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Chemical bonding is the process by which two or more atoms combine to form a molecule or compound. This occurs when the atoms are attracted to each other by electrostatic forces and share or transfer electrons to achieve a stable electronic configuration.
There are three main types of chemical bonds: ionic bonds, covalent bonds, and metallic bonds. In ionic bonding, one or more electrons are transferred from one atom to another, resulting in positively charged cations and negatively charged anions that are attracted to each other. In covalent bonding, electrons are shared between atoms, resulting in a stable molecule with a shared electron pair. In metallic bonding, electrons are shared between a sea of metal atoms, resulting in a stable structure with a delocalized electron cloud. Students can check here NCERT class 11Chemistry Solution to score good marks in the exam. Those preparing for entrance exams can refer NCERT Solutions Chemistry class 12.
NCERT Chemistry Class11th Chemical Bonding and Molecular Structure Chapter Covers
- Nature of Chemical Bonding
- Structure of Molecules
- Properties of Molecules
NCERT Chemistry Class11th Solution PDF for Chemical Bonding and Molecular Structure
The properties of molecules and compounds are determined by their molecular structure, which is the arrangement of the atoms in space. The molecular structure is determined by the type of bond and the number and arrangement of atoms in the molecule. The shape of the molecule affects its polarity, reactivity, and physical properties such as boiling point, melting point, and solubility.
The valence shell electron pair repulsion (VSEPR) theory is a model used to predict the molecular structure of covalent molecules based on the repulsion between valence electron pairs. This theory states that the electron pairs in the valence shell of an atom repel each other and try to occupy positions that minimize repulsion. The VSEPR theory is used to predict the geometry and shape of molecules, such as linear, trigonal planar, tetrahedral, and octahedral.
In summary, chemical bonding and molecular structure are important concepts in chemistry that help us understand the behavior and properties of molecules and compounds. The type of bond and molecular structure determine the properties of a substance, including its reactivity, polarity, and physical properties. The VSEPR theory is a useful tool for predicting the molecular structure of covalent molecules based on the repulsion between valence electron pairs.
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Chemical Bonding and Molecular Structure Additional Question
4.1. Assertion: The dipole moment in case of BeF2 is zero.
Reason: There are two F atoms in a molecule of BeF2.
Answer:(b) Reason do not explain the assertion, although both are correct statements independently. The dipole moment in case of BeF2 is zero because the two equal bond dipolespoint in opposite directions and cancel theeffect of each other.
4.2. Assertion: In BF3, the dipole moment is zero.
Reason: The three bond moments give anet sum of zero as the resultant of any two isequal and opposite to the third.
Answer:(a) In tetra-atomic molecule like BF3, the dipole moment is zero although the B – F bonds are oriented at an angle of 120o to one another, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
4.3. Assertion: Sigma bond is weaker as compared to the pi bond.
Reason: The overlapping of orbitals takes place to a larger extent in sigma bond while the extent of overlapping occurs to a smaller extent in case of pi bond.
Answer: (d) Basically the strength of a bond depends uponthe extent of overlapping. In case of sigma bond,the overlapping of orbitals takes place to alarger extent. Hence, it is stronger as comparedto the pi bond where the extent of overlappingoccurs to a smaller extent. Further, it isimportant to note that in the formation ofmultiple bonds between two atoms of amolecule, pi bond(s) is formed in addition to asigma bond.
4.4.Assertion: A bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it.
Reason: Electrons placed in a bonding molecular orbital tend to destabilise a molecule.
Answer: (c) R is an incorrect statement. Electrons placed in a bonding molecular orbital tend to hold the nuclei together and stabilise a molecule. Therefore, a bonding molecular orbital alwayspossesses lower energy than either of the atomic orbitals that have combined to form it.In contrast, the electrons placed in the antibonding molecular orbital destabilise the molecule. This is because the mutual repulsion of the electrons in this orbital is more than the attraction between the electrons and the nuclei, which causes a net increase in energy.
4.5. Assertion: In PCl5, the axial bonds are slightly longer and hence slightly weaker than the equatorial bonds.
Reason: The axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.
Answer: (a) It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl5 the five sp3d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P–Cl sigma bonds. Three P–Cl bond lie in one plane and make an angle of 120° with each other; these bonds are termed as equatorial bonds.
The remaining two P–Cl bonds–one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane.These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds which makes PCl5 molecule more reactive.
MCQs
4.1. A co-ordinate bond is formed by:
(a)sharing of electrons contributed by both the atoms
(b)complete transfer of electrons
(c)sharing of electrons contributed by one atom only
(d) none of these
Answer: (c)sharing of electrons contributed by one atom only
4.2. The axial overlap between the two orbitals leads to the formation of a:
(a) sigma bond (b) pi bond (c) multiple bond (d) none of these
Answer: (a)sigma bond
4.3. Which one is diamagnetic among NO+, NO and NO?
(a) NO+ (b) NO (c) NO– (d) None of these
Answer: (a) NO+
4.4. Out of the following, intramolecular hydrogen bonding exists in:
(a) water (b)H2S (c) 4-nitrophenol (d) 2-nitrophenol
Answer: (d) 2-nitrophenol
4.5. Relation between bond order and stability of a molecule:
(a) Lower the bond order, greater is the stability
(b) not related to each other
(c) Higher the bond order, greater is the stability
(d) none of the above
Answer: (c) Higher the bond order, greater is the stability.
Questions and Answers:
4.1. Write the type of hybridisation involved in CH4, C2H4 and C2H2.
Answer: CH4= sp3
C2H4 = sp2
C2H2 = sp
4.2. Predict the shapes of the following molecules using VSEPR theory?
(i) BeCl2 (ii) SiCl4
Answer: (i) Linear
(ii) Tetrahedral
4.3. What is meant by bond pairs of electrons?
Answer: The electron pairs involved in the bond formation are known as bond pairs or shared pairs.
4.4. Why ethyl alcohol is completely miscible with water?
Answer: This is because ethyl alcohol forms H-bonds with water.
4.5. Which is more polar CO2 or N2O? Give reason.
Answer: N2O is more polar than CO2.
This is because CO2 is linear and symmetrical. Its net dipole moment is zero.
N2O is linear but not symmetrical.
4.6. Why N2 is more stable than O2? Explain on the basis of molecular orbital theory.
Answer: Bond order of N2 (= 3) is greater than that of O2 (= 2).
4.7. Out of bonding and antibonding molecular orbitals, which one has lower energy and which one has higher stability?
Answer: Bonding molecular orbital has lower energy and higher stability.
4.8. Name the two conditions which must be satisfied for hydrogen bonding to take place in a molecule.
Answer: (i) The molecule should contain highly electronegative atom like hydrogen atom. (ii) The size of electronegative atom should be small.
4.9. What are Lewis structures? Write the Lewis structure of H2, BeF2 and H2O.
Answer: The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures.
4.10. Define Lattice energy. How is Lattice energy influenced by
(i) Charge on the ions (ii) Size of the ions?
Answer: Lattice energy is defined as the energy released when one mole of crystalline solid is formed by the combination of oppositely charged ions.
(i) As the magnitude of charge on an ion increases there will be greater force of interionic attraction and hence greater will be the value of Lattice energy,
(ii) Smaller the. size of the ions> lower will be the internuclear distance and thus greater will be the Lattice energy
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Chemical Bonding and Molecular Structure Solutions and FAQs
Q 4.1. Explain the formation of a chemical bond.
A 4.1 Chemical bond is an attractive force which binds atoms, ions etc. together in a compound. According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the configuration of the nearest stable inert gas. This can occur in two ways; by transfer of one or more electrons from one atom to other or by sharing of electrons between two or more atoms. The chemical bond formed by sharing of electrons is called a covalent bond. In this process a chemical bond is formed.
Q 4.2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
A 4.2
Q 4.3 Write Lewis symbols for the following atoms and ions:
S and S2–; Al and Al3+; H and H–.
A 4.3
Q 4.4 Draw the Lewis structures for the following molecules and ions:
H2S, SiCl4, BeF2, CO32-, HCOOH
A 4.4
Q 4.5. Define Octet rule. Write its significance and limitations.
A 4.5 Kössel and Lewis developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this rule, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells. This is known as octet rule.
Significance: It helps to explain why different atoms combine with each other to form ionic compounds or covalent compounds.
Limitations of Octet rule:
- According to Octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements. However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: XeF2, XeF4 Therefore, validity of the octet rule has been challenged. This theory does not account for shape of molecules.
- Stability of odd electron species: Molecules like NO and NO2 do not satisfy the octet rule and yet they exist as stable molecules. The octet rule failed to explain the reason for this.
- Expansion of octet: The rule could not explain the stability of compounds which have more than 8 valence electrons around the central atom. For eg SF6, PF5, IF7
- Electron deficient compounds: The octet rule could not explain the stability of electron deficient compounds like LiCl, AlCl3, BH3
Q 4.6. Write the favourable factors for the formation of ionic bond.
A 4.6 The favourable factors for the formation of ionic bond are:
- Low ionization enthalpy of metal atoms.
- High electron gain enthalpy of non-metal atoms.
- High lattice enthalpy of compound formed.
Q 4.7. Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3.
A 4.7
| Molecule | Number of electron pairs around central atom | Type | Molecular geometry | Bond angles |
| BeCl2 | 2 | AB2 | Linear | 180o |
| BCl3 | 3 | AB3 | trigonal planar | 120 o |
| SiCl4 | 4 | AB4 | tetrahedral | 109.5o |
| AsF5 | 5 | AB5 | trigonal bipyramidal | three 120 o, two 90 o |
| H2S | 6 | AB2L2 | V-shaped/bent | 92 o |
| PH3 | 5 | AB3L | trigonal pyramidal | 93.5o |
Q 4.8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
A 4.8
Because of two lone pairs of electrons on O-atom, repulsion on bond pairs is greater in H2O in comparison to NH3. Thus, the bond angle is less in H2O molecules.
Q 4.9 How do you express the bond strength in terms of bond order?
A 4.9 Bond strength is directly proportional to the bond order. Greater the bond order more is the bond strength.
Q 4.10. Define the bond-length.
A 4.10 Bond-length: It is the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond-lengths are measured by spectroscopic methods.
Q 4.11. Explain the important aspects of resonance with reference to the CO32-ion.
A 4.11
- In these structures, the position of nuclei is the same.
- The Lewis dot structure has two single bonds and one double bond.
- All three forms have almost equal energy.
- The three forms have same number of paired and unpaired electrons; they differ only in their position.
Q 4.12. H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.
A 4.12 No, these cannot be taken as canonical forms because the positions of atoms have been changed.
Q 4.13. Write the resonance structures for SO3, NO2 and NO3.
A 4.13
Q 4.14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions (a) K and S (b) Ca and O (c) Al and N.
A 4.14
4.15. Although both CO2 and H2O are tri atomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
A 4.15
In CO2, there are two C=O bonds. Each C=O bond is a polar bond. The net dipole moment of CO2 molecule is zero. This is possible only if CO2 is a linear molecule. (O=C=O). The bond dipoles of two C=O bonds cancel the moment of each other. Whereas, H2O has a net dipole moment of 1.84 D. H2O molecule has a bent structure because here the O—H bonds are oriented at an angle of 104.5° and do not cancel the bond moments of each other.
Q 4.16. Write the significance/applications of dipole moment.
A 4.16 Significance of dipole moment are:
- In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those of zero dipole moments are non-polar in nature.
- In the determination of shapes of molecules.
- In calculating the percentage ionic character.
Q 4.17. Define electronegativity. How does it differ from electron gain enthalpy?
A 4.17 Electronegativity: Electronegativity is the tendency of an atom to attract shared pair of electrons towards itself. Electronegativity of any given element is never the same. It depends on the element it is bonded with in a compound. Electronegativity is a relative quantity and cannot be measured.
Whereas, electron gain enthalpy is the change in enthalpy when an electron is added to a neutral gaseous atom to form an anionic specie. It can have a positive or negative value. Every element has a specific electron gain enthalpy value.
Q 4.18 Explain with the help of suitable example polar covalent bond
A 4.18 When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons is not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom. As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of a bond is called a polar covalent bond. For example, in hydrogen fluoride molecule (HF), flouride has greater electronegativity than hydrogen. Thus, the shared electron pair is displaced more towards flourine atom; the later will acquire a partial negative charge (∂–). At the same time hydrogen atom will have a partial positive charge (∂+). Such a covalent bond is known as polar covalent bond or simply polar bond.
It is represented as
Q 4.19 Arrange the bonds in order of increasing ionic character in the molecules:
LiF, K2O, N2, SO2 and ClF3.
A 4.19 N2 < SO2 < ClF3 < K2O
Q 4.20. The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
A 4.20
Q 4.21. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the comers of the square and the C atom at its centre. Explain why CH4 is not square planar?
A 4.21 According to VSEPR theory, if CH4 were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28′. Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less. Moreover, the orbitals of carbon in ground and excited states look as shown below. In the excited state, the electrons from 4 H atoms occupy one 2s orbital and three 2p orbitals. This results in an sp3 hybridisation, which is the hybridisation of a tetrahedral geometry.
Q 4.22. Explain why BeH2 molecule has a zero dipole moment although the Be—H bonds are polar.
A 4.22 BeH2is a linear molecular (H—Be—H), the bond angle = 180°. Be—H bonds are polar due to difference in their electronegativity but the bond polarities cancel each other. Thus, molecule has resultant dipole moment of zero.
Q 4.23. Which out of NH3 and NF3 has higher dipole moment and why?
A 4.23 The dipole moment of NH3 (1.47D) is higher than the dipole moment of NF3 (0.24D), even though the molecular geometry is pyramidal for both the molecules. In e
ach molecule, N atom has one lone pair. F is more electronegative than H and N−F bond is more polar than N−H bond. Hence, NF3 is expected to have much larger dipole moment than NH3. However, reverse is true. In case of ammonia, the direction of the lone pair dipole moment and the bond pair dipole moment is same whereas in case of NF3, it is opposite. Thus, in ammonia molecule, individual dipole moment vectors add whereas in NF3, they cancel each other.
Q 4.24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
A s4.24 Hybridisation: It is defined as the process of intermixing of atomic orbitals of slightly different energies to give rise to new hybridized orbitals having equivalent energy and identical shapes.
Shapes of Orbitals:
sp hybridisation: When one s-and one p-orbital intermix then it is called sp-hybridisation. For example, in BeF2, Be atom undergoes sp-hybridisation. It has linear shape. Bond angle is 180°.
sp2 hybridisation: One s-and two p-orbitals get hybridised to form three equivalent hybrid orbitals. The three hybrid orbitals directed towards three corners of an equilateral triangle. It is, therefore, known as trigonal hybridisation.
sp3 hybridisation: One s-and three p-orbitals get hybridised to form four equivalent hybrid orbitals. These orbitals are directed towards the four corners of a regular tetrahedron.
Q 4.25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl– ——> AlCl4-
A 4.25 Electronic configuration of 13Al in ground state= 1s2 2s2 2p6 3s2 3p1 Electronic configuration of 13Al in excited state = 1s2 2s2 2p6 3s1 3px13py1
Hence, hybridisation will be sp2 and this makes the geometry to be Trigonal planar (in case of AlCl3).
In AlCl–4, the empty 3pz orbital is also involved. So, the hybridisation is sp3 and the shape is tetrahedral.
Q 4.26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 ——-> F3B.NH3
A 4.26 In BF3, B atom is sp2 hybridised. In NH3, N is sp3 hybridised.
After the reaction, hybridisation of B changes from sp2 to sp3, while the hybridisation of N remains the same.
Q 4.27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
A 4.27
Q 4.28. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4
A 4.28 (a) Acetylene C2H2 contains 3σ bonds and 2π bonds.
(b) Ethylene C2H4 contains 5σ bonds and 1π bond.
Q 4.29 Considering x-axis as the inter-nuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s
(b) 1s and 2px
(c) 2py and 2py
(d) 1s and 2s
A 4.29
(c) It will not form a sigma bond because taking x-axis as the inter-nuclear axis, there will be lateral overlap between the two 2py orbitals forming a -bond.
Q 4.30. Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3-CH3 (b) CH3-CH=CH2 (c) CH3-CH2-OH (d) CH3-CHO (e) CH3COOH.
A 4.30
Q 4.31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
A 4.31 The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.
For example: CH4 has 4 bond pairs but H2O has 2 bond pairs and 2 lone pairs.
Q 4.32. Distinguish between a sigma bond and a pi bond.
A 4.32
| Sigma bond |
Pi bond |
| It is formed by axial overlap of the atomic orbitals. |
It is formed by the sidewise overlap of the atomic orbitals. |
| The bond is quite strong. |
It is a comparatively weaker bond. |
| Only one lobe of the p-orbitals is involved in the overlap. |
Both lobes of the p-orbitals are involved in the overlap. |
| Electron cloud of the molecular orbital is symmetrical around the inter-nuclear axis. |
The electron cloud is not symmetrical |
Q 4.33. Explain the formation of H2 molecule on the basis of valence bond theory.
A 4.33 Let us consider the combination between atoms of hydrogen HA and HB and eA and eB be their respective electrons. As they tend to come closer, two different forces operate between the nucleus and the electron of the other and vice versa. The nuclei of the atoms as well as their electrons repel each other. Energy is needed to overcome the force of repulsion. Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other, the overall potential energy of the system decreases. Thus, a stable molecule of hydrogen is formed.
Q 4.34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
A 4.34
- The combining atomic orbitals should have comparable energies like, 1s orbital of one atom can combine with 1s atomic orbital of another atom, 2s can combine with 2s.
- The combining atomic orbitals must have proper orientations so that they are able to overlap to a considerable extent.
- The extent of overlapping should be large.
Q 4.35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
A 4.35
Since the bond order is zero, which means that the molecule is unstable. So, Be2 does not exist.
Q 4.36. Compare the relative stability of the following species and indicate their magnetic properties:
O2, O2, O2– (Superoxide), O22- (peroxide)
A 4.36
O2: Bond order = 2, paramagnetic
O2+: Bond order = 2.5, paramagnetic
O2–: Bond order = 1.5, paramagnetic
O22-: Bond order = 1, diamagnetic
Order of relative stability is
O2+ > O2 > O2– > O22-
(2.5) (2.0) (1.5) (1.0)
Q 4.37. Write the significance of plus and minus sign in representing the orbitals.
A 4.37 Molecular orbitals are represented by wavefunctions. A positive sign in an orbital indicates a positive wavefunction while a negative sign in an orbital shows a negative wavefunction.
Q 4.38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
A 4.38 P has ground state valence shell electronic configuration 3s23p3 and the first excited state valence shell electronic configuration is 3s13p33d1. P undergoes sp3d hybridization and has trigonal bipyramidal geometry. Because axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than equatorial bonds.
Q 4.39. Define hydrogen bonds. Is it weaker or stronger than the van der Waals forces?
A 4.39 When hydrogen is attached with highly electronegative element in a covalent bonding the electrons of the covalent bond are shifted towards the more electronegative atom. Thus, a partially positively charged hydrogen atom forms a bond with the other more electronegative atom. This bond is known as a hydrogen bond. Hydrogen bond is stronger than the van der Waals forces.
Q 4.40. What is meant by the term bond order? Calculate the bond order of N2, O2, O2+ and O2.
A 4.40 Bond order is defined as one half the difference between the number of electrons present in the bonding and the antibonding orbitals i.e.
Bond order (B.O.) = ½ (Nb–Na)
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