Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of ClF3 . The ions from the following with 'n' number of unpaired electrons are : A. V3 + B. Ti3+ C. Cu2 + D. Ni2+ E. Ti2+ Choose the correct answer from the options given below:

ClF3 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. Let us compute the number of unpaired electrons for each configuration: - **A. V3+:** [Ar] 3d2 arrow 2 unpaired electrons. - **B. Ti3+:** [Ar] 3d1 arrow 1 unpaired electron. - **C. Cu2+:** [Ar] 3d9 arrow 1 unpaired electron. - **D. Ni2+:** [Ar] 3d8 arrow 2 unpaired electrons. - **E. Ti2+:** [Ar] 3d2 arrow 2 unpaired electrons. Thus, A, D, and E have exactly n=2 unpaired electrons. Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember ClF3 is T-shaped with 2 equatorial lone pairs. Look for d2 or d8 configurations among the transition metal ions. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of ClF3 . The ions from the following with 'n' number of unpaired electrons are : A. V3 + B. Ti3+ C. Cu2 + D. Ni2+ E. Ti2+ Choose the correct answer from the options given below:

ClF3 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. Let us compute the number of unpaired electrons for each configuration: - **A. V3+:** [Ar] 3d2 arrow 2 unpaired electrons. - **B. Ti3+:** [Ar] 3d1 arrow 1 unpaired electron. - **C. Cu2+:** [Ar] 3d9 arrow 1 unpaired electron. - **D. Ni2+:** [Ar] 3d8 arrow 2 unpaired electrons. - **E. Ti2+:** [Ar] 3d2 arrow 2 unpaired electrons. Thus, A, D, and E have exactly n=2 unpaired electrons. Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember ClF3 is T-shaped with 2 equatorial lone pairs. Look for d2 or d8 configurations among the transition metal ions. Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of…

Year	2026	2025	2024	Total
Questions	12	14	16	42

Year

2026

2025

2024

Total

Questions

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

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

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^+

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

Step 1: Determine 'n'

Step 2: Find ions with 2 unpaired electrons

Pattern Recognition

Chapter Mix

Reference Study Guides

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

Solution

Core Logic

Pattern Recognition

Chapter Mix

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

Chapter Mix

More Chemical Bonding and Molecular Structure Questions — jee_main_2025_28_jan_morning

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

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