \(BrO_3^{-}\) corresponds to \(II\) (Pyramidal). - Compound (D): \(NH_4^{+}\)
- Valence Electrons Count: Nitrogen has 5 valence electrons, and each hydrogen contributes 1 electron. Total = \(5 + (4 \times 1) = 9\) electrons, minus 1 for the positive charge, gives 8 electrons.
- Geometry: With 4 bonded pairs and no lone pairs, the shape is tetrahedral.
- Thus, \(NH_4^{+}\) corresponds to \(I\) (Tetrahedral).
Hence, the correct matching is: A-III, B-IV, C-II, D-I.
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