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) $2 : 2$ $2 : 2$

Q89 jee_main_2024_27_jan_morning Molecular Orbital Theory

Sum of bond order of

textCO

and

textNO^+

is

textquadquad

.

Numerical Answer. Answer: 6 to 6

Solution

### Step 1: Determine the bond order of

textCO

Carbon monoxide (

textCO

) contains

6 + 8 = 14

total electrons. Its structural representation is

textCequivtextO

, matching a bond order value of 3. ### Step 2: Determine the bond order of

textNO^+

The nitrosonium ion (

textNO^+

) contains

7 + 8 - 1 = 14

total electrons. Since it is isoelectronic with

textN_2

and

textCO

(

14text electrons

), its corresponding bond order value is also 3. ### Step 3: Sum the results

textSum = 3 + 3 = 6

### Pattern Recognition Isoelectronic species possessing 14 total electrons consistently demonstrate a bond order value of 3. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q81 jee_main_2024_29_jan_morning VSEPR Theory

Number of compounds with one lone pair of electrons on central atom amongst following is

O_3

,

H_2O

,

SF_4

,

ClF_3

,

NH_3

,

BrF_5

,

XeF_4

Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Let us determine the steric number (

Z

) and number of lone pairs (

LP

) for the central atom in each given molecule. Formula:

Z = frac12 (V + M - C + A)

Where

V

= valence electrons on central atom,

M

= number of monovalent atoms,

C

= cationic charge,

A

= anionic charge.

LP = Z - textBond Pairs (B.P.)

1. **

O_3

**: Central atom O (

V=6

). It forms one double bond and one dative bond. It has 1 lone pair remaining. 2. **

H_2O

**: Central atom O (

V=6

).

Z = frac12(6 + 2) = 4

.

LP = 4 - 2 = 2

. 3. **

SF_4

**: Central atom S (

V=6

).

Z = frac12(6 + 4) = 5

.

LP = 5 - 4 = 1

(See-saw shape). 4. **

ClF_3

**: Central atom Cl (

V=7

).

Z = frac12(7 + 3) = 5

.

LP = 5 - 3 = 2

(T-shape). 5. **

NH_3

**: Central atom N (

V=5

).

Z = frac12(5 + 3) = 4

.

LP = 4 - 3 = 1

(Pyramidal). 6. **

BrF_5

**: Central atom Br (

V=7

).

Z = frac12(7 + 5) = 6

.

LP = 6 - 5 = 1

(Square Pyramidal). 7. **

XeF_4

**: Central atom Xe (

V=8

).

Z = frac12(8 + 4) = 6

.

LP = 6 - 4 = 2

(Square Planar). ### Step 1: Final Counting

VSEPR Theory diagram for Q81 - JEE Main 2024 Morning

The compounds containing exactly ONE lone pair on the central atom are

O_3

,

SF_4

,

NH_3

, and

BrF_5

. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q88 jee_main_2024_29_jan_morning Molecular Orbital Theory

The number of species from the following which are paramagnetic and with bond order equal to one is

mathrm H_2, mathrmHe_2^+, mathrmO_2^+, mathrmN_2^2-, mathrmO_2^2-, mathrmF_2, mathrmNe_2^+, mathrmB_2

Numerical Answer. Answer: 1 to 1

Solution

### Core Logic Using Molecular Orbital (MO) Theory, we evaluate the bond order (

BO = fracN_b - N_a2

) and magnetic nature (unpaired electrons = paramagnetic, all paired = diamagnetic) for each species:

Species	Magnetic behaviour	Bond order
$H_2$ $H_2$	Diamagnetic	1
$He_2^+$ $He_2^+$	Paramagnetic	0.5
$O_2^+$ $O_2^+$	Paramagnetic	2.5
$N_2^2-$ $N_2^{2-}$	Paramagnetic	2
$O_2^2-$ $O_2^{2-}$	Diamagnetic	1
$F_2$ $F_2$	Diamagnetic	1
$Ne_2^+$ $Ne_2^+$	Paramagnetic	0.5
$B_2$ $B_2$	Paramagnetic	1

### Step 1: Final Selection We need the species that satisfies BOTH conditions: 1. Paramagnetic 2. Bond Order = 1 Looking at the table,

B_2

is the only molecule that is paramagnetic (it has 2 unpaired electrons in degenerate

pi_2p

orbitals) and has a bond order of 1. Total number of such species = 1. ### Pattern Recognition

B_2

(10 electrons) and

O_2

(16 electrons) are the classic exceptions in MO theory that are paramagnetic despite having an even number of electrons.

B_2

has BO = 1, and

O_2

has BO = 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q69 jee_main_2024_30_january_evening VSEPR Theory and Molecular Shapes

The molecule/ion with square pyramidal shape is:

A. $text[Ni(CN)_4]^2-$ $\text{[Ni(CN)}_4]^{2-}$
B. $textPCl_5$ $\text{PCl}_5$
C. $textBrF_5$ $\text{BrF}_5$
D. $textPF_5$ $\text{PF}_5$

Solution

### Core Logic According to VSEPR theory: 1.

[Ni(CN)_4]^2-

:

dsp^2

hybridization

rightarrow

Square Planar. 2.

PCl_5

:

sp^3d

hybridization with 0 lone pairs

rightarrow

Trigonal Bipyramidal. 3.

BrF_5

: Bromine has 7 valence electrons. It forms 5 single bonds with Fluorine, leaving 1 lone pair.

sp^3d^2

hybridization

rightarrow

geometry is octahedral, but shape is Square Pyramidal. 4.

PF_5

:

sp^3d

hybridization with 0 lone pairs

rightarrow

Trigonal Bipyramidal.

Square Pyramidal structure of BrF5 diagram for Q69 - JEE Main 2024 Evening

### Pattern Recognition

AX_5E_1

configuration (5 bond pairs + 1 lone pair) typically yields a Square Pyramidal shape. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds

Q75 jee_main_2024_30_january_evening Dipole Moment

Given below are two statements: Statement-I: Since fluorine is more electronegative than nitrogen, the net dipole moment of

NF_3

is greater than

NH_3

. Statement-II: In

NH_3

, the orbital dipole due to lone pair and the dipole moment of NH bonds are in opposite direction, but in

NF_3

the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction. In the light of the above statements. Choose the most appropriate from the options given below.

A. $textStatement I is true but Statement II is false.$
B. $textBoth Statement I and Statement II are false.$
C. $textBoth statement I and Statement II is are true.$
D. $textStatement I is false but Statement II is are true.$

Solution

### Core Logic Statement I: The net dipole moment of

NH_3

(

1.47\, D

) is actually greater than that of

NF_3

(

0.23\, D

). Therefore, Statement I is false. Statement II: In

NH_3

, the

N-H

bond dipole moments (pointing towards the more electronegative N) reinforce the orbital dipole moment of the lone pair. In

NF_3

, the

N-F

bond dipole moments point away from N (towards the more electronegative F), opposing the orbital dipole moment of the lone pair. This partial cancellation in

NF_3

makes its net dipole moment lower. Therefore, Statement II is also false, as it reverses the correct orientations.

Dipole moments in NH3 and NF3 diagram for Q75 - JEE Main 2024 Evening

### Step 1: Final Conclusion Since both statements assert the opposite of established facts regarding

NH_3

and

NF_3

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

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