Chemical Bonding and Molecular Structure appeared 42 times across 3 years — 4.9% of Chemistry.
This question is from 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 ClF₃$\mathrm{ClF}_3$ . The ions from the following with 'n' number of unpaired electrons are :
A. V3 +$\mathrm{V}^{3 + }$
B. Ti³⁺$\mathrm{Ti}^{3+}$
C. Cu2 +$\mathrm{Cu}^{2 + }$
D. Ni²⁺$\mathrm{Ni}^{2+}$
E. Ti²⁺$\mathrm{Ti}^{2+}$
Choose the correct answer from the options given below:
A.A and C only$\text{A and C only}$
B.A, D and E only$\text{A, D and E only}$
C.B and C only$\text{B and C only}$
D.B and D only$\text{B and D only}$
Solution & Explanation
Step 1: Determine 'n'
ClF₃$\mathrm{ClF}_3$ 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$n = 2$.
Step 2: Find ions with 2 unpaired electrons
Let us compute the number of unpaired electrons for each configuration:
A. V³⁺$\mathrm{V}^{3+}$:[Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.
B. Ti³⁺$\mathrm{Ti}^{3+}$:[Ar] 3d¹ arrow 1$[\mathrm{Ar}] 3d^1 \rightarrow 1$ unpaired electron.
C. Cu²⁺$\mathrm{Cu}^{2+}$:[Ar] 3d⁹ arrow 1$[\mathrm{Ar}] 3d^9 \rightarrow 1$ unpaired electron.
D. Ni²⁺$\mathrm{Ni}^{2+}$:[Ar] 3d⁸ arrow 2$[\mathrm{Ar}] 3d^8 \rightarrow 2$ unpaired electrons.
E. Ti²⁺$\mathrm{Ti}^{2+}$:[Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.
Thus, A, D, and E have exactly n=2$n=2$ unpaired electrons.
Pattern Recognition
Sees: Number of equatorial lone pairs linked to unpaired electrons.
Shortcut: Remember ClF₃$\mathrm{ClF}_3$ is T-shaped with 2 equatorial lone pairs. Look for d²$d^2$ or d⁸$d^8$ configurations among the transition metal ions.
Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Class 12 Chemistry: The d-and f-Block Elements
Keywords:#lone pair of electrons present in the equatorial position#JEE Main 2025 Morning Q31#Chemical Bonding JEE Main 2025#VSEPR Theory JEE Main 2025
More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 5
Qjee_main_2025_04_april_eveningVSEPR Theory
Given below are two statements:
Statement (I) : for C F₃$\mathrm{C}\ell \mathrm{F}_{3}$ , all three possible structures may be drawn as follows.
The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.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 p-$\ell \mathfrak{p}-$ bp repulsion is minimum.
In the light of the above statements, choose the most appropriate answer from the options given below:
The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.
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
Steric Number for ClF₃ = (7+3)/(2) = 5 sp³d hybridization (Trigonal Bipyramidal geometry)$$\text{Steric Number for } ClF_3 = \frac{7+3}{2} = 5 \implies sp^3d \text{ hybridization (Trigonal Bipyramidal geometry)}$$
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 sp³d$sp^3d$ hybridization, lone pairs must occupy equatorial positions to minimize strong 90^°$90^\circ$ lone pair-bond pair (p-bp$\ell p-bp$) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable.
Pattern Recognition
For sp³d$sp^3d$ configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience 120^circ$120^circ$ interactions, minimizing severe 90^circ$90^circ$ structural strains. This results in the classic stable T-shaped configuration for ClF₃$ClF_3$.
Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q28jee_main_2025_04_april_morningMolecular Orbital Theory
Which of the following molecules(s) show/s paramagnetic behavior?
(A) O₂$O_2$
(B) N₂$N_2$
(C) F₂$F_2$
(D) S₂$S_2$
(E) Cl₂$Cl_2$
Choose the correct answer from the options given below:
A.B only$\text{B only}$
B.A & C only$\text{A \& C only}$
C.A & E only$\text{A \& E only}$
D.A & D only$\text{A \& D only}$
Solution
Related Formula
Paramagnetism $\implies$ Presence of at least one unpaired electron in the molecular orbitals.
Core Logic
According to Molecular Orbital Theory (MOT):
O₂$O_2$ has 16 electrons. Its outer configuration contains two unpaired electrons in the anti-bonding orbitals: π2pₓ = π2py$\pi^{}_{2p_x} = \pi^{}_{2p_y}$. Thus, it is paramagnetic.
S₂$S_2$ belongs to the same oxygen family group and shares an analogous valence configuration with two unpaired electrons in its anti-bonding π^*$\pi^*$ orbitals. Hence, it is also paramagnetic.
N₂$N_2$ (14e-), F₂$F_2$ (18e-), and Cl₂$Cl_2$ (34e-) have completely paired electronic systems and behave diamagnetically.
Pattern Recognition
Both O₂$O_2$ and S₂$S_2$ contain 2 unpaired electrons in their highest occupied molecular orbitals, making them classic examples of paramagnetic diatomic species.
Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q39jee_main_2025_07_april_eveningHybridization
In SO₂$\text{SO}_2$, NO₂^-$\text{NO}_2^-$ and N₃^-$\text{N}_3^-$ the hybridizations at the central atom are respectively:
A.sp², sp² and sp$sp^2\text{, } sp^2 \text{ and } sp$
B.sp², sp and sp$sp^2\text{, } sp \text{ and } sp$
C.sp², sp² and sp²$sp^2\text{, } sp^2 \text{ and } sp^2$
D.sp, sp² and sp$sp\text{, } sp^2 \text{ and } sp$
Solution
Related Formula
Steric Number (Steric count) = Number of lone pairs on central atom + Number of σ-bonds$$\text{Steric Number (Steric count)} = \text{Number of lone pairs on central atom} + \text{Number of } \sigma\text{-bonds}$$
Core Logic
Let's perform steric calculations for each species:
SO₂$\text{SO}_2$: Central sulfur atom has 6 valence electrons, forms 2 σ$2\,\sigma$-bonds (and 2 π$2\,\pi$-bonds) with oxygen, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
NO₂^-$\text{NO}_2^-$: Central nitrogen atom has 5 valence electrons + 1$+ 1$ from negative charge = 6$= 6$. It forms 2 σ$2\,\sigma$-bonds, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
N₃^-$\text{N}_3^-$ (Azide ion): Linear configuration structure can be drawn as:
The central nitrogen has 2 σ$2\,\sigma$-bonds and 0 lone pairs. Steric number = 2 + 0 = 2 sp$= 2 + 0 = 2 \implies sp$.
Step 1: Geometry Outlines
The individual orbital fields are represented visually:
Hybridization diagram for Q39 - JEE Main 2025 Evening
Hence, hybridizations follow the order: sp²$sp^2$, sp²$sp^2$, and sp$sp$.
Pattern Recognition
Steric short tracking: Species with linear structures like CO₂, N₂O, N₃^-$\text{CO}_2, \text{N}_2O, \text{N}_3^-$ possess central atoms that are always sp$sp$ hybridized due to the requirement of two opposing σ$\sigma$-bonds.
Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q32jee_main_2025_24_jan_eveningResonance and Bond Parameters
Given below are two statements:
Statement (I) : Experimentally determined oxygen-oxygen bond lengths in the O₃$\mathrm{O}_{3}$ are found to be same and the bond length is greater than that of a O=O$\mathrm{O=O}$ (double bond) but less than that of a single (O-O)$(\mathrm{O-O})$ bond.
Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O=O)$(\mathrm{O=O})$ but more than that of a single bond (O-O)$(\mathrm{O-O})$.
In the light of the above statements, choose the correct answer from the options given below:
A. \text{Statement I is true but Statement II is false}
B. \text{Both Statement I and Statement II are true}
C. \text{Both Statement I and Statement II are false}
D. \text{Statement I is false but Statement II is true}
Solution
Core Logic
Analysis of Statement I:
Ozone (O₃$\mathrm{O}_3$) exhibits resonance. The two major canonical forms contribute equally to the resonance hybrid, meaning both oxygen-oxygen bonds are identical. Their bond order is 1.5$1.5$, making the bond length intermediate between a true single bond and a true double bond. Thus, Statement I is completely true.
Analysis of Statement II:
Statement II claims that lone pair-lone pair repulsion is solely responsible for this intermediate bond length. This is incorrect. The intermediate bond parameter is fundamentally a direct consequence of resonance delocalization, not lone-pair repulsions. Thus, Statement II is false.
Pattern Recognition
Whenever a molecule has identical intermediate bond lengths instead of distinct single and double bonds, resonance delocalization is almost always the core underlying reason.
Chapter Mix
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q30jee_main_2025_24_jan_morningMolecular Orbital Theory
Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction]?
A. 2pz$2p_{z}$ and 2pₓ$2p_{x}$
B. 2s and 2pₓ$2p_{x}$
C. 3dxy$3d_{xy}$ and 3dx²-y²$3d_{x^{2}-y^{2}}$
D. 2s and 2pz$2p_{z}$
E. 2pz$2p_{z}$ and 3dx²-y²$3d_{x^{2}-y^{2}}$
Choose the correct answer from the options given below:
A. E Only
B. A and B Only
C. D Only
D. C and D Only
Solution
Core Logic
For atomic orbitals to successfully combine into molecular orbitals, they must share appropriate spatial symmetry relative to the internuclear axis (z$z$-axis).
Combination A, B, C, and E involve orbitals with mismatching symmetry planes, resulting in a net zero overlap integral.
Combination D (2s$2s$ and 2pz$2p_z$) preserves continuous spatial alignment along the z$z$-axis, allowing effective frontal overlap to synthesize a stable sigma molecular orbital.
Visual symmetry breakdowns:
Molecular Orbital Theory diagram 1 for Q30 (Symmetry mismatch for A)
Molecular Orbital Theory diagram 1 for Q30 (Symmetry mismatch for C)
Molecular Orbital Theory diagram 1 for Q30 (Valid overlapping leading to Sigma molecular orbital for D)
Molecular Orbital Theory diagram 1 for Q30 (Symmetry mismatch for E)
Pattern Recognition
Verify orbital symmetry signs across the designated internuclear reference line. Mismatched symmetries cancel out completely (Ioverlap = 0$I_{\text{overlap}} = 0$).
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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