Given below are two statements: Statement I: The number of species among mathrmSF_4, mathrmNH_4^+, [mathrmNiCl_4]^2-, mathrmXeF_4, [mathrmPtCl_4]^2-, mathrmSeF_4 and [mathrmNi(CN)_4]^2-, that have tetrahedral geometry is 3. Statement II: In the set [NO_2, BeH_2, BF_3, AlCl_3], all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Evaluating Statement I: - mathrmSF_4: sp^3d (1 lone pair) rightarrow See-saw - mathrmXeF_4: sp^3d^2 (2 lone pairs) rightarrow Square planar - [mathrmPtCl_4]^2-: dsp^2 rightarrow Square planar - [mathrmNiCl_4]^2-: sp^3 rightarrow Tetrahedral - [mathrmNi(CN)_4]^2-: dsp^2 rightarrow Square planar - mathrmSeF_4: sp^3d (1 lone pair) rightarrow See-saw - mathrmNH_4^+: sp^3 (0 lone pairs) rightarrow Tetrahedral Total tetrahedral species = 2 ([mathrmNiCl_4]^2- and mathrmNH_4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO_2: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). - BeH_2: Central Be has 4 electrons (incomplete octet). - BF_3: Central B has 6 electrons (incomplete octet). - AlCl_3 (monomer): Central Al has 6 electrons (incomplete octet). Therefore, all molecules have incomplete octets. Statement II is true. ### Step 1: Final Conclusion Statement I is false, Statement II is true. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Reference Study Guides

Chemical Bonding And Molecular Structure

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

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