JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31

Download JEE MCQs for JEE Chemistry: Chemical Bonding and Molecular Structure

Access targeted multiple-choice questions for Chemical Bonding and Molecular Structure designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Chemical Bonding and Molecular Structure

View or download the dedicated Chemical Bonding and Molecular Structure MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: Which of the following molecular structures is NOT possible?
(a) \( \ce{OF2} \)
(b) \( \ce{SF2} \)
(c) \( \ce{SF4} \)
(d) \( \ce{OF4} \)
Show Answer & Explanation

Answer: (d) \( \ce{OF4} \)

Explanation:
1. Oxygen is in period 2 and has no d orbitals, so it cannot expand its octet to form \( \ce{OF4} \)
2. Sulphur can (period 3), so \( \ce{SF4} \) is possible
3. \( \ce{OF2} \) and \( \ce{SF2} \) obey the octet rule.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 2: Which of the molecules has the highest dipole moment?
(a) \( \ce{NH3} \)
(b) \( \ce{CO2} \)
(c) \( \ce{CH4} \)
(d) \( \ce{NF3} \)
Show Answer & Explanation

Answer: (a) \( \ce{NH3} \)

Explanation:
1. \( \ce{CO2} \) and \( \ce{CH4} \) are symmetric with zero dipole
2. In \( \ce{NH3} \) the N-H bond dipoles and the lone pair point the same way
3. In \( \ce{NF3} \) the lone pair opposes the N-F dipoles, so its moment is small
4. So \( \ce{NH3} \) has the highest.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 3: The correct order of stability of the species \( \ce{Li2} \), \( \ce{He2^+} \), \( \ce{O2^+} \) and \( \ce{C2} \) is
(a) \( \ce{Li2} \) < \( \ce{He2^+} \) < \( \ce{O2^+} \) < \( \ce{C2} \)
(b) \( \ce{C2} \) < \( \ce{O2^+} \) < \( \ce{Li2} \) < \( \ce{He2^+} \)
(c) \( \ce{He2^+} \) < \( \ce{Li2} \) < \( \ce{C2} \) < \( \ce{O2^+} \)
(d) \( \ce{O2^+} \) < \( \ce{C2} \) < \( \ce{Li2} \) < \( \ce{He2^+} \)
Show Answer & Explanation

Answer: (c) \( \ce{He2^+} \) < \( \ce{Li2} \) < \( \ce{C2} \) < \( \ce{O2^+} \)

Explanation:
1. Bond orders: \( \ce{He2^+} \) 0.5, \( \ce{Li2} \) 1, \( \ce{C2} \) 2, \( \ce{O2^+} \) 2.5
2. Stability increases with bond order, so He2+ < Li2 < C2 < O2+.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 4: The hybridisation of boron in the stable borane of lowest molecular weight is
(a) sp2
(b) sp
(c) sp3
(d) sp3d
Show Answer & Explanation

Answer: (c) sp3

Explanation:
1. The simplest stable borane is diborane \( \ce{B2H6} \)
2. Each boron forms four bonds: two terminal B-H and two bridging B-H-B. So boron is sp3 hybridised.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 5: For \( \ce{SF4} \), the molecular geometry and hybridisation of the central atom are respectively
(a) see-saw, sp3d
(b) square planar, sp3d
(c) tetrahedral, sp3
(d) square pyramidal, sp3d
Show Answer & Explanation

Answer: (a) see-saw, sp3d

Explanation:
1. \( \ce{SF4} \) has 4 bond pairs and 1 lone pair, so 5 electron pairs and sp3d hybridisation
2. With the lone pair equatorial, the shape is see-saw.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 6: Which of the following species has the highest bond energy?
(a) \( \ce{O2} \)
(b) \( \ce{O2^+} \)
(c) \( \ce{O2^-} \)
(d) \( \ce{O2^2-} \)
Show Answer & Explanation

Answer: (b) \( \ce{O2^+} \)

Explanation:
1. Bond energy increases with bond order: \( \ce{O2^+} \) 2.5, \( \ce{O2} \) 2, \( \ce{O2^-} \) 1.5, \( \ce{O2^2-} \) 1
2. So \( \ce{O2^+} \) has the highest bond energy.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 7: The structure of the molecule \( \ce{N(SiMe3)3} \) (with Me = \( \ce{CH3} \)) is
(a) pyramidal with angle close to 110 degrees
(b) trigonal planar with a bond angle close to 120 degrees
(c) T-shaped with angle 90 degrees
(d) bent T-shaped with angle close to 89 degrees
Show Answer & Explanation

Answer: (b) trigonal planar with a bond angle close to 120 degrees

Explanation:
1. The lone pair on nitrogen is delocalised into the empty d orbitals of silicon (p-pi to d-pi bonding)
2. The nitrogen becomes sp2-like, so the molecule is planar with about 120 degrees, unlike \( \ce{NMe3} \).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 8: The extent of p-pi to d-pi interaction in bonds from Si, P, S and Cl to oxygen increases in the order
(a) P > Si > Cl > S
(b) S < Cl < P < Si
(c) Si > P > S > Cl
(d) Si < P < S < Cl
Show Answer & Explanation

Answer: (d) Si < P < S < Cl

Explanation:
1. The d orbitals contract and become more effective for overlap as the effective nuclear charge rises across the period
2. So the p-pi to d-pi interaction increases from Si to P to S to Cl.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 9: The pair that is isostructural (same shape and hybridisation) is
(a) \( \ce{BF4^-} \) and \( \ce{NH4^+} \)
(b) \( \ce{NF3} \) and \( \ce{BF3} \)
(c) \( \ce{BCl3} \) and \( \ce{BrCl3} \)
(d) \( \ce{NH3} \) and \( \ce{NO3^-} \)
Show Answer & Explanation

Answer: (a) \( \ce{BF4^-} \) and \( \ce{NH4^+} \)

Explanation:
1. \( \ce{BF4^-} \) and \( \ce{NH4^+} \) both have four bond pairs and no lone pair, so both are sp3 and tetrahedral
2. \( \ce{NF3} \) is pyramidal and \( \ce{BF3} \) planar
3. \( \ce{BCl3} \) is planar and \( \ce{BrCl3} \) T-shaped
4. \( \ce{NH3} \) is pyramidal and \( \ce{NO3^-} \) planar.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 10: The number of P-S single bonds and P=S double bonds in \( \ce{P4S10} \) are respectively
(a) 10 and 6
(b) 16 and 0
(c) 12 and 4
(d) 14 and 2
Show Answer & Explanation

Answer: (c) 12 and 4

Explanation:
1. \( \ce{P4S10} \) has the adamantane-like \( \ce{P4} \) cage structure
2. Six bridging S atoms give 12 P-S single bonds, and four terminal S atoms give 4 P=S bonds
3. Total sulphur = 6 + 4 = 10.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 11: In the Lewis structure of ozone, the formal charge on the central oxygen atom is
(a) +1
(b) -1
(c) 0
(d) -2
Show Answer & Explanation

Answer: (a) +1

Explanation:
1. The central oxygen has one double bond, one single bond and one lone pair: 3 bonds
2. Formal charge = 6 - 2 - 3 = +1.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 12: Which of the following represents the correct order of dipole moments?
(a) \( \ce{NH3} \) > \( \ce{H2O} \) > \( \ce{NF3} \)
(b) \( \ce{NH3} \) > \( \ce{NF3} \) > \( \ce{H2O} \)
(c) \( \ce{H2O} \) > \( \ce{NF3} \) > \( \ce{NH3} \)
(d) \( \ce{H2O} \) > \( \ce{NH3} \) > \( \ce{NF3} \)
Show Answer & Explanation

Answer: (d) \( \ce{H2O} \) > \( \ce{NH3} \) > \( \ce{NF3} \)

Explanation:
1. The observed dipole moments are \( \ce{H2O} \) about 1.85 D, \( \ce{NH3} \) about 1.47 D and \( \ce{NF3} \) about 0.24 D. In \( \ce{NF3} \) the lone pair moment opposes the N-F bond moments
2. Hence H2O > NH3 > NF3.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Question 13: Which of the following species is nonlinear?
(a) \( \ce{N3^-} \)
(b) \( \ce{BrCl2^+} \)
(c) \( \ce{ClF2^-} \)
(d) \( \ce{Br3^-} \)
Show Answer & Explanation

Answer: (b) \( \ce{BrCl2^+} \)

Explanation:
1. \( \ce{N3^-} \) has 16 valence electrons and is linear (sp)
2. \( \ce{ClF2^-} \) and \( \ce{Br3^-} \) have 3 lone pairs on the central halogen, giving a linear shape
3. \( \ce{BrCl2^+} \) has 2 lone pairs and 2 bond pairs on Br, so it is bent.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 14: Which of the following is an example of a super-octet (hypervalent) molecule?
(a) \( \ce{SF6} \) only
(b) all of \( \ce{SF6} \), \( \ce{PCl5} \) and \( \ce{IF7} \)
(c) \( \ce{PCl5} \) only
(d) \( \ce{IF7} \) only
Show Answer & Explanation

Answer: (b) all of \( \ce{SF6} \), \( \ce{PCl5} \) and \( \ce{IF7} \)

Explanation:
1. In \( \ce{SF6} \) sulphur has 12 electrons, in \( \ce{PCl5} \) phosphorus has 10 and in \( \ce{IF7} \) iodine has 14
2. Each exceeds the octet, so all three are hypervalent.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 15: The correct order of C-O bond energy in \( \ce{CO3^2-} \), \( \ce{CO2} \) and \( \ce{CO} \) is
(a) \( \ce{CO3^2-} \) > \( \ce{CO2} \) > \( \ce{CO} \)
(b) \( \ce{CO2} \) > \( \ce{CO} \) > \( \ce{CO3^2-} \)
(c) \( \ce{CO2} \) > \( \ce{CO3^2-} \) > \( \ce{CO} \)
(d) \( \ce{CO} \) > \( \ce{CO2} \) > \( \ce{CO3^2-} \)
Show Answer & Explanation

Answer: (d) \( \ce{CO} \) > \( \ce{CO2} \) > \( \ce{CO3^2-} \)

Explanation:
1. Bond energy rises with bond order
2. The C-O bond order is 3 in \( \ce{CO} \), 2 in \( \ce{CO2} \) and 1.33 in \( \ce{CO3^2-} \)
3. So the order is CO > CO2 > CO3 2-.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Question 16: The structure of \( \ce{F2SeO} \) is analogous to that of
(a) \( \ce{SO3} \) only
(b) \( \ce{ClO3^-} \) only
(c) both \( \ce{ClO3^-} \) and \( \ce{XeO3} \)
(d) \( \ce{XeO3} \) only
Show Answer & Explanation

Answer: (c) both \( \ce{ClO3^-} \) and \( \ce{XeO3} \)

Explanation:
1. \( \ce{F2SeO} \) has 3 sigma bonds and one lone pair on selenium, giving a pyramidal shape
2. \( \ce{ClO3^-} \) and \( \ce{XeO3} \) are also pyramidal with 3 bond pairs and 1 lone pair, while \( \ce{SO3} \) is planar with no lone pair.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 17: In which of the following is the nitrogen sp2 hybridised?
(a) \( \ce{(CH3)3N} \)
(b) \( \ce{CH3CN} \)
(c) \( \ce{CH2=NH} \)
(d) \( \ce{NO2^+} \)
Show Answer & Explanation

Answer: (c) \( \ce{CH2=NH} \)

Explanation:
1. In \( \ce{CH2=NH} \) nitrogen has 2 sigma bonds (one to C and one to H) and a lone pair, so sp2. \( \ce{(CH3)3N} \) is sp3. \( \ce{CH3CN} \) and \( \ce{NO2^+} \) have nitrogen with 2 sigma bonds and no lone pair, so sp.
2. Only option (c) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 18: In the \( \ce{H2SO4} \) molecule
(a) S is sp3 hybridised and there are 8 lone pairs on the oxygen atoms
(b) S is sp3d2 hybridised and there are 4 lone pairs on the oxygen atoms
(c) S is sp3 hybridised and there is no lone pair in the molecule
(d) S is sp2 hybridised and there are 8 lone pairs on the oxygen atoms
Show Answer & Explanation

Answer: (a) S is sp3 hybridised and there are 8 lone pairs on the oxygen atoms

Explanation:
1. Sulphur has 4 sigma bonds (two S=O and two S-OH) and no lone pair, so sp3. Counting lone pairs: each of the two terminal O atoms has 2 and each of the two OH oxygens has 2, giving 8 in all.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 19: In which of the following pairs is the hybridisation of the central atoms different?
(a) \( \ce{ClF3} \) and \( \ce{ClF3O} \)
(b) \( \ce{ClF3O} \) and \( \ce{ClF3O2} \)
(c) \( \ce{[ClF4O]^-} \) and \( \ce{XeOF4} \)
(d) \( \ce{[ClF2O]^+} \) and \( \ce{[ClF4O]^-} \)
Show Answer & Explanation

Answer: (d) \( \ce{[ClF2O]^+} \) and \( \ce{[ClF4O]^-} \)

Explanation:
1. Count sigma bonds plus lone pairs
2. \( \ce{ClF3} \) has 3+2 = 5 and \( \ce{ClF3O} \) 4+1 = 5 (sp3d)
3. \( \ce{ClF3O2} \) has 5+0 = 5 (sp3d)
4. \( \ce{[ClF2O]^+} \) has 3+1 = 4 (sp3) while \( \ce{[ClF4O]^-} \) has 5+1 = 6 (sp3d2)
5. \( \ce{XeOF4} \) has 5+1 = 6, same as \( \ce{[ClF4O]^-} \).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 20: Which of the following has the highest bond angle?
(a) \( \ce{NH3} \)
(b) \( \ce{CH4} \)
(c) \( \ce{PH3} \)
(d) \( \ce{H2O} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH4} \)

Explanation:
1. The angles are \( \ce{CH4} \) 109.5 degrees, \( \ce{NH3} \) about 107, \( \ce{H2O} \) about 104.5 and \( \ce{PH3} \) about 93.5
2. So \( \ce{CH4} \) has the highest.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 21: Phosphorus pentachloride in the solid state exists as \( \ce{PCl4^+} \) and \( \ce{PCl6^-} \). The hybridisation of P in these two ions is respectively
(a) sp3d, sp3d2
(b) sp3d2, sp3d
(c) sp3, sp3d
(d) sp3, sp3d2
Show Answer & Explanation

Answer: (d) sp3, sp3d2

Explanation:
1. \( \ce{PCl4^+} \) has 4 bond pairs and no lone pair (sp3, tetrahedral)
2. \( \ce{PCl6^-} \) has 6 bond pairs (sp3d2, octahedral).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 22: Arrange \( \ce{P4} \), \( \ce{PH3} \) and \( \ce{H2O} \) in increasing order of deviation from the normal tetrahedral angle.
(a) \( \ce{H2O} \) < \( \ce{PH3} \) < \( \ce{P4} \)
(b) \( \ce{P4} \) < \( \ce{PH3} \) < \( \ce{H2O} \)
(c) \( \ce{PH3} \) < \( \ce{H2O} \) < \( \ce{P4} \)
(d) \( \ce{P4} \) < \( \ce{H2O} \) < \( \ce{PH3} \)
Show Answer & Explanation

Answer: (a) \( \ce{H2O} \) < \( \ce{PH3} \) < \( \ce{P4} \)

Explanation:
1. The tetrahedral angle is 109.5 degrees
2. \( \ce{H2O} \) (104.5) deviates by 5, \( \ce{PH3} \) (93.5) by 16 and \( \ce{P4} \) (60) by 49.5 degrees
3. So the order is H2O < PH3 < P4.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 23: Which of the following species contains banana (3-centre 2-electron) bonds?
(a) \( \ce{BF3} \)
(b) \( \ce{(AlCl3)2} \)
(c) \( \ce{(BeH2)_n} \)
(d) \( \ce{(BeCl2)_n} \)
Show Answer & Explanation

Answer: (c) \( \ce{(BeH2)_n} \)

Explanation:
1. In polymeric \( \ce{BeH2} \) the hydrogens bridge two Be atoms using only 2 electrons for 3 centres, which is a banana bond
2. In \( \ce{Al2Cl6} \) and \( \ce{(BeCl2)_n} \) the chlorine bridges use lone pairs (3-centre 4-electron)
3. \( \ce{BF3} \) has no bridge.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Question 24: What is the nature of the bond between B and O in \( \ce{(C2H5)2O.BH3} \)?
(a) ionic
(b) purely covalent (normal)
(c) coordinate covalent
(d) banana-shaped
Show Answer & Explanation

Answer: (c) coordinate covalent

Explanation:
1. The oxygen of the ether donates a lone pair to the empty orbital of boron in \( \ce{BH3} \)
2. This is a coordinate (dative) covalent bond.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Free study material for Chemistry

Chemical Bonding and Molecular Structure Objective Questions & Solutions for JEE Chemistry

Download Multiple Choice Questions: Chemical Bonding and Molecular Structure (JEE Chemistry)

Explore reliable practice questions for Chemical Bonding and Molecular Structure tailored for JEE Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Concept Clarification for Chemical Bonding and Molecular Structure

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Additional Study Resources for JEE Chemistry

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31?

You can get most exhaustive JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 31 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.