/ Preparation Chemistry NCERT Exemplar Solutions Class 11th Chapter Four
Chemistry NCERT Exemplar Solutions Class 11th Chapter Four Chemical Bonding and Molecular Structure offers comprehensive practice material for how atoms combine to form molecules. The chapter is crucial in chemistry and it explains the types, nature, and geometry of chemical bonds. The solutions are given in a step-by-step format that helps students solve the numerical, theoretical, and conceptual problems with ease.
Students can also download the Chemical Bonding NCERT PDF for practicing the solutions of MCQs, Shorty type questions, Long Type Questions, Matching Type and Assertion and Reason questions. The PDF is prepared by Shiksha's subject matter experts. The students can rely on these solutions, and practice them offline or they can also take a printout and practice on paper.
The students should also read the Class 11th Chapter Four Chemical Bonding and Molecular Structure NCERT Solutions.
- Download PDF of NCERT Exemplar Class 11 Chemistry Chapter Four Chemical Bonding and Molecular Structure
- Chemical Bonding and Molecular Structure Class 11 Short Answer Type Questions
- Chemistry Class 11 Chemical Bonding and Molecular Structure Long Answer Type Questions
- Class 11 Chemical Bonding and Molecular Structure Matching Type Questions
- NCERT Exemplar Chemical Bonding and Molecular Structure Class 11 Assertion and Reason Type
- Chemical Bonding and Molecular Structure NCERT Exemplar Class 11 Objective Type Questions
- Common Mistakes and Tips for NCERT Chemistry Exemplar Chapter Four
- JEE Mains 2020
- 01 Oct 2025
Download PDF of NCERT Exemplar Class 11 Chemistry Chapter Four Chemical Bonding and Molecular Structure
The students can download the Chemical Bonding PDF NCERT from this page. There are numerous benefits of this PDF for exam preparation and improving learning efficiency. Once the students download the PDF, they can use it for quick revision and practice on the go, even when they do not have any internet connectivity. Practicing the exemplar questions helps student in deepening their concept clarity. The chapter deals with critical topics like hybridization, bond formation, bond parameters, and molecular geometry.
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NCERT Class 11 Chemistry
Chemical Bonding and Molecular Structure Class 11 Short Answer Type Questions
SA questions involve small derivations, brief explanations, or the application of basic theories like the octet rule, Lewis structures, bond parameters, VSEPR theory, and electronegativity. It requires students to give answers in a concise format. Practicing short-answer type questions helps students to build accuracy in presenting precise chemical reasoning, and also in clarifying the concepts. Students can also read NCERT Class 11 Notes here.
SA Questions
Explain the non linear shape of H2S and non planar shape of PCl3 using valence shell electron pair repulsion theory.
This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: In the Lewis structure of PCl3, the phosphorus atom is surrounded by three bond pairs (chlorine atoms) and one lone pair. These four electron pairs are arranged in a tetrahedral geometry around the central phosphorus atom. Due to the presence of lone pair of electron on the phosphorus atom, PCl3 will have a distorted tetrahedral geometry. Thus, it will form a pyramidal shape and is non-linear in structure.
In H2S, the central sulphur atom is surrounded by two bond pairs and two lone pairs of electrons. It can be said that these four electron pairs are arranged in a tetrahedral geometry. Due to the presence of two lone pairs on the central sulphur atom, the lone pair-bond pair repulsion happens and due to this the molecule H2S has a V-shaped geometry and is non-linear in structure.
Using molecular orbital theory, compare the bond energy and magnetic character of O2+ and O2 – species
This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: The electronic configurations of O2+ and O2 - according to molecular orbital theory is:
O2+: σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2, π2p2y , π2p2x, π*2p1x
O2- : σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2, π2p2y , π2p2x, π*2p2y π*2p1x
The bond order of O2+
Bond order = ( Nb- Na)
Bond order = (10-5)=2.5
The bond order of O2-
(10-7)= = 1.5
As the bond order of O2+ is higher, it is more stable than O2 -, because higher the bond order more stable is the bond. Both the molecular species have the presence of unpaired electrons. So, they both are paramagnetic in nature.
Commonly asked questions
Explain the shape of BrF5.
Structures of molecules of two compounds are given below :
Why does type of overlap given in the following figure not result in bond formation?
Chemistry Class 11 Chemical Bonding and Molecular Structure Long Answer Type Questions
LA questions require an in-depth understanding of the concepts. Students need to provide detailed explanations. These solutions help students to prepare for the descriptive questions in board exams and other entrance exams. LA questions for the Chemical Bonding may include comparing ionic vs. covalent bonds, predicting molecular geometry using VSEPR theory, deriving or explaining hybridization in molecules, drawing and analyzing molecular orbital diagrams, and explaining bond polarity and dipole moment.
LA Questions
| 1. (i) Discuss the significance/ applications of dipole moment. |
| This is a Long Answer Type Questions as classified in NCERT Exemplar Ans: The significance of dipole moment is as follows: (a) Dipole moment is used to predict the nature of the molecules. A molecule having zero dipole moment is considered as a nonpolar molecule and a molecule having a specific dipole moment is considered as polar in nature. (b) Dipole moment is important in determining the shape of a molecule. The shape of a molecule will not be symmetrical if it has a specific dipole moment, it may be bent or angular. A linear molecular which is symmetrical has a zero dipole moment. (c) Dipole moment is helpful in the comparison of the polarities of the molecules. If the value of the dipole moment is higher, more will be the polarity of the molecule and vice-versa. (d) Dipole moment also determines if a molecule is in cis- or trans- position, because cis-isomers have higher dipole moment than trans-isomers. (e) Dipole moment measurements help in differentiating in the ortho-, meta-, and para-isomers. The dipole moment of para-isomers is zero and the dipole moment of ortho-isomers is more than meta-isomers. |
| 1. (ii) Represent diagrammatically the bond moments and the resultant dipole moment in CO2 , NF3 and CHCl3. |
| This is a Long Answer Type Questions as classified in NCERT Exemplar Ans: According to the below shown diagrams, the bond moments and dipole moment of the following molecules is: For CO2 molecule: Carbon dioxide has a linear geometry, hence in the above structure the dipole of one C-Ois cancelled by another C-O . So, the dipole moment of CO2 is zero. For NF3 molecule: In nitrogen trifluoride, fluorine is more electronegative than hydrogen. The lone pair of electrons on nitrogen atoms also influences the dipole moment of the structure. The dipole moment of NF3 was found out to be 0.23D. In the above molecule, the dipole in F-N weakens the dipole induced by the lone pair of electrons. Thus the vector sum of the four dipole moments is low. For CHCl3 molecule: The two terminal chlorine atoms are cancelled with each other, the net dipole moment for chloroform was found out to be 1.04D. 
|
| This is a Long Answer Type Questions as classified in NCERT Exemplar Ans: Introduced by Hietler and London in 1927, valence bond theory was further develop dy Pauling. Valence bond theory (VBT) describes the information about the atomic orbitals, it tells about the electronic configuration of elements, the overlap criteria of atomic orbitals and also the hybridization of atomic orbitals is depicted by the valence bond theory. A and B are two hydrogen atoms which are approaching each other and these hydrogen atoms have the nuclei , NA NB respectively, the electrons in the two hydrogen atoms are represented as , eA eB respectively. When the distance between these two hydrogen atoms is large, then there will be no interaction between them. New attractive and repulsive forces begin to generate as these two hydrogen atoms approach each other. Attractive forces arise between these two hydrogen atoms. As a result of these attractive forces, the two atoms come closer to each other. It has been found experimentally that the magnitude of new attractive forces was more than the new repulsive forces. Due to this when the two atoms approach each other the potential energy decreases. Therefore, a stage occurs, where the net force of attraction balances the net repulsive forces and a state of minimum energy is acquired by the system. At this stage the two hydrogen atoms form a bond together, thus forming a stable bond. When the two hydrogen atoms form a bond, the energy is released, the hydrogen molecule formed is more stable than the individual hydrogen atom. The energy released during the formation of the hydrogen molecule is called bond enthalpy. |
| 4. Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and equatorial bonds have the same bond length. Explain. |
| This is a Long Answer Type Questions as classified in NCERT Exemplar Ans: Hybridisation in the case of PCl5 : 
The hybridisation of phosphorus PCl5 is sp3d and it has a trigonal bipyramidal geometry. The axial bonds in PCl5 are slightly longer as compared to the equatorial bonds because axial bonds experience greater repulsive forces from other bonds as compared to the equatorial bonds. Hybridisation in the case of SF6: 
The hybridisation of sulphur in SF6 is sp3d2 and the molecule has an octahedral geometry. The bond length of axial as well as the equatorial bonds are similar because all the bonds in octahedral geometry experience similar repulsive forces from other bonds. |
Commonly asked questions
Use the molecular orbital energy level diagram to show that N2 would be expected to have a triple bond, F2, a single bond and Ne2, no bond.
Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and equatorial bonds have the same bond length. Explain.
(i) Discuss the concept of hybridisation. What are its different types in a carbon atom.
(ii) What is the type of hybridisation of carbon atoms marked with star.
Class 11 Chemical Bonding and Molecular Structure Matching Type Questions
These questions are to test students' ability to link the related concepts and chemical terms, such as bond types with examples, hybridizations with molecular geometries, molecules with their shapes, and bond angles with specific molecular shapes. These questions help in reinforcing the inter-conceptual understanding, clarity, and memory.
Matching Type Questions
Match the species in Column I with the type of hybrid orbitals in Column II.
This is a Matching Type Questions as classified in NCERT Exemplar
Ans:
(i) SF4 (c) sp3d
(ii) IF5 (a) sp3d2
(iii) NO2+ (e) sp
(iv) NH4+ (d) sp3
Explanation :-
(i) SF4 has a total of 34 valence electrons. Sulphur has 6 valence electrons and each fluoride has 7 valence electrons. The central atom sulphur will thus share one electron with all the four fluoride ions and thus make four bonds. A lone pair of electrons (two electrons) is left on the sulphur atom. Thus, the sulphur atom is surrounded by five electron pairs, four bond pairs and one lone pair. Having five electron pairs surrounding sulphur, it will have a trigonal bipyramidal geometry with sp3d hybridization
(ii) IF5 has the total of 42 valence shell electrons, 7 valence electrons in iodine and 7 in each of the fluoride ions. The central iodine atom will thus make five sigma bonds with chlorine and a lone pair of electrons will be left on the iodine atom. The six electron pairs thus will have a hybridization of sp3d2 and octahedral geometry.
(iii) In NO2+, nitrogen forms two sigma bonds and two pi-bonds with the oxygen atom. The oxygen atoms are arranged in an angle of 180° with the nitrogen atoms. The s-orbital of the nitrogen forms a hybrid with the p-orbital of the nitrogen atom, thereby forming two sp hybridized hybrid orbitals, having a linear geometry.
(iv) NH4+ has the total of 8 valence electrons. Due to the presence of positive charge, nitrogen has 4 valence electrons and hydrogen has one in every four hydrogen atoms. The four valence electrons in nitrogen make sigma bonds with the four hydrogen atoms, thereby making a tetrahedral geometry and having 3 sp hybridized orbitals.
Match the species in Column I with the geometry/shape in Column II.
| Column I |
Column II |
| (i) H3O+ |
(a) Linear |
| (ii) HC ≡ CH |
(b) Angular |
| (iii) ClO2- |
(c) Tetrahedral |
| (iv) NH4+ |
(d) Trigonal bipyramidal |
|
|
(e) Pyramidal |
This is a Matching Type Questions as classified in NCERT Exemplar
Ans:
(i) H3O+ (e) Pyramidal
(ii) HC? CH (a) Linear
(iii) ClO2- (b) Angular
(iv) NH4+ (c) Tetrahedral
Explanation :-
(i) The shape of H3O+ is pyramidal this structure has the total 8 valence electrons, six in oxygen atom, three in each hydrogen atoms and one electron is lost due to positive charge. Out of five valence electrons in oxygen three form a sigma bond with the hydrogen atom and a lone pair is left on the oxygen atom. Although the four electron pairs make a tetrahedral geometry but to the lone pair of electrons, lone-pair bond-pair repulsion occurs which distorts the shape and makes it pyramidal.
(ii) The ethyne molecule has a linear geometry, because the two carbon atoms make two pi bonds with each other and one sigma bond with the hydrogen atom. The carbon and hydrogen atoms are sp hybridized.
(iii) ClO2 - has 20 valence electrons in its structure. Chlorine atoms make one pibond and two sigma bonds with the oxygen atoms. The pi-bond is in resonance and hence ClO2 - has two resonating structures. Two lone pairs of electrons are left on the oxygen at om and hence, due to lone pair-lone pair repulsion the bond angle reduces and the shape becomes linear.
(iv) NH4+ has the total of 8 valence electrons. Due to the presence of positive charge, nitrogen has 4 valence electrons and hydrogen has one in every four hydrogen atoms. The four valence electrons in nitrogen make sigma bonds with the four hydrogen atoms, thereby making a tetrahedral geometry and having sp3 hybridized orbitals.
Commonly asked questions
Match the species in Column I with the bond order in Column II.
|
Column I |
Column II |
|
(i) NO |
1.5 |
|
(ii) CO |
2.0 |
|
(iii) O22- |
2.5 |
|
(iv) O2 |
3.0 |
Match the items given in Column I with examples given in Column II.
|
Column I |
Column II |
|
(i) Hydrogen bond |
(a) C |
|
(ii) Resonance |
(b) LiF |
|
(iii) Ionic solid |
(c) H2 |
|
(iv) Covalent solid |
(d) HF |
|
|
(e) O3 |
Match the shape of molecules in Column I with the type of hybridisation in Column II.
|
Column I |
Column II |
|
(i) Tetrahedral |
(a) sp2 |
|
(ii) Trigonal |
(b) sp |
|
(iii) Linear |
(c) sp3 |
NCERT Exemplar Chemical Bonding and Molecular Structure Class 11 Assertion and Reason Type
These evaluate the student's ability to see the logical relationship between two statements. Assertion and Reason Type Questions cover areas like factors affecting bond angles, the cause-and-effect relation in molecular shape, exceptions to the octet rule, and stability of molecules based on bond formation. The students need to analyze whether the reason correctly explains the assertion and whether both the assertion and the reason are true. Practicing these types of questions helps in building conceptual accuracy and logical thinking.
Assertion and Reason Type
Assertion (A) : Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R) : This is because sodium and chloride ions acquire octet in sodium chloride formation.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false
This is a Assertion Type Questions as classified in NCERT Exemplar
Ans: Option (i)
Both the statements given in assertion and reason are correct and the reason given is the correct explanation for the assertion. The sodium chloride formed by the action of chlorine gas on the sodium metal has a complete octet. The valence shells of both the atoms are fulfilled, that is why attains an octet.
Assertion (A) : Though the central atom of both NH3 and H2O molecules are sp3 hybridised, yet H–N–H bond angle is greater than that of H–O–H.
Reason (R) : This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(i) A and R both are correct, and R is the correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
This is a Assertion Type Questions as classified in NCERT Exemplar
Ans: Option (i)
The statements given in the assertion and reason, both are correct and the reason given is the correct explanation for the assertion. In NH3 only one lone pair of electron is present and due to that lone pair-bond pair repulsion occurs which forms a distorted tetrahedral geometry (pyramidal) .
In H2O two lone pairs of electrons are present on the oxygen atom and due to this lone pairlone pair repulsion occurs which shifts the bonds and forms a bent geometry. The lone pair-lone pair repulsion is more than the lone pair-bond pair repulsion. Thus, bond angle in H-N-H is more than that of H-O-H.
Commonly asked questions
Assertion (A): Among the two O–H bonds in H2O molecule, the energy required to break the first O–H bond and the other O–H bond is the same.
Reason (R) : This is because the electronic environment around oxygen is the same even after breakage of one O–H bond.
(i) A and R both are correct, and R is correct explanation of A.
(ii) A and R both are correct, but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A and R both are false.
Chemical Bonding and Molecular Structure NCERT Exemplar Class 11 Objective Type Questions
These mostly include multiple-choice questions. MCQs test the quick recall, concept understanding, and application of theories. They cover theories like VSEPR, Lewis theory, and MOT, bond characteristics (length, angle, energy), shape and hybridization of molecules, and fundamental definitions. Practicing the MCQs is especially beneficial for the NEET and JEE exams, where speed and accuracy matter.
Objective Type Questions
(i) [NF3 and BF3 ]
(ii) [BF4 and NH4+ ]
(iii) [BCl3 and BrCl3]
(iv) [NH3 and NO3– ]
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (ii)
In a molecular structure if a central atom has the same hybridization as the central atom of another molecular species, then the two structures are known as isostructural species.
(i) In NF3 the central nitrogen atom is sp3 hybridized and in BF3 the central boron atom is sp2 . So, these two molecules are not isostructural pairs.
(ii) In BF4 the central boron atom is sp3 hybridized and in NH4 + the central nitrogen atom is also sp3 So, these two molecules are identified as isostructural pairs.
(iii) In BCl3 the central boron atom is sp2 hybridized and in BrCl3 the central bromine atom is sp3d hybridized. So, these two molecules are not isostructural pairs.
(iv) In NH3 the central nitrogen atom is sp3 and in NO3 the central nitrogen atom is sp2 So, these two molecules are not isostructural pairs.
Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(i) CO2
(ii) HI
(iii) H2O
(iv) SO2
This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Option (iii)
The dipole moment of the molecule depends on the difference in the electronegativity of the atoms present in the structure. The dipole moment of CO2 is 0, HI is 0.38, H2O is 1.84 and SO2 is 1.62.
As the oxygen atom is highly electronegative and hydrogen is least electronegative, the difference in electronegativity will be the highest for water molecules. Therefore, water molecules will have the highest dipole moment.
Commonly asked questions
In NO3 – ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(i) 2, 2
(ii) 3, 1
(iii) 1, 3
(iv) 4, 0
Which molecule/ion out of the following does not contain unpaired electrons?
(i) N2+
(ii) O2
(iii) O22–
(iv) B2
If the electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2, the four electrons involved in chemical bond formation will be_____.
(i) 3p6
(ii) 3p6, 4s2
(iii) 3p6, 3d2
(iv) 3d2, 4s2
Common Mistakes and Tips for NCERT Chemistry Exemplar Chapter Four
These are the common mistakes that students make in Chapter Four Chemical Bonding and Molecular Structure:
- Students make mistakes in the Octet rule by assuming that all atoms must strictly follow the octet rule. They overlook the exceptions like molecules with incomplete octets (e.g., BF₃) or expanded octets (e.g., SF₆, PCl₅).
- Many students mix up the molecular shapes (like tetrahedral, and trigonal planar) with hybridization types (like sp³, sp²) leading to incorrect predictions of geometry.
- They make errors in the Lewis Dot Structures such as misplaced lone pairs, incorrect counting of valence electrons, or forming invalid bonds—especially for resonance structures or polyatomic ions.
- In VSEPR Theory, they may fail to correctly count bonding and lone pairs or may also ignore lone pair repulsion effects.
- They get confused between the bond length and bond order by incorrectly assuming that the bond order increases bond length. The fact is higher bond order = shorter and stronger bonds.
- They make mistakes in the Molecular Orbital Theory (MOT) by misplacing the electrons in the MO diagrams.
The following are the tips to avoid common mistakes:
- Read and understand the Lewis structures, valence electrons, bonding theories, and hybridization.
- Use tables and flowcharts to apply concepts correctly as they make the concepts easier to remember. The students should create a quick reference table for bond angles, hybridizations, shapes, and examples.
- Visual practice is important for improving spatial understanding and reducing errors. Hence, students should draw the Lewis structures, MO diagrams, and VSEPR models.
- Before memorizing the concepts, try to understand them first.
- Practice the NCERT exemplar to gain more accuracy and confidence for the exam.
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| NCERT Solutions | NCERT Solutions for Class 11 Chemistry |
| NCERT Solutions for Class 12 Chemistry | NCERT Notes |
JEE Mains 2020
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01 Oct 2025
Chemistry NCERT Exemplar Solutions Class 11th Chapter Four Exam
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