Given below are two statements: Statement I: The number of species among SF4, NH4^+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3. Statement II: In the set [NO2, BeH2, BF3, AlCl3], 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:

Evaluating Statement I: - SF4: sp3d (1 lone pair) arrow See-saw - XeF4: sp3d2 (2 lone pairs) arrow Square planar - [PtCl4]2-: dsp2 arrow Square planar - [NiCl4]2-: sp3 arrow Tetrahedral - [Ni(CN)4]2-: dsp2 arrow Square planar - SeF4: sp3d (1 lone pair) arrow See-saw - NH4^+: sp3 (0 lone pairs) arrow Tetrahedral Total tetrahedral species = 2 ([NiCl4]2- and NH4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO2: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). - BeH2: Central Be has 4 electrons (incomplete octet). - BF3: Central B has 6 electrons (incomplete octet). - AlCl3 (monomer): Central Al has 6 electrons (incomplete octet). Therefore, all molecules have incomplete octets. Statement II is true. Statement I is false, Statement II is true. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Given below are two statements: Statement I: The number of species among SF4, NH4^+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3. Statement II: In the set [NO

Evaluating Statement I: - SF4: sp3d (1 lone pair) arrow See-saw - XeF4: sp3d2 (2 lone pairs) arrow Square planar - [PtCl4]2-: dsp2 arrow Square planar - [NiCl4]2-: sp3 arrow Tetrahedral - [Ni(CN)4]2-: dsp2 arrow Square planar - SeF4: sp3d (1 lone pair) arrow See-saw - NH4^+: sp3 (0 lone pairs) arrow Tetrahedral Total tetrahedral species = 2 ([NiCl4]2- and NH4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO2: Central N has 7 valence electrons (odd-electron molecule, incomple

Given below are two statements: Statement I: The number of species among SF4, NH4^+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3. Statement II: In the set [NO2, BeH2, BF3, AlCl3], 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:

Evaluating Statement I: - SF4: sp3d (1 lone pair) arrow See-saw - XeF4: sp3d2 (2 lone pairs) arrow Square planar - [PtCl4]2-: dsp2 arrow Square planar - [NiCl4]2-: sp3 arrow Tetrahedral - [Ni(CN)4]2-: dsp2 arrow Square planar - SeF4: sp3d (1 lone pair) arrow See-saw - NH4^+: sp3 (0 lone pairs) arrow Tetrahedral Total tetrahedral species = 2 ([NiCl4]2- and NH4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO2: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). - BeH2: Central Be has 4 electrons (incomplete octet). - BF3: Central B has 6 electrons (incomplete octet). - AlCl3 (monomer): Central Al has 6 electrons (incomplete octet). Therefore, all molecules have incomplete octets. Statement II is true. Statement I is false, Statement II is true. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Given below are two statements: Statement I: The number of species among SF4, NH4^+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3. Statement II: In the set [NO

Evaluating Statement I: - SF4: sp3d (1 lone pair) arrow See-saw - XeF4: sp3d2 (2 lone pairs) arrow Square planar - [PtCl4]2-: dsp2 arrow Square planar - [NiCl4]2-: sp3 arrow Tetrahedral - [Ni(CN)4]2-: dsp2 arrow Square planar - SeF4: sp3d (1 lone pair) arrow See-saw - NH4^+: sp3 (0 lone pairs) arrow Tetrahedral Total tetrahedral species = 2 ([NiCl4]2- and NH4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO2: Central N has 7 valence electrons (odd-electron molecule, incomple

Given below are two statements: Statement I: The number of species among SF4, NH4^+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4…

Year	2026	2025	2024	Total
Questions	12	14	16	42

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Evaluating Statement I:

$SF₄$ $\mathrm{SF}_4$: $sp³d$ $sp^3d$ (1 lone pair) $arrow$ $\rightarrow$ See-saw
$XeF₄$ $\mathrm{XeF}_4$: $sp³d²$ $sp^3d^2$ (2 lone pairs) $arrow$ $\rightarrow$ Square planar
$[PtCl₄]²⁻$ $[\mathrm{PtCl}_4]^{2-}$: $dsp²$ $dsp^2$ $arrow$ $\rightarrow$ Square planar
$[NiCl₄]²⁻$ $[\mathrm{NiCl}_4]^{2-}$: $sp³$ $sp^3$ $arrow$ $\rightarrow$ Tetrahedral
$[Ni(CN)₄]²⁻$ $[\mathrm{Ni(CN)}_4]^{2-}$: $dsp²$ $dsp^2$ $arrow$ $\rightarrow$ Square planar
$SeF₄$ $\mathrm{SeF}_4$: $sp³d$ $sp^3d$ (1 lone pair) $arrow$ $\rightarrow$ See-saw
$NH₄^+$ $\mathrm{NH}_4^+$: $sp³$ $sp^3$ (0 lone pairs) $arrow$ $\rightarrow$ Tetrahedral

Total tetrahedral species = 2 ( $[NiCl₄]²⁻$ $[\mathrm{NiCl}_4]^{2-}$ and $NH₄^+$ $\mathrm{NH}_4^+$). Statement I says 3, so it is false.

Evaluating Statement II:

$NO₂$ $NO_2$: Central N has 7 valence electrons (odd-electron molecule, incomplete octet).
$BeH₂$ $BeH_2$: Central Be has 4 electrons (incomplete octet).
$BF₃$ $BF_3$: Central B has 6 electrons (incomplete octet).
$AlCl₃$ $AlCl_3$ (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.

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Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 9

Q77 jee_main_2024_31_jan_morning Molecular Orbital Theory

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:

A. $A, B, C only$ $\text{A, B, C only}$
B. $A and C only$ $\text{A and C only}$
C. $B, C, D only$ $\text{B, C, D only}$
D. $B and D only$ $\text{B and D only}$

Solution

Core Logic

Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:

The combining atomic orbitals must have the same or nearly the same energy.
The combining atomic orbitals must have the same symmetry about the molecular axis.
The combining atomic orbitals must overlap to the maximum extent (not minimum).

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_31_jan_morning Hybridization

The number of species from the following in which the central atom uses

sp³

hybrid orbitals in its bonding is

NH₃, SO₂, SiO₂, BeCl₂, CO₂, H₂O, CH₄, BF₃

Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Analyzing the hybridization of the central atom in each species:

$NH₃$ $NH_3$: 3 bp + 1 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$SO₂$ $SO_2$: 2 bp + 1 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$
$SiO₂$ $SiO_2$: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally $arrow sp³$ $\rightarrow sp^3$
$BeCl₂$ $BeCl_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$CO₂$ $CO_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$H₂O$ $H_2O$: 2 bp + 2 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$CH₄$ $CH_4$: 4 bp + 0 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$BF₃$ $BF_3$: 3 bp + 0 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$

Total species with $sp³$ $sp^3$ hybridization: $NH₃, SiO₂, H₂O, CH₄$ $NH_3, SiO_2, H_2O, CH_4$. Total count = 4.

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Step 1: Final Conclusion

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