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_2024_27_jan_morning Dipole Moment

Choose the polar molecule from the following:

A. $CCl₄$ $CCl_4$
B. $CO₂$ $CO_2$
C. $CH₂=CH₂$ $CH_2=CH_2$
D. $CHCl₃$ $CHCl_3$

Solution

Core Logic

CCl₄ arrow μ = 0 (Symmetrical tetrahedral)

CO₂ arrow μ = 0 (Linear structure)

CH₂=CH₂ arrow μ = 0 (Planar symmetrical structure)

For $CHCl₃$ $CHCl_3$, the individual dipole vectors do not cancel due to differing electronegativities of H and Cl, leading to a permanent non-zero dipole moment ( $μ \neq 0$ $\mu \neq 0$).

Dipole vectors structural cancellation schema for Q69 - JEE Main 2024 Morning

Pattern Recognition

Symmetry yields vector cancellation $arrow μ=0$ $\rightarrow \mu=0$. Asymmetry in $CHCl₃$ $CHCl_3$ prevents cancellation $arrow$ $\rightarrow$ polar.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q89 jee_main_2024_27_jan_morning Molecular Orbital Theory

Sum of bond order of

CO

and

NO^+

is

.

Numerical Answer. Answer: 6 to 6

Solution

Step 1: Determine the bond order of $CO$ $\text{CO}$

Carbon monoxide ( $CO$ $\text{CO}$) contains $6 + 8 = 14$ $6 + 8 = 14$ total electrons. Its structural representation is $C$ $\text{C}\equiv\text{O}$, matching a bond order value of 3.

Step 2: Determine the bond order of $NO^+$ $\text{NO}^+$

The nitrosonium ion ( $NO^+$ $\text{NO}^+$) contains $7 + 8 - 1 = 14$ $7 + 8 - 1 = 14$ total electrons. Since it is isoelectronic with $N₂$ $\text{N}_2$ and $CO$ $\text{CO}$ ( $14 electrons$ $14\text{ electrons}$), its corresponding bond order value is also 3.

Step 3: Sum the results

Sum = 3 + 3 = 6

Pattern Recognition

Isoelectronic species possessing 14 total electrons consistently demonstrate a bond order value of 3.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q81 jee_main_2024_29_jan_morning VSEPR Theory

Number of compounds with one lone pair of electrons on central atom amongst following is

O₃

,

H₂O

,

SF₄

,

ClF₃

,

NH₃

,

BrF₅

,

XeF₄

Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Let us determine the steric number ( $Z$ $Z$) and number of lone pairs ( $LP$ $LP$) for the central atom in each given molecule. Formula: $Z = (1)/(2) (V + M - C + A)$ $Z = \frac{1}{2} (V + M - C + A)$ Where $V$ $V$ = valence electrons on central atom, $M$ $M$ = number of monovalent atoms, $C$ $C$ = cationic charge, $A$ $A$ = anionic charge. $LP = Z - Bond Pairs (B.P.)$ $LP = Z - \text{Bond Pairs (B.P.)}$

$O₃$ $O_3$: Central atom O ( $V=6$ $V=6$). It forms one double bond and one dative bond. It has 1 lone pair remaining.
$H₂O$ $H_2O$: Central atom O ( $V=6$ $V=6$). $Z = (1)/(2)(6 + 2) = 4$ $Z = \frac{1}{2}(6 + 2) = 4$. $LP = 4 - 2 = 2$ $LP = 4 - 2 = 2$.
$SF₄$ $SF_4$: Central atom S ( $V=6$ $V=6$). $Z = (1)/(2)(6 + 4) = 5$ $Z = \frac{1}{2}(6 + 4) = 5$. $LP = 5 - 4 = 1$ $LP = 5 - 4 = 1$ (See-saw shape).
$ClF₃$ $ClF_3$: Central atom Cl ( $V=7$ $V=7$). $Z = (1)/(2)(7 + 3) = 5$ $Z = \frac{1}{2}(7 + 3) = 5$. $LP = 5 - 3 = 2$ $LP = 5 - 3 = 2$ (T-shape).
$NH₃$ $NH_3$: Central atom N ( $V=5$ $V=5$). $Z = (1)/(2)(5 + 3) = 4$ $Z = \frac{1}{2}(5 + 3) = 4$. $LP = 4 - 3 = 1$ $LP = 4 - 3 = 1$ (Pyramidal).
$BrF₅$ $BrF_5$: Central atom Br ( $V=7$ $V=7$). $Z = (1)/(2)(7 + 5) = 6$ $Z = \frac{1}{2}(7 + 5) = 6$. $LP = 6 - 5 = 1$ $LP = 6 - 5 = 1$ (Square Pyramidal).
$XeF₄$ $XeF_4$: Central atom Xe ( $V=8$ $V=8$). $Z = (1)/(2)(8 + 4) = 6$ $Z = \frac{1}{2}(8 + 4) = 6$. $LP = 6 - 4 = 2$ $LP = 6 - 4 = 2$ (Square Planar).

Step 1: Final Counting

VSEPR Theory diagram for Q81 - JEE Main 2024 Morning

The compounds containing exactly ONE lone pair on the central atom are $O₃$ $O_3$, $SF₄$ $SF_4$, $NH₃$ $NH_3$, and $BrF₅$ $BrF_5$.

Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q88 jee_main_2024_29_jan_morning Molecular Orbital Theory

The number of species from the following which are paramagnetic and with bond order equal to one is

H₂, He₂^+, O₂^+, N₂²⁻, O₂²⁻, F₂, Ne₂^+, B₂

Numerical Answer. Answer: 1 to 1

Solution

Core Logic

Using Molecular Orbital (MO) Theory, we evaluate the bond order ( $BO = (Nb - Nₐ)/(2)$ $BO = \frac{N_b - N_a}{2}$) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Species	Magnetic behaviour	Bond order
$H₂$ $H_2$	Diamagnetic	1
$He₂^+$ $He_2^+$	Paramagnetic	0.5
$O₂^+$ $O_2^+$	Paramagnetic	2.5
$N₂²⁻$ $N_2^{2-}$	Paramagnetic	2
$O₂²⁻$ $O_2^{2-}$	Diamagnetic	1
$F₂$ $F_2$	Diamagnetic	1
$Ne₂^+$ $Ne_2^+$	Paramagnetic	0.5
$B₂$ $B_2$	Paramagnetic	1

Step 1: Final Selection

We need the species that satisfies BOTH conditions:

Paramagnetic
Bond Order = 1

Looking at the table, $B₂$ $B_2$ is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate $π₂ₚ$ $\pi_{2p}$ orbitals) and has a bond order of 1.

Total number of such species = 1.

Pattern Recognition

$B₂$ $B_2$ (10 electrons) and $O₂$ $O_2$ (16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons. $B₂$ $B_2$ has BO = 1, and $O₂$ $O_2$ has BO = 2.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2024_30_january_evening VSEPR Theory and Molecular Shapes

The molecule/ion with square pyramidal shape is:

A. $[Ni(CN)₄]²⁻$ $\text{[Ni(CN)}_4]^{2-}$
B. $PCl₅$ $\text{PCl}_5$
C. $BrF₅$ $\text{BrF}_5$
D. $PF₅$ $\text{PF}_5$

Solution

Core Logic

According to VSEPR theory:

$[Ni(CN)₄]²⁻$ $[Ni(CN)_4]^{2-}$: $dsp²$ $dsp^2$ hybridization $arrow$ $\rightarrow$ Square Planar.
$PCl₅$ $PCl_5$: $sp³d$ $sp^3d$ hybridization with 0 lone pairs $arrow$ $\rightarrow$ Trigonal Bipyramidal.
$BrF₅$ $BrF_5$: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine, leaving 1 lone pair. $sp³d²$ $sp^3d^2$ hybridization $arrow$ $\rightarrow$ geometry is octahedral, but shape is Square Pyramidal.
$PF₅$ $PF_5$: $sp³d$ $sp^3d$ hybridization with 0 lone pairs $arrow$ $\rightarrow$ Trigonal Bipyramidal.

Square Pyramidal structure of BrF5 diagram for Q69 - JEE Main 2024 Evening

Pattern Recognition

$AX₅E₁$ $AX_5E_1$ configuration (5 bond pairs + 1 lone pair) typically yields a Square Pyramidal shape.

Chapter Mix

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

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 7

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Solution

Step 1: Determine the bond order of CO$\text{CO}$

Step 2: Determine the bond order of NO^+$\text{NO}^+$

Step 3: Sum the results

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Counting

Chapter Mix

Solution

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

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Step 1: Determine the bond order of $CO$ $\text{CO}$

Step 2: Determine the bond order of $NO^+$ $\text{NO}^+$