Match List-I with List-II. List-I (Molecule) List-II (Shape) (A) BrF5 (I) T-shape (B) H2O (II) See saw (C) ClF3 (III)…

Q: Match List-I with List-II. List-I (Molecule) List-II (Shape) (A) BrF5 (I) T-shape (B) H2O (II) See saw (C) ClF3 (III) Bent (D) SF4 (IV) Square pyramidal

### Core Logic Using VSEPR theory: (A) BrF5: Br has 7 valence electrons. 5 form bonds with F, leaving 1 lone pair. (5 bp + 1 lp) arrow sp3d2 hybridization arrow Square pyramidal shape. (B) H2O: O has 6 valence electrons. 2 form bonds with H, leaving 2 lone pairs. (2 bp + 2 lp) arrow sp3 hybridization arrow Bent shape. (C) ClF3: Cl has 7 valence electrons. 3 form bonds with F, leaving 2 lone pairs. (3 bp + 2 lp) arrow sp3d hybridization arrow T-shape. (D) SF4: S has 6 valence electrons. 4 form bo

Year	2026	2025	2024	Total
Questions	12	14	16	42

List-I (Molecule)	List-II (Shape)
(A) $BrF₅$ $BrF_5$	(I) T-shape
(B) $H₂O$ $H_2O$	(II) See saw
(C) $ClF₃$ $ClF_3$	(III) Bent
(D) $SF₄$ $SF_4$	(IV) Square pyramidal

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

Match List-I with List-II. List-I (Molecule)List-II (Shape)(A) BrF₅$BrF_5$(I) T-shape(B) H₂O$H_2O$(II) See saw(C) ClF₃$ClF_3$(III) Bent(D) SF₄$SF_4$(IV) Square pyramidal

Solution & Explanation

Core Logic

Step 1: Matching

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Match List-I with List-II.
List-I (Molecule) List-II (Shape)
(A) $BrF₅$ $BrF_5$ (I) T-shape
(B) $H₂O$ $H_2O$ (II) See saw
(C) $ClF₃$ $ClF_3$ (III) Bent
(D) $SF₄$ $SF_4$ (IV) Square pyramidal