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 2

Q50 jee_main_2025_29_jan_evening Lewis Structures and Valence Electrons
Total number of non bonded electrons present in NO2^- ion based on Lewis theory is ________.
Numerical Answer. Answer: 12 to 12

Solution

### Core Logic Let's compute the total valence electrons for the nitrite ion (NO_2^-): textValence electrons = 5text (from N) + 2 times 6text (from O) + 1text (negative charge) = 18text electrons In the valid Lewis structural representation: * The central nitrogen atom forms one single bond and one double bond with the terminal oxygens, consuming 2 + 4 = 6 bonding electrons. * Remaining non-bonded valence electrons = 18 - 6 = 12 electrons. ### Step 1: Account for Lone Pairs Distribution of non-bonded electrons across the individual atoms: * Central Nitrogen atom has 1 lone pair (2 electrons). * Single-bonded Oxygen atom has 3 lone pairs (6 electrons). * Double-bonded Oxygen atom has 2 lone pairs (4 electrons). textTotal non-bonded electrons = 2 + 6 + 4 = 12 ### Pattern Recognition Non-bonded electrons can always be obtained directly by subtracting total bonding electrons from total valence electrons. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q31 jee_main_2025_28_jan_morning VSEPR Theory and d-Electron Configurations
Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of mathrmClF_3 . The ions from the following with 'n' number of unpaired electrons are : A. mathrmV^3 + B. mathrmTi^3+ C. mathrmCu^2 + D. mathrmNi^2+ E. mathrmTi^2+ Choose the correct answer from the options given below:
  • A. textA and C only
  • B. textA, D and E only
  • C. textB and C only
  • D. textB and D only

Solution

### Step 1: Determine 'n' mathrmClF_3 has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus, n = 2. ### Step 2: Find ions with 2 unpaired electrons Let us compute the number of unpaired electrons for each configuration: - **A. mathrmV^3+:** [mathrmAr] 3d^2 rightarrow 2 unpaired electrons. - **B. mathrmTi^3+:** [mathrmAr] 3d^1 rightarrow 1 unpaired electron. - **C. mathrmCu^2+:** [mathrmAr] 3d^9 rightarrow 1 unpaired electron. - **D. mathrmNi^2+:** [mathrmAr] 3d^8 rightarrow 2 unpaired electrons. - **E. mathrmTi^2+:** [mathrmAr] 3d^2 rightarrow 2 unpaired electrons. Thus, A, D, and E have exactly n=2 unpaired electrons. ### Pattern Recognition Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember mathrmClF_3 is T-shaped with 2 equatorial lone pairs. Look for d^2 or d^8 configurations among the transition metal ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements
Q35 jee_main_2025_28_jan_morning Molecular Geometry and VSEPR
The molecules having square pyramidal geometry are
  • A. mathrmBrF_5 text \& mathrmXeOF_4
  • B. mathrmSbF_5 text \& mathrmXeOF_4
  • C. mathrmSbF_5 text \& mathrmPCl_5
  • D. mathrmBrF_5 text \& mathrmPCl_5

Solution

### 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
Q43 jee_main_2025_03_april_morning Hybridisation
Match the LIST-I with LIST-II.
LIST-I (Molecules/ion)LIST-II (Hybridisation of central atom)A. PF*5I. dsp^2B. SF*6II. sp^3dC. Ni(CO)*4III. sp^3d^2D. [PtCl*4]^2-IV. sp^3
Choose the correct answer from the options given below:
  • A. A-II, B-III, C-IV, D-I
  • B. A-IV, B-I, C-II, D-III
  • C. A-I, B-II, C-III, D-IV
  • D. A-III, B-I, C-IV, D-II

Solution

### Core Logic Let us evaluate each central atom configuration systematically: * A. PF_5: Phosphorus has 5 valence electrons, forming 5sigma bonds with zero lone pairs. Steric number = 5 implies sp^3d hybridisation. * B. SF_6: Sulfur has 6 valence electrons, forming 6sigma bonds with zero lone pairs. Steric number = 6 implies sp^3d^2 hybridisation. * C. Ni(CO)_4: Nickel is in a 0 oxidation state (3d^8 4s^2). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled 3d^10 state. The vacant 4s and three 4p orbitals hybridise to give an sp^3 configuration.
Orbital configuration matrix for Q43 - JEE Main 2025 Morning
Orbital configuration matrix for Q43 - JEE Main 2025 Morning
* D. [PtCl_4]^2-: Platinum is in the +2 oxidation state (5d^8). Since it belongs to the 5d transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar dsp^2 hybridisation state.
Orbital configuration matrix for Q43 - JEE Main 2025 Morning
Orbital configuration matrix for Q43 - JEE Main 2025 Morning
### Pattern Recognition Shortcut: Match structural main elements first: PF_5 ightarrow sp^3d (II), SF_6 ightarrow sp^3d^2 (III). This isolates option (1) instantly without checking d-block coordinate fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds
Q44 jee_main_2025_04_april_evening VSEPR Theory
Given below are two statements: Statement (I) : for mathrmCell mathrmF_3 , all three possible structures may be drawn as follows.
ClF3 structure variant I for Q44
The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.
ClF3 structure variant I for Q44
The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.
ClF3 structure variant I for Q44
The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.
Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the ell mathfrakp- bp repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. Statement I is incorrect but statement II is correct.
  • B. Statement I is correct but statement II is incorrect.
  • C. Both Statement I and statement II are correct.
  • D. Both Statement I and statement II are incorrect.

Solution

### Related Formula textSteric Number for ClF_3 = frac7+32 = 5 implies sp^3d text hybridization (Trigonal Bipyramidal geometry) ### Core Logic - **Statement I is correct:** The three structural arrangements represent the different ways to place three bond pairs and two lone pairs within a trigonal bipyramidal grid. - **Statement II is incorrect:** According to VSEPR theory and Bent's rule, in sp^3d hybridization, **lone pairs must occupy equatorial positions** to minimize strong 90^circ lone pair-bond pair (ell p-bp) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable. ### Pattern Recognition For sp^3d configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience 120^circ interactions, minimizing severe 90^circ structural strains. This results in the classic stable T-shaped configuration for ClF_3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Questions — jee_main_2026_21_jan_morning

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