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

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

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

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

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

Q64 jee_main_2026_21_jan_morning VSEPR Theory

Given below are two statements: Statement I: The number of species among

SF₄

,

NH₄^+

,

[NiCl₄]²⁻

,

XeF₄

,

[PtCl₄]²⁻

,

SeF₄

and

[Ni(CN)₄]²⁻

, that have tetrahedral geometry is 3. Statement II: In the set

[NO₂, BeH₂, BF₃, AlCl₃]

, all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below:

A. $Statement I is true but Statement II is false$
B. $Both Statement I and Statement II are false$
C. $Statement I is false but Statement II is true$
D. $Both Statement I and Statement II are true$

Solution

Core Logic

Evaluating Statement I:

$SF₄$ $\mathrm{SF}_4$: $sp³d$ $sp^3d$ (1 lone pair) $arrow$ $\rightarrow$ See-saw
$XeF₄$ $\mathrm{XeF}_4$: $sp³d²$ $sp^3d^2$ (2 lone pairs) $arrow$ $\rightarrow$ Square planar
$[PtCl₄]²⁻$ $[\mathrm{PtCl}_4]^{2-}$: $dsp²$ $dsp^2$ $arrow$ $\rightarrow$ Square planar
$[NiCl₄]²⁻$ $[\mathrm{NiCl}_4]^{2-}$: $sp³$ $sp^3$ $arrow$ $\rightarrow$ Tetrahedral
$[Ni(CN)₄]²⁻$ $[\mathrm{Ni(CN)}_4]^{2-}$: $dsp²$ $dsp^2$ $arrow$ $\rightarrow$ Square planar
$SeF₄$ $\mathrm{SeF}_4$: $sp³d$ $sp^3d$ (1 lone pair) $arrow$ $\rightarrow$ See-saw
$NH₄^+$ $\mathrm{NH}_4^+$: $sp³$ $sp^3$ (0 lone pairs) $arrow$ $\rightarrow$ Tetrahedral

Total tetrahedral species = 2 ( $[NiCl₄]²⁻$ $[\mathrm{NiCl}_4]^{2-}$ and $NH₄^+$ $\mathrm{NH}_4^+$). Statement I says 3, so it is false.

Evaluating Statement II:

$NO₂$ $NO_2$: Central N has 7 valence electrons (odd-electron molecule, incomplete octet).
$BeH₂$ $BeH_2$: Central Be has 4 electrons (incomplete octet).
$BF₃$ $BF_3$: Central B has 6 electrons (incomplete octet).
$AlCl₃$ $AlCl_3$ (monomer): Central Al has 6 electrons (incomplete octet).

Therefore, all molecules have incomplete octets. Statement II is true.

Step 1: Final Conclusion

Statement I is false, Statement II is true.

Chapter Mix

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

Q64 jee_main_2026_21_jan_evening Bond Length Trends

The correct increasing order of

C-H (A)

,

C-O (B)

,

C=O (C)

and

C (D)

bonds in terms of covalent bond length is: (1)

A < B < C < D

(2)

A < D < C < B

(3)

D < C < B < A

(4)

D < C < A < B

A. $(1) A < B < C < D$ $(1) \ A < B < C < D$
B. $(2) A < D < C < B$ $(2) \ A < D < C < B$
C. $(3) D < C < B < A$ $(3) \ D < C < B < A$
D. $(4) D < C < A < B$ $(4) \ D < C < A < B$

Solution

Core Logic

Comparing bond lengths:

C–H (A): $107 pm$ $\sim 107 \text{ pm}$
C≡N (D): $116 pm$ $\sim 116 \text{ pm}$
C=O (C): $121 pm$ $\sim 121 \text{ pm}$
C–O (B): $143 pm$ $\sim 143 \text{ pm}$

Step 1: Final Conclusion

Thus, the increasing order is $A < D < C < B$ $A < D < C < B$, corresponding to option (2).

Pattern Recognition

Sees: covalent bond length comparison across bond orders and atomic radii. Trap: Assuming triple bonds are always longer or shorter without accounting for smaller atomic radii like hydrogen.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q69 jee_main_2026_21_jan_evening Bond Dissociation Enthalpy and Fajan's Rules

Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is

Cl₂ > Br₂ > F₂ > I₂

. Statement II: The correct trend in the covalent character of the metal halides is

[SnCl₄ > SnCl₂]

,

[PbCl₄ > PbCl₂]

and

[UF₄ > UF₆]

(or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:

A. $(1) Statement I is true but Statement II is false$
B. $(2) Both Statement I and Statement II are true$
C. $(3) Statement I is false but Statement II is true$
D. $(4) Both Statement I and Statement II are false$

Solution

Core Logic

Statement I: Bond dissociation energy order for halogens is $Cl₂ > Br₂ > F₂ > I₂$ due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true.
Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so $UF₆ > UF₄$ $\text{UF}_6 > \text{UF}_4$ (higher oxidation state has greater covalent character), making the statement II claim regarding $UF₄ > UF₆$ $\text{UF}_4 > \text{UF}_6$ false.

Step 1: Final Conclusion

Statement I is true but Statement II is false, corresponding to option (1).

Pattern Recognition

Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q52 jee_main_2026_22_january_morning Lewis Structures and Formal Charge

The formal changes on the atoms marked as (1) to (4) in the Lewis representation of

HNO₃

molecule respectively are

Lewis structure of HNO3 diagram for Q52 - JEE Main 2026 Morning — The image shows the Lewis structure of HNO3 with four atoms numbered 1 through 4.

A. $+1, 0, 0, -1$ $\text{+1, 0, 0, -1}$
B. $0, -1, 0, +1$ $\text{0, -1, 0, +1}$
C. $0, +1, 0, -1$ $\text{0, +1, 0, -1}$
D. $0, 0, -1, +1$ $\text{0, 0, -1, +1}$

Solution

Related Formula

Formal charge = (Valence e⁻) - (Non-bonding e⁻) - Bonding e⁻2

Core Logic

Evaluate the structure of $HNO₃$ $HNO_{3}$ shown in the solution image:

Detailed Lewis structure of HNO3 diagram — The image shows the Lewis structure of HNO3 with four atoms numbered 1 through 4.

Atom 1 (Oxygen with H and N): $F.C. = 6 - 4 - (4)/(2) = 0$ $F.C. = 6 - 4 - \frac{4}{2} = 0$

Atom 2 (Nitrogen): $F.C. = 5 - 0 - (8)/(2) = +1$ $F.C. = 5 - 0 - \frac{8}{2} = +1$

Atom 3 (Double bonded Oxygen): $F.C. = 6 - 4 - (4)/(2) = 0$ $F.C. = 6 - 4 - \frac{4}{2} = 0$

Atom 4 (Single bonded Oxygen): $F.C. = 6 - 6 - (2)/(2) = -1$ $F.C. = 6 - 6 - \frac{2}{2} = -1$

Step 1: Final Conclusion

The formal charges on atoms (1), (2), (3), and (4) are respectively 0, +1, 0, -1.

Pattern Recognition

In nitro groups ( $-NO₂$ $-NO_2$), the central nitrogen is always +1, the single bonded oxygen is -1, and the double-bonded oxygen is 0.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q68 jee_main_2026_22_january_morning Hybridization and VSEPR Theory

Two p-block elements X and Y form fluroides of the type

EF₃

. The fluoride compound

XF₃

is a Lewis acid and

YF₃

is a Lewis base. The hybridization of the central atoms of

XF₃

and

YF₃

respectively are

A. $Both sp³$ $\text{Both } sp^{3}$
B. $sp² and sp³$ $sp^{2} \text{ and } sp^{3}$
C. $sp³ and sp²$ $sp^{3} \text{ and } sp^{2}$
D. $Both sp²$ $\text{Both } sp^{2}$

Solution

Core Logic

$XF₃$ $XF_3$ acts as a Lewis acid, meaning it is an electron-deficient species capable of accepting an electron pair. A common example from the p-block is $BF₃$ $BF_3$. The central Boron atom has 3 bond pairs and 0 lone pairs. Therefore, its steric number is 3, corresponding to $sp²$ $sp^2$ hybridization.

$YF₃$ $YF_3$ acts as a Lewis base, meaning it has an available lone pair to donate. A common example is $NF₃$ $NF_3$. The central Nitrogen atom has 3 bond pairs and 1 lone pair. Its steric number is 4, corresponding to $sp³$ $sp^3$ hybridization.

Step 1: Final Conclusion

The hybridization of X and Y respectively are $sp²$ $sp^2$ and $sp³$ $sp^3$.

Pattern Recognition

Electron deficient central atoms (Group 13) form $sp²$ $sp^2$ planar molecules. Atoms with a lone pair (Group 15) form $sp³$ $sp^3$ pyramidal molecules.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 11 Chemistry: p-Block Elements

Solution & Explanation

Step 1: Determine 'n'

Step 2: Find ions with 2 unpaired electrons

Pattern Recognition

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