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 4

Q jee_main_2025_07_april_morning VSEPR Theory

Match the LIST-I with LIST-II.

LIST-I (Molecule/ion)	LIST-II (Bond pair : lone pair on the central atom)
(A) $ICl₂^-$ $\mathrm{ICl}_2^-$	(I) $4 : 2$ $4 : 2$
(B) $H₂O$ $\mathrm{H}_2\mathrm{O}$	(II) $4 : 1$ $4 : 1$
(C) $SO₂$ $\mathrm{SO}_2$	(III) $2 : 3$ $2 : 3$
(D) $XeF₄$ $\mathrm{XeF}_4$	(IV) $2 : 2$ $2 : 2$

Choose the correct answer from the options given below:

A. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
B. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
D. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

Core Logic

Let's find the number of bond pairs ( $σ$ $\sigma$-bonds or regions) and lone pairs on the central atom of each species:

$ICl₂^-$ $\mathrm{ICl}_2^-$:
Central atom Iodine has $7$ $7$ valence electrons $+ 1$ $+ 1$ negative charge $= 8$ $= 8$ electrons.
Forms $2$ $2$ single bonds (bond pairs $= 2$ $= 2$).
Remaining $6$ $6$ electrons form $3$ $3$ lone pairs.
Ratio is $2 : 3$ $2 : 3$ (Matches LIST-II, III).

$H₂O$ $\mathrm{H}_2\mathrm{O}$:
Oxygen has $6$ $6$ valence electrons.
Forms $2$ $2$ bond pairs with Hydrogens.
Remaining $4$ $4$ electrons form $2$ $2$ lone pairs.
Ratio is $2 : 2$ $2 : 2$ (Matches LIST-II, IV).

$SO₂$ $\mathrm{SO}_2$:
Sulfur has $6$ $6$ valence electrons.
Forms $2$ $2$ double bonds (which are counted as $4$ $4$ bonding pairs of electrons/bond pairs in typical VSEPR representations here).
Remaining $2$ $2$ electrons form $1$ $1$ lone pair.
Ratio is $4 : 1$ $4 : 1$ (Matches LIST-II, II).

$XeF₄$ $\mathrm{XeF}_4$:
Xenon has $8$ $8$ valence electrons.
Forms $4$ $4$ bond pairs with Fluorines.
Remaining $4$ $4$ electrons form $2$ $2$ lone pairs.
Ratio is $4 : 2$ $4 : 2$ (Matches LIST-II, I).

Thus, the correct mapping is: A-III, B-IV, C-II, D-I.

Pattern Recognition

For match-the-column with VSEPR structures:

Always find steric number: $Steric Number = (1)/(2)(V + M - C + A)$ .
Water is $2$ $2$ bond pairs, $2$ $2$ lone pairs ( $sp³$ $sp^3$) $arrow$ $\rightarrow$ B-IV. This alone helps eliminate multiple incorrect options immediately.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q50 jee_main_2025_29_jan_evening Lewis Structures and Valence Electrons

Total number of non bonded electrons present in

NO2⁻

ion based on Lewis theory is ________.

Numerical Answer. Answer: 12 to 12

Solution

Core Logic

Let's compute the total valence electrons for the nitrite ion ( $NO₂^-$ $NO_2^-$):

Valence electrons = 5 (from N) + 2 \times 6 (from O) + 1 (negative charge) = 18 electrons

In the valid Lewis structural representation:

The central nitrogen atom forms one single bond and one double bond with the terminal oxygens, consuming $2 + 4 = 6$ $2 + 4 = 6$ bonding electrons.
Remaining non-bonded valence electrons = $18 - 6 = 12$ $18 - 6 = 12$ electrons.

Step 1: Account for Lone Pairs

Distribution of non-bonded electrons across the individual atoms:

Central Nitrogen atom has $1$ $1$ lone pair ( $2$ $2$ electrons).
Single-bonded Oxygen atom has $3$ $3$ lone pairs ( $6$ $6$ electrons).
Double-bonded Oxygen atom has $2$ $2$ lone pairs ( $4$ $4$ electrons).

Total non-bonded electrons = 2 + 6 + 4 = 12

Pattern Recognition

Non-bonded electrons can always be obtained directly by subtracting total bonding electrons from total valence electrons.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q35 jee_main_2025_28_jan_morning Molecular Geometry and VSEPR

The molecules having square pyramidal geometry are

A. $BrF₅ & XeOF₄$
B. $SbF₅ & XeOF₄$
C. $SbF₅ & PCl₅$
D. $BrF₅ & PCl₅$

Solution

Core Logic

Let us check the steric details using VSEPR theory:

$BrF₅$ $\mathrm{BrF}_5$: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 ( $sp³d²$ $sp^3d^2$), geometry is square pyramidal.
$XeOF₄$ $\mathrm{XeOF}_4$: 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 ( $sp³d²$ $sp^3d^2$), geometry is square pyramidal.
$SbF₅$ $\mathrm{SbF}_5$ & $PCl₅$ $\mathrm{PCl}_5$: Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 ( $sp³d$ $sp^3d$), geometry is trigonal bipyramidal.

Visual representations of geometries:

Geometry structure diagram 1 for Q35 - JEE Main 2025 Morning

Pattern Recognition

Sees: Steric count 6 with 5 bonded segments + 1 lone pair $arrow$ $\rightarrow$ always square pyramidal geometry.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q jee_main_2025_03_april_morning Hybridisation

Match the LIST-I with LIST-II. Choose the correct answer from the options given below:

A. A-II, B-III, C-IV, D-I
B. A-IV, B-I, C-II, D-III
C. A-I, B-II, C-III, D-IV
D. A-III, B-I, C-IV, D-II

Solution

Core Logic

Let us evaluate each central atom configuration systematically:

A. $PF₅$ $\text{PF}_5$: Phosphorus has 5 valence electrons, forming $5σ$ $5\sigma$ bonds with zero lone pairs. Steric number $= 5 sp³d$ $= 5 \implies sp^3d$ hybridisation.
B. $SF₆$ $\text{SF}_6$: Sulfur has 6 valence electrons, forming $6σ$ $6\sigma$ bonds with zero lone pairs. Steric number $= 6 sp³d²$ $= 6 \implies sp^3d^2$ hybridisation.
C. $Ni(CO)₄$ $\text{Ni}(\text{CO})_4$: Nickel is in a 0 oxidation state ( $3d⁸ 4s²$ $3d^8 4s^2$). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled $3d¹⁰$ $3d^{10}$ state. The vacant $4s$ $4s$ and three $4p$ $4p$ orbitals hybridise to give an $sp³$ $sp^3$ configuration.
Orbital configuration matrix for Q43 - JEE Main 2025 Morning
D. $[PtCl₄]²⁻$ $[\text{PtCl}_4]^{2-}$: Platinum is in the $+2$ $+2$ oxidation state ( $5d⁸$ $5d^8$). Since it belongs to the $5d$ $5d$ transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar $dsp²$ $dsp^2$ hybridisation state.
Orbital configuration matrix for Q43 - JEE Main 2025 Morning

Pattern Recognition

Shortcut: Match main-group species first: $PF₅ arrow sp³d (II)$ , $SF₆ arrow sp³d² (III)$ . This immediately isolates Option (A) without needing to evaluate coordination fields.

Evaluation Rubric / Model Answer

Option (A)

Chapter Mix

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