The molecules having square pyramidal geometry are

Solution & Explanation

### Core Logic Let us check the steric details using VSEPR theory: - **mathrmBrF_5:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (sp^3d^2), geometry is square pyramidal. - **mathrmXeOF_4:** Xenon has 8 valence electrons. It forms 1 double bond with Oxygen, 4 single bonds with Fluorine, and retains 1 lone pair. Steric number = 6 (sp^3d^2), geometry is square pyramidal. - **mathrmSbF_5 & mathrmPCl_5:** Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 (sp^3d), geometry is trigonal bipyramidal. Visual representations of geometries:
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning
### Pattern Recognition Sees: Steric count 6 with 5 bonded segments + 1 lone pair rightarrow always square pyramidal geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp^3 hybrid orbitals in its bonding is NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Analyzing the hybridization of the central atom in each species: - NH_3: 3 bp + 1 lp = 4 electron domains rightarrow sp^3 - SO_2: 2 bp + 1 lp = 3 electron domains rightarrow sp^2 - SiO_2: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally rightarrow sp^3 - BeCl_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - CO_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - H_2O: 2 bp + 2 lp = 4 electron domains rightarrow sp^3 - CH_4: 4 bp + 0 lp = 4 electron domains rightarrow sp^3 - BF_3: 3 bp + 0 lp = 3 electron domains rightarrow sp^2 Total species with sp^3 hybridization: NH_3, SiO_2, H_2O, CH_4. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Questions — jee_main_2025_28_jan_morning

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