The molecules having square pyramidal geometry are

Q: The molecules having square pyramidal geometry are

Let us check the steric details using VSEPR theory: - **BrF5:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (sp3d2), geometry is square pyramidal. - **XeOF4:** 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 (sp3d2), geometry is square pyramidal. - **SbF5 & PCl5:** Central element has 5 valence electrons, forming 5 bonds with no lone pa

Year	2026	2025	2024	Total
Questions	12	14	16	42

Q jee_main_2025_04_april_evening VSEPR Theory

Given below are two statements: Statement (I) : for

C F₃

, 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.

Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the

p-

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

Steric Number for ClF₃ = (7+3)/(2) = 5 sp³d 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

Q28 jee_main_2025_04_april_morning Molecular Orbital Theory

Which of the following molecules(s) show/s paramagnetic behavior? (A)

O₂

(B)

N₂

(C)

F₂

(D)

S₂

(E)

Cl₂

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: $π 2pₓ = π 2py$ $\pi^{}_{2p_x} = \pi^{}_{2p_y}$. 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

Q39 jee_main_2025_07_april_evening Hybridization

In

SO₂

,

NO₂^-

and

N₃^-

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

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:

N= +N= N

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 $CO₂, N₂O, N₃^-$ 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

Q32 jee_main_2025_24_jan_evening Resonance and Bond Parameters

Given below are two statements: Statement (I) : Experimentally determined oxygen-oxygen bond lengths in the

O₃

are found to be same and the bond length is greater than that of a

O=O

(double bond) but less than that of a single

(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)

but more than that of a single bond

(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

Q30 jee_main_2025_24_jan_morning Molecular 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

and

2pₓ

B. 2s and

2pₓ

C.

3dxy

and

3dx²-y²

D. 2s and

2pz

E.

2pz

and

3dx²-y²

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

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Geometry Outlines

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

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