Question

Match List-I with the List-II

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

Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is B

(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.

Answer 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:

1. 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).
2. 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).
3. 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).
4. 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.