Chemical Bonding & Molecular Structure: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (2021-2026): 
Chemical Bonding & Molecular Structure 
 
 
Q1: Match List - I with List - II according to shape. 
 
Choose the correct answer from the options given below : 
A: A-II, B-III, C-I, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-III, C-IV, D-I 
D: A-II, B-I, C-III, D-IV 
Answer: C 
Explanation: 
XeF2 and I3
?
 have 2 bond pairs and 3 lone pairs on the central atom (total 5 electron 
pairs). In this case, the molecular shape becomes linear. 
XeOF4 and BrF5 have 5 bond pairs and 1 lone pair on the central atom (total 6 electron 
pairs). This gives a square pyramidal shape. 
XeO2F2 and SF4 have 4 bond pairs and 1 lone pair on the central atom (total 5 electron 
pairs). This leads to a see-saw shape. 
XeO3 and NH3 have 3 bond pairs and 1 lone pair on the central atom (total 4 electron 
pairs). So, the shape is pyramidal. 
 
Q2: Given below are two statements : 
Statement I : The number of species among BF4?, SiF4, XeF4 and SF4, that have unequal 
E - F bond lengths is two. Here, E is the central atom. 
Statement II : Among O2?, O2²?, F2 and O2?, O2? has the highest bond order. 
In the light of the above statements, choose the correct answer from the options given 
below : 
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JEE Main Previous Year Questions (2021-2026): 
Chemical Bonding & Molecular Structure 
 
 
Q1: Match List - I with List - II according to shape. 
 
Choose the correct answer from the options given below : 
A: A-II, B-III, C-I, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-III, C-IV, D-I 
D: A-II, B-I, C-III, D-IV 
Answer: C 
Explanation: 
XeF2 and I3
?
 have 2 bond pairs and 3 lone pairs on the central atom (total 5 electron 
pairs). In this case, the molecular shape becomes linear. 
XeOF4 and BrF5 have 5 bond pairs and 1 lone pair on the central atom (total 6 electron 
pairs). This gives a square pyramidal shape. 
XeO2F2 and SF4 have 4 bond pairs and 1 lone pair on the central atom (total 5 electron 
pairs). This leads to a see-saw shape. 
XeO3 and NH3 have 3 bond pairs and 1 lone pair on the central atom (total 4 electron 
pairs). So, the shape is pyramidal. 
 
Q2: Given below are two statements : 
Statement I : The number of species among BF4?, SiF4, XeF4 and SF4, that have unequal 
E - F bond lengths is two. Here, E is the central atom. 
Statement II : Among O2?, O2²?, F2 and O2?, O2? has the highest bond order. 
In the light of the above statements, choose the correct answer from the options given 
below : 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: A 
Explanation: 
Statement I 
BF4? 
Central atom B has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four B - F bonds are equivalent ? equal bond lengths. 
SiF4 
Central atom Si has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four Si - F bonds equivalent ? equal bond lengths. 
XeF4 
Xe has 4 bond pairs and 2 lone pairs (AX4E2) ? square planar. 
All four Xe - F bonds equivalent by symmetry ? equal bond lengths. 
SF4 
S has 4 bond pairs and 1 lone pair (AX4E) ? see-saw shape. 
There are two axial and two equatorial S - F bonds, and they are not equivalent (axial 
bonds are longer). 
? unequal bond lengths. 
So, number of species with unequal E - F bond lengths = 1 (only SF4), not 2. 
? Statement I is false. 
Statement II 
Bond order (NCERT MO theory): 
 
Using known MO results for these diatomic species: 
 
? Statement II is false. 
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JEE Main Previous Year Questions (2021-2026): 
Chemical Bonding & Molecular Structure 
 
 
Q1: Match List - I with List - II according to shape. 
 
Choose the correct answer from the options given below : 
A: A-II, B-III, C-I, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-III, C-IV, D-I 
D: A-II, B-I, C-III, D-IV 
Answer: C 
Explanation: 
XeF2 and I3
?
 have 2 bond pairs and 3 lone pairs on the central atom (total 5 electron 
pairs). In this case, the molecular shape becomes linear. 
XeOF4 and BrF5 have 5 bond pairs and 1 lone pair on the central atom (total 6 electron 
pairs). This gives a square pyramidal shape. 
XeO2F2 and SF4 have 4 bond pairs and 1 lone pair on the central atom (total 5 electron 
pairs). This leads to a see-saw shape. 
XeO3 and NH3 have 3 bond pairs and 1 lone pair on the central atom (total 4 electron 
pairs). So, the shape is pyramidal. 
 
Q2: Given below are two statements : 
Statement I : The number of species among BF4?, SiF4, XeF4 and SF4, that have unequal 
E - F bond lengths is two. Here, E is the central atom. 
Statement II : Among O2?, O2²?, F2 and O2?, O2? has the highest bond order. 
In the light of the above statements, choose the correct answer from the options given 
below : 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: A 
Explanation: 
Statement I 
BF4? 
Central atom B has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four B - F bonds are equivalent ? equal bond lengths. 
SiF4 
Central atom Si has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four Si - F bonds equivalent ? equal bond lengths. 
XeF4 
Xe has 4 bond pairs and 2 lone pairs (AX4E2) ? square planar. 
All four Xe - F bonds equivalent by symmetry ? equal bond lengths. 
SF4 
S has 4 bond pairs and 1 lone pair (AX4E) ? see-saw shape. 
There are two axial and two equatorial S - F bonds, and they are not equivalent (axial 
bonds are longer). 
? unequal bond lengths. 
So, number of species with unequal E - F bond lengths = 1 (only SF4), not 2. 
? Statement I is false. 
Statement II 
Bond order (NCERT MO theory): 
 
Using known MO results for these diatomic species: 
 
? Statement II is false. 
Correct Option 
Both Statement I and Statement II are false. 
Option A 
 
Q3: Pair of species among the following having same bond order as well as 
paramagnetic character will be- 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
Q4: Given below are statements about some molecules/ions. 
Identify the CORRECT statements. 
A. The dipole moment value of NF3 is higher than that of NH3. 
B. The dipole moment value of BeH2 is zero. 
C. The bond order of O2²? and F2 is same. 
D. The formal charge on the central oxygen atom of ozone is -1. 
E. In NO2, all the three atoms satisfy the octet rule, hence it is very stable. 
Choose the correct answer from the options given below: 
A: B & C Only 
B: B, C & D Only 
C: A, C & D Only 
D: A, B, C, D & E 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Chemical Bonding & Molecular Structure 
 
 
Q1: Match List - I with List - II according to shape. 
 
Choose the correct answer from the options given below : 
A: A-II, B-III, C-I, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-III, C-IV, D-I 
D: A-II, B-I, C-III, D-IV 
Answer: C 
Explanation: 
XeF2 and I3
?
 have 2 bond pairs and 3 lone pairs on the central atom (total 5 electron 
pairs). In this case, the molecular shape becomes linear. 
XeOF4 and BrF5 have 5 bond pairs and 1 lone pair on the central atom (total 6 electron 
pairs). This gives a square pyramidal shape. 
XeO2F2 and SF4 have 4 bond pairs and 1 lone pair on the central atom (total 5 electron 
pairs). This leads to a see-saw shape. 
XeO3 and NH3 have 3 bond pairs and 1 lone pair on the central atom (total 4 electron 
pairs). So, the shape is pyramidal. 
 
Q2: Given below are two statements : 
Statement I : The number of species among BF4?, SiF4, XeF4 and SF4, that have unequal 
E - F bond lengths is two. Here, E is the central atom. 
Statement II : Among O2?, O2²?, F2 and O2?, O2? has the highest bond order. 
In the light of the above statements, choose the correct answer from the options given 
below : 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: A 
Explanation: 
Statement I 
BF4? 
Central atom B has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four B - F bonds are equivalent ? equal bond lengths. 
SiF4 
Central atom Si has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four Si - F bonds equivalent ? equal bond lengths. 
XeF4 
Xe has 4 bond pairs and 2 lone pairs (AX4E2) ? square planar. 
All four Xe - F bonds equivalent by symmetry ? equal bond lengths. 
SF4 
S has 4 bond pairs and 1 lone pair (AX4E) ? see-saw shape. 
There are two axial and two equatorial S - F bonds, and they are not equivalent (axial 
bonds are longer). 
? unequal bond lengths. 
So, number of species with unequal E - F bond lengths = 1 (only SF4), not 2. 
? Statement I is false. 
Statement II 
Bond order (NCERT MO theory): 
 
Using known MO results for these diatomic species: 
 
? Statement II is false. 
Correct Option 
Both Statement I and Statement II are false. 
Option A 
 
Q3: Pair of species among the following having same bond order as well as 
paramagnetic character will be- 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
Q4: Given below are statements about some molecules/ions. 
Identify the CORRECT statements. 
A. The dipole moment value of NF3 is higher than that of NH3. 
B. The dipole moment value of BeH2 is zero. 
C. The bond order of O2²? and F2 is same. 
D. The formal charge on the central oxygen atom of ozone is -1. 
E. In NO2, all the three atoms satisfy the octet rule, hence it is very stable. 
Choose the correct answer from the options given below: 
A: B & C Only 
B: B, C & D Only 
C: A, C & D Only 
D: A, B, C, D & E 
Answer: A 
Explanation: 
Check each statement: 
 
Both are trigonal pyramidal. In NH3, the bond dipoles and lone pair dipole add to give a 
larger net dipole moment. In NF3, the N - F bond dipoles oppose the lone pair direction, 
so net dipole is small. 
So, µ(NF3) < µ(NH3). 
A is incorrect. 
B. Dipole moment of BeH2 is zero 
BeH2 is linear: H - Be - H (NCERT: sp hybridised). The two Be - H bond dipoles are 
equal and opposite, hence cancel. 
So, µ = 0. 
B is correct. 
C. Bond order of O2²? and F2 is same. 
Using MO theory: 
O2²? has 14 valence electrons (like F2). For peroxide ion, bond order becomes 
B.O. = 1 
For F2, bond order is also 
B.O. = 1 
So they are the same. 
C is correct. 
D. Formal charge on central oxygen of ozone is -1 
In one resonance form: O = O - O (central O has one lone pair, one double bond, one 
single bond) 
Formal charge on central O: 
F.C. = 6 - (2 + 6/2) = 6 - (2 + 3) = +1 
So it is +1, not -1. 
D is incorrect. 
E. In NO2, all atoms satisfy octet rule, hence very stable 
NO
2
 has an odd number of electrons (17 valence electrons), so it is an odd-electron 
molecule and nitrogen does not complete its octet. Hence it is quite reactive (dimerises 
to (N
2
O
4
).  
E is incorrect. 
Correct statements: B and C only 
 
Q5: Among the following, the CORRECT combinations are 
A. IF3 ? T -shaped (sp³d) 
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JEE Main Previous Year Questions (2021-2026): 
Chemical Bonding & Molecular Structure 
 
 
Q1: Match List - I with List - II according to shape. 
 
Choose the correct answer from the options given below : 
A: A-II, B-III, C-I, D-IV 
B: A-III, B-II, C-IV, D-I 
C: A-II, B-III, C-IV, D-I 
D: A-II, B-I, C-III, D-IV 
Answer: C 
Explanation: 
XeF2 and I3
?
 have 2 bond pairs and 3 lone pairs on the central atom (total 5 electron 
pairs). In this case, the molecular shape becomes linear. 
XeOF4 and BrF5 have 5 bond pairs and 1 lone pair on the central atom (total 6 electron 
pairs). This gives a square pyramidal shape. 
XeO2F2 and SF4 have 4 bond pairs and 1 lone pair on the central atom (total 5 electron 
pairs). This leads to a see-saw shape. 
XeO3 and NH3 have 3 bond pairs and 1 lone pair on the central atom (total 4 electron 
pairs). So, the shape is pyramidal. 
 
Q2: Given below are two statements : 
Statement I : The number of species among BF4?, SiF4, XeF4 and SF4, that have unequal 
E - F bond lengths is two. Here, E is the central atom. 
Statement II : Among O2?, O2²?, F2 and O2?, O2? has the highest bond order. 
In the light of the above statements, choose the correct answer from the options given 
below : 
A: Both Statement I and Statement II are false 
B: Statement I is true but Statement II is false 
C: Statement I is false but Statement II is true 
D: Both Statement I and Statement II are true 
Answer: A 
Explanation: 
Statement I 
BF4? 
Central atom B has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four B - F bonds are equivalent ? equal bond lengths. 
SiF4 
Central atom Si has 4 bond pairs and 0 lone pairs ? tetrahedral. 
All four Si - F bonds equivalent ? equal bond lengths. 
XeF4 
Xe has 4 bond pairs and 2 lone pairs (AX4E2) ? square planar. 
All four Xe - F bonds equivalent by symmetry ? equal bond lengths. 
SF4 
S has 4 bond pairs and 1 lone pair (AX4E) ? see-saw shape. 
There are two axial and two equatorial S - F bonds, and they are not equivalent (axial 
bonds are longer). 
? unequal bond lengths. 
So, number of species with unequal E - F bond lengths = 1 (only SF4), not 2. 
? Statement I is false. 
Statement II 
Bond order (NCERT MO theory): 
 
Using known MO results for these diatomic species: 
 
? Statement II is false. 
Correct Option 
Both Statement I and Statement II are false. 
Option A 
 
Q3: Pair of species among the following having same bond order as well as 
paramagnetic character will be- 
A:  
B:  
C:  
D:  
Answer: D 
Explanation: 
 
 
Q4: Given below are statements about some molecules/ions. 
Identify the CORRECT statements. 
A. The dipole moment value of NF3 is higher than that of NH3. 
B. The dipole moment value of BeH2 is zero. 
C. The bond order of O2²? and F2 is same. 
D. The formal charge on the central oxygen atom of ozone is -1. 
E. In NO2, all the three atoms satisfy the octet rule, hence it is very stable. 
Choose the correct answer from the options given below: 
A: B & C Only 
B: B, C & D Only 
C: A, C & D Only 
D: A, B, C, D & E 
Answer: A 
Explanation: 
Check each statement: 
 
Both are trigonal pyramidal. In NH3, the bond dipoles and lone pair dipole add to give a 
larger net dipole moment. In NF3, the N - F bond dipoles oppose the lone pair direction, 
so net dipole is small. 
So, µ(NF3) < µ(NH3). 
A is incorrect. 
B. Dipole moment of BeH2 is zero 
BeH2 is linear: H - Be - H (NCERT: sp hybridised). The two Be - H bond dipoles are 
equal and opposite, hence cancel. 
So, µ = 0. 
B is correct. 
C. Bond order of O2²? and F2 is same. 
Using MO theory: 
O2²? has 14 valence electrons (like F2). For peroxide ion, bond order becomes 
B.O. = 1 
For F2, bond order is also 
B.O. = 1 
So they are the same. 
C is correct. 
D. Formal charge on central oxygen of ozone is -1 
In one resonance form: O = O - O (central O has one lone pair, one double bond, one 
single bond) 
Formal charge on central O: 
F.C. = 6 - (2 + 6/2) = 6 - (2 + 3) = +1 
So it is +1, not -1. 
D is incorrect. 
E. In NO2, all atoms satisfy octet rule, hence very stable 
NO
2
 has an odd number of electrons (17 valence electrons), so it is an odd-electron 
molecule and nitrogen does not complete its octet. Hence it is quite reactive (dimerises 
to (N
2
O
4
).  
E is incorrect. 
Correct statements: B and C only 
 
Q5: Among the following, the CORRECT combinations are 
A. IF3 ? T -shaped (sp³d) 
B. IF5 ? Square pyramidal (sp³d²) 
C. IF7 ? Pentagonal bipyramidal (sp³d³) 
D. ClO4? ? Square planar (sp²d) 
Choose the correct answer from the options given below: 
A: B, C and D Only 
B: A, B, C and D 
C: A, B and C Only 
D: A and B Only 
Answer: C 
Explanation:  
 
 
Q6: Which statements are NOT TRUE about XeO2F2? 
A. It has a see-saw shape. 
B. Xe has 5 electron pairs in its valence shell in XeO2F2. 
C. The O — Xe — O bond angle is close to 180°. 
D. The F — Xe — F bond angle is close to 180°. 
E. Xe has 16 valence electrons in XeO2F2. 
Choose the correct answer from the options given below: 
A: B, D and E Only 
B: B and D Only 
C: B, C and E Only