Match the LIST-I with LIST-II. LIST-I (Molecule/ion) LIST-II (Bond pair : lone pair on the central atom) (A) ICl2^- (I) 4 : 2 (B) H2O (II) 4 : 1 (C) SO2 (III) 2 : 3 (D) XeF4 (IV) 2 : 2 Choose the correct answer from the options given below:

### Core Logic Let's find the number of bond pairs (sigma-bonds or regions) and lone pairs on the central atom of each species: 1. **ICl2^-**: - Central atom Iodine has 7 valence electrons + 1 negative charge = 8 electrons. - Forms 2 single bonds (bond pairs = 2). - Remaining 6 electrons form 3 lone pairs. - Ratio is 2 : 3 (Matches LIST-II, III). 2. **H2O**: - Oxygen has 6 valence electrons. - Forms 2 bond pairs with Hydrogens. - Remaining 4 electrons form 2 lone pairs. - Ratio is 2 : 2 (Matches LIST-II, IV). 3. **SO2**: - Sulfur has 6 valence electrons. - Forms 2 double bonds (which are counted as 4 bonding pairs of electrons/bond pairs in typical VSEPR representations here). - Remaining 2 electrons form 1 lone pair. - Ratio is 4 : 1 (Matches LIST-II, II). 4. **XeF4**: - Xenon has 8 valence electrons. - Forms 4 bond pairs with Fluorines. - Remaining 4 electrons form 2 lone pairs. - Ratio is 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 bond pairs, 2 lone pairs (sp3) arrow B-IV. This alone helps eliminate multiple incorrect options immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Match the LIST-I with LIST-II. LIST-I (Molecule/ion) LIST-II (Bond pair : lone pair on the central atom) (A) ICl2^- (I) 4 : 2 (B) H2O (II) 4 : 1 (C) SO2 (III) 2 : 3 (D) XeF4 (IV) 2 : 2 Choose the correct answer from the options given below:

### Core Logic Let's find the number of bond pairs (sigma-bonds or regions) and lone pairs on the central atom of each species: 1. **ICl2^-**: - Central atom Iodine has 7 valence electrons + 1 negative charge = 8 electrons. - Forms 2 single bonds (bond pairs = 2). - Remaining 6 electrons form 3 lone pairs. - Ratio is 2 : 3 (Matches LIST-II, III). 2. **H2O**: - Oxygen has 6 valence electrons. - Forms 2 bond pairs with Hydrogens. - Remaining 4 electrons form 2 lone pairs. - Ratio is 2 : 2 (Matches LIST-II, IV). 3. **SO2**: - Sulfur has 6 valence electrons. - Forms 2 double bonds (which are counted as 4 bonding pairs of electrons/bond pairs in typical VSEPR representations here). - Remaining 2 electrons form 1 lone pair. - Ratio is 4 : 1 (Matches LIST-II, II). 4. **XeF4**: - Xenon has 8 valence electrons. - Forms 4 bond pairs with Fluorines. - Remaining 4 electrons form 2 lone pairs. - Ratio is 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 bond pairs, 2 lone pairs (sp3) arrow B-IV. This alone helps eliminate multiple incorrect options immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Match the LIST-I with LIST-II. LIST-I (Molecule/ion) LIST-II (Bond pair : lone pair on the central atom) (A) ICl2^- (I)…

LIST-I (Molecule/ion)	LIST-II (Bond pair : lone pair on the central atom)
(A) $mathrmICl_2^-$ $\mathrm{ICl}_2^-$	(I) $4 : 2$ $4 : 2$
(B) $mathrmH_2mathrmO$ $\mathrm{H}_2\mathrm{O}$	(II) $4 : 1$ $4 : 1$
(C) $mathrmSO_2$ $\mathrm{SO}_2$	(III) $2 : 3$ $2 : 3$
(D) $mathrmXeF_4$ $\mathrm{XeF}_4$	(IV) $2 : 2$ $2 : 2$

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 (

NO_2^-

):

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

bonding electrons. * Remaining non-bonded valence electrons =

18 - 6 = 12

electrons. ### Step 1: Account for Lone Pairs Distribution of non-bonded electrons across the individual atoms: * Central Nitrogen atom has

1

lone pair (

2

electrons). * Single-bonded Oxygen atom has

3

lone pairs (

6

electrons). * Double-bonded Oxygen atom has

2

lone pairs (

4

electrons).

textTotal 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q31 jee_main_2025_28_jan_morning VSEPR Theory and d-Electron Configurations

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

mathrmClF_3

. The ions from the following with 'n' number of unpaired electrons are : A.

mathrmV^3 +

B.

mathrmTi^3+

C.

mathrmCu^2 +

D.

mathrmNi^2+

E.

mathrmTi^2+

Choose the correct answer from the options given below:

A. $textA and C only$ $\text{A and C only}$
B. $textA, D and E only$ $\text{A, D and E only}$
C. $textB and C only$ $\text{B and C only}$
D. $textB and D only$ $\text{B and D only}$

Solution

### Step 1: Determine 'n'

mathrmClF_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

. ### Step 2: Find ions with 2 unpaired electrons Let us compute the number of unpaired electrons for each configuration: - **A.

mathrmV^3+

:**

[mathrmAr] 3d^2 rightarrow 2

unpaired electrons. - **B.

mathrmTi^3+

:**

[mathrmAr] 3d^1 rightarrow 1

unpaired electron. - **C.

mathrmCu^2+

:**

[mathrmAr] 3d^9 rightarrow 1

unpaired electron. - **D.

mathrmNi^2+

:**

[mathrmAr] 3d^8 rightarrow 2

unpaired electrons. - **E.

mathrmTi^2+

:**

[mathrmAr] 3d^2 rightarrow 2

unpaired electrons. Thus, A, D, and E have exactly

n=2

unpaired electrons. ### Pattern Recognition Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember

mathrmClF_3

is T-shaped with 2 equatorial lone pairs. Look for

d^2

or

d^8

configurations among the transition metal ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

Q35 jee_main_2025_28_jan_morning Molecular Geometry and VSEPR

The molecules having square pyramidal geometry are

A. $mathrmBrF_5 text \& mathrmXeOF_4$
B. $mathrmSbF_5 text \& mathrmXeOF_4$
C. $mathrmSbF_5 text \& mathrmPCl_5$
D. $mathrmBrF_5 text \& mathrmPCl_5$

Solution

### Core Logic Let us check the steric details using VSEPR theory: - **

mathrmBrF_5

:** Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine and retains 1 lone pair. Steric number = 6 (

sp^3d^2

), geometry is square pyramidal. - **

mathrmXeOF_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 (

sp^3d^2

), geometry is square pyramidal. - **

mathrmSbF_5

&

mathrmPCl_5

:** Central element has 5 valence electrons, forming 5 bonds with no lone pairs. Steric number = 5 (

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

rightarrow

always square pyramidal geometry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q43 jee_main_2025_03_april_morning Hybridisation

Match the LIST-I with LIST-II.

LIST-I (Molecules/ion)	LIST-II (Hybridisation of central atom)	A. $PF5$ $PF{5}$	I. $dsp^2$ $dsp^{2}$	B. $SF6$ $SF{6}$	II. $sp^3d$ $sp^{3}d$	C. $Ni(CO)4$ $Ni(CO){4}$	III. $sp^3d^2$ $sp^{3}d^{2}$	D. $[PtCl4]^2-$ $[PtCl{4}]^{2-}$	IV. $sp^3$ $sp^{3}$

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.

PF_5

: Phosphorus has 5 valence electrons, forming

5sigma

bonds with zero lone pairs. Steric number

= 5 implies sp^3d

hybridisation. * B.

SF_6

: Sulfur has 6 valence electrons, forming

6sigma

bonds with zero lone pairs. Steric number

= 6 implies sp^3d^2

hybridisation. * C.

Ni(CO)_4

: Nickel is in a 0 oxidation state (

3d^8 4s^2

). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled

3d^10

state. The vacant

4s

and three

4p

orbitals hybridise to give an

sp^3

configuration.

* D.

[PtCl_4]^2-

: Platinum is in the

+2

oxidation state (

5d^8

). Since it belongs to the

5d

transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar

dsp^2

hybridisation state.

### Pattern Recognition Shortcut: Match structural main elements first:

PF_5 ightarrow sp^3d

(II),

SF_6 ightarrow sp^3d^2

(III). This isolates option (1) instantly without checking d-block coordinate fields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Q44 jee_main_2025_04_april_evening VSEPR Theory

Given below are two statements: Statement (I) : for

mathrmCell mathrmF_3

, all three possible structures may be drawn as follows.

ClF3 structure variant I for Q44 — The prompt displays three configurations of chlorine trifluoride with differing spatial positions for its two lone electron pairs.

Statement (II) : Structure III is most stable, as the orbitals having the lone pairs are axial, where the

ell mathfrakp-

bp repulsion is minimum. In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is incorrect but statement II is correct.
B. Statement I is correct but statement II is incorrect.
C. Both Statement I and statement II are correct.
D. Both Statement I and statement II are incorrect.

Solution

### Related Formula

textSteric Number for ClF_3 = frac7+32 = 5 implies sp^3d text hybridization (Trigonal Bipyramidal geometry)

### Core Logic - **Statement I is correct:** The three structural arrangements represent the different ways to place three bond pairs and two lone pairs within a trigonal bipyramidal grid. - **Statement II is incorrect:** According to VSEPR theory and Bent's rule, in

sp^3d

hybridization, **lone pairs must occupy equatorial positions** to minimize strong

90^circ

lone pair-bond pair (

ell p-bp

) repulsions. Placing them axially maximizes repulsions, making that structure the least stable, not the most stable. ### Pattern Recognition For

sp^3d

configurations (Trigonal Bipyramidal), lone pairs ALWAYS prefer equatorial sites where they experience

120^circ

interactions, minimizing severe

90^circ

structural strains. This results in the classic stable T-shaped configuration for

ClF_3

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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