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) $2 : 2$ $2 : 2$

Q28 jee_main_2025_04_april_morning Molecular Orbital Theory

Which of the following molecules(s) show/s paramagnetic behavior? (A)

O_2

(B)

N_2

(C)

F_2

(D)

S_2

(E)

Cl_2

Choose the correct answer from the options given below:

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

Solution

### Related Formula Paramagnetism

implies

Presence of at least one unpaired electron in the molecular orbitals. ### Core Logic According to Molecular Orbital Theory (MOT): *

O_2

has 16 electrons. Its outer configuration contains two unpaired electrons in the anti-bonding orbitals:

pi^*_2p_x = pi^*_2p_y

. Thus, it is paramagnetic. *

S_2

belongs to the same oxygen family group and shares an analogous valence configuration with two unpaired electrons in its anti-bonding

pi^*

orbitals. Hence, it is also paramagnetic. *

N_2

(14e-),

F_2

(18e-), and

Cl_2

(34e-) have completely paired electronic systems and behave diamagnetically. ### Pattern Recognition Both

O_2

and

S_2

contain 2 unpaired electrons in their highest occupied molecular orbitals, making them classic examples of paramagnetic diatomic species. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q39 jee_main_2025_07_april_evening Hybridization

In

textSO_2

,

textNO_2^-

and

textN_3^-

the hybridizations at the central atom are respectively:

A. $sp^2text, sp^2 text and sp$ $sp^2\text{, } sp^2 \text{ and } sp$
B. $sp^2text, sp text and sp$ $sp^2\text{, } sp \text{ and } sp$
C. $sp^2text, sp^2 text and sp^2$ $sp^2\text{, } sp^2 \text{ and } sp^2$
D. $sptext, sp^2 text and sp$ $sp\text{, } sp^2 \text{ and } sp$

Solution

### Related Formula

textSteric Number (Steric count) = textNumber of lone pairs on central atom + textNumber of sigmatext-bonds

### Core Logic Let's perform steric calculations for each species: -

textSO_2

: Central sulfur atom has 6 valence electrons, forms

2\,sigma

-bonds (and

2\,pi

-bonds) with oxygen, leaving 1 lone pair. Steric number

= 2 + 1 = 3 implies sp^2

. -

textNO_2^-

: Central nitrogen atom has 5 valence electrons

+ 1

from negative charge

= 6

. It forms

2\,sigma

-bonds, leaving 1 lone pair. Steric number

= 2 + 1 = 3 implies sp^2

. -

textN_3^-

(Azide ion): Linear configuration structure can be drawn as:

overlinetextN=overset+textN=overlinetextN

The central nitrogen has

2\,sigma

-bonds and 0 lone pairs. Steric number

= 2 + 0 = 2 implies sp

. ### Step 1: Geometry Outlines The individual orbital fields are represented visually:

Hybridization diagram for Q39 - JEE Main 2025 Evening

Hence, hybridizations follow the order:

sp^2

,

sp^2

, and

sp

. ### Pattern Recognition Steric short tracking: Species with linear structures like

textCO_2, textN_2O, textN_3^-

possess central atoms that are always

sp

hybridized due to the requirement of two opposing

sigma

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

Q32 jee_main_2025_24_jan_evening Resonance and Bond Parameters

Given below are two statements: Statement (I) : Experimentally determined oxygen-oxygen bond lengths in the

mathrmO_3

are found to be same and the bond length is greater than that of a

mathrmO=O

(double bond) but less than that of a single

(mathrmO-O)

bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond

(mathrmO=O)

but more than that of a single bond

(mathrmO-O)

. In the light of the above statements, choose the correct answer from the options given below:

A. \text{Statement I is true but Statement II is false}
B. \text{Both Statement I and Statement II are true}
C. \text{Both Statement I and Statement II are false}
D. \text{Statement I is false but Statement II is true}

Solution

### Core Logic Analysis of Statement I: Ozone (

mathrmO_3

) exhibits resonance. The two major canonical forms contribute equally to the resonance hybrid, meaning both oxygen-oxygen bonds are identical. Their bond order is

1.5

, making the bond length intermediate between a true single bond and a true double bond. Thus, Statement I is completely true. Analysis of Statement II: Statement II claims that lone pair-lone pair repulsion is solely responsible for this intermediate bond length. This is incorrect. The intermediate bond parameter is fundamentally a direct consequence of resonance delocalization, not lone-pair repulsions. Thus, Statement II is false. ### Pattern Recognition Whenever a molecule has identical intermediate bond lengths instead of distinct single and double bonds, resonance delocalization is almost always the core underlying reason. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q30 jee_main_2025_24_jan_morning Molecular Orbital Theory

Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction]? A.

2p_z

and

2p_x

B. 2s and

2p_x

C.

3d_xy

and

3d_x^2-y^2

D. 2s and

2p_z

E.

2p_z

and

3d_x^2-y^2

Choose the correct answer from the options given below:

A. E Only
B. A and B Only
C. D Only
D. C and D Only

Solution

### Core Logic For atomic orbitals to successfully combine into molecular orbitals, they must share appropriate spatial symmetry relative to the internuclear axis (

z

-axis). - Combination A, B, C, and E involve orbitals with mismatching symmetry planes, resulting in a net zero overlap integral. - Combination D (

2s

and

2p_z

) preserves continuous spatial alignment along the

z

-axis, allowing effective frontal overlap to synthesize a stable sigma molecular orbital. Visual symmetry breakdowns: -

Molecular Orbital Theory diagram 1 for Q30

(Symmetry mismatch for A) -

(Symmetry mismatch for C) -

(Valid overlapping leading to Sigma molecular orbital for D) -

(Symmetry mismatch for E) ### Pattern Recognition Verify orbital symmetry signs across the designated internuclear reference line. Mismatched symmetries cancel out completely (

I_textoverlap = 0

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

Q43 jee_main_2025_24_jan_morning Hybridization and Molecular Geometry

Which of the following statement is true with respect to

mathrmH_2mathrmO, mathrmNH_3

and

mathrmCH_4

? A. The central atoms of all the molecules are

mathfraksp^3

hybridized. B. The H-O-H, H-N-H and H-C-H angles in the above molecules are

104.5^circ

,

107.5^circ

and

109.5^circ

respectively. C. The increasing order of dipole moment is

mathrmCH_4 < mathrmNH_3 < mathrmH_2mathrmO

. D. Both

mathrmH_2mathrmO

and

mathrmNH_3

are Lewis acids and

mathrmCH_4

is a Lewis base E. A solution of

mathrmNH_3

in

mathrmH_2mathrmO

is basic. In this solution

mathrmNH_3

and

mathrmH_2mathrmO

act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below:

A. A, B and C only
B. C, D and E only
C. A, D and E only
D. A, B, C and E only

Solution

### Core Logic Analyzing each statement individually: - **Statement A is true**: The central atoms (

O, N, C

) all possess an electron steric number equal to 4, indicating

mathfraksp^3

hybridization state pathways. - **Statement B is true**: Due to valence shell electron pair repulsions, the bond angles decrease from the ideal tetrahedral angle (

109.5^circ

in

CH_4

,

Water molecule structural bond configuration shape representation

) as lone pairs are added (

107.5^circ

in

NH_3

with 1 lone pair,

;

104.5^circ

in

H_2O

with 2 lone pairs,

). - **Statement C is true**: The dipole moment increases alongside central atom electronegativity and asymmetric lone pair configurations, following the sequence

mathrmCH_4 (0text D) < mathrmNH_3 (1.47text D) < mathrmH_2mathrmO (1.85text D)

. ### Pattern Recognition Lone pairs repel bonding electron pairs more strongly than bonding pairs repel each other, systematically compressing adjacent bond angles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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