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NCERT Class 11 Chemistry Chapter 4 Chemical Bonding and molecular Structure
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Chemical Bonding and molecular Structure
Chapter: 4
| Part – I |
1. Explain the formation of a chemical bond.
Ans: A chemical bond is a formation of bonds due to the electrostatic forces of attraction due to sharing of electrons or donating electrons. The atoms of different elements combine with each other in order to complete their respective octets (i.e., 8 electrons in their outermost shell) or diplet (i.e., outermost shell having 2 electrons) in case of H, Li and Be to attain stable nearest noble gas configuration.
2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Ans:
3. Write Lewis symbols for the following atoms and ions:
S and S2–; Al and Al3+; H and H–
Ans: Here are the Lewis symbols for the given atoms and ions:
4. Draw the Lewis structures for the following molecules and ions:
PH3, H2S, SiCl4, BeF2, CO3 2−, HCOOH
Ans: The lewis structures of the given molecules or ions are shown below:
5. Define octet rule. Write its significance and limitations.
Ans: According to octet rule, atom tends to have 8 electrons in the valence shell to attain stable configuration.
Significance:
(i) It is useful for understanding the structures of most of the organic compounds.
(ii) It explains the reason behind the chemical combination of different atoms to form ionic or covalent bond.
Limitations:
(i) The incomplete octet of the central atom: In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons. Examples are LiCI, BeH, and BC13
(ii) Odd-electron molecules: In molecules with an odd number of electrons like nitric oxides, NO and nitrogen dioxide NO₂, the octet rule is not satisfied for all the atoms.
(iii) The expanded octet: Elements in and beyond the third period of the periodic table have, apart from 3s and 3p orbitals also available for bonding. In a number of compounds of these elements there are more than eight valence electrons around the central atom. This is termed as the expanded octet. The octet rule does not apply in such cases.
Some of the examples of such compounds : PF5, SF6, H₂SO₄ and a number of coordination compounds.
10 electron around the proton
12 electrons around the S atom
12 electrons around the S atom
6. Write the favourable factors for the formation of ionic bond.
Ans: Favourable Factors for Ionic Bond Formation:
(i) Low Ionisation Energy: Atoms with low ionisation energy easily lose electrons to form positive ions. This is common in alkali metals and alkaline earth metals.
(ii) High Electron Affinity: Atoms with high electron affinity readily gain electrons to form negative ions. Non-metals like halogens have high electron affinities.
(iii) High lattice energy:The lattice energy should be high so that it overcomes the apparent deficit of energy of absorption, that is, sublimation energy, ionisation energy, dissociation energy and second electron affinity. Higher the lattice energy, greater will be the case of formation of the ionic compound.
The magnitude of lattice energy gives an idea about the interionic forces. It depends upon the following factors:
(a) Size of the ions: Smaller the size of the ions, lesser is the internuclear distance. Consequently, the interionic attractions will be high and the lattice energy will also be large.
(b) Charge on the ions: Larger the magnitude of charge on the ions, greater will be the attractive forces between the ions. Consequently, the lattice energy will be high.
7. Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans: BeCl₂: A linear molecule. Be atom has 2 electrons in its outermost orbit. Each chlorine atom has seven valence electrons. The Lewis structure of BeCl₂ is
There are two electron pairs and to minimise the repulsion, these electron pairs tend to keep themselves far away from each other, ie., 180° apart. This gives BeCL a linear structure.
BCl3: In BCI3 molecule, the three bond pairs of electrons are located around B in a triangular arrangement. Thus, the molecule BCI3 has a triangular planar geometry.
SiCl4: A tetrahedral molecule. Si has 4 electrons in its outermost shell. Due to mutual sharing of electrons with Cl there are 4 electron pairs around Si. To keep the repulsion at the minimum, these 4 electron pairs should be arranged in a tetrahedral manner around Si. Thus, SiCl4 is a tetrahedral molecule.
AsF5: Trigonal bipyramidal molecule: As has five electrons in its outermost orbit. Due to sharing of 5 electrons from 5 F-atoms, are in all 5 electrons pairs. These are distributed in space to form a trigonal by pyramid.
H₂S: Bent (V-shaped) structure: S has 6 electrons in its outermost shell. 2H-atoms contribute 2 electrons during bonding. Thus, there are 8 electrons and 4 electron pairs around S. This gives a tetrahedral distribution of electron pairs around S. The two corners of the tetrahedron are occupied by H-atoms and the other two by the lone-pairs of electrons. Thus, H₂S has a bent structure.
PH3: Trigonal pyramidal: Phosphorus atom has 5 electrons in its outermost orbit. H-atoms contribute one electron each to make in all 8 electrons around P-atom. Thus, 4 pairs of electrons would be distributed in a tetrahedral fashion around the central atom. Three pairs from three P-H bonds while the fourth pair remains unused. Due to repulsion between the bonding and lone pairs of electrons, the angle HPH is not exactly tetrahedral (109°28′). The actual HPH angle is 93.3°. Thus PH3 is a trigonal pyramidal molecule
8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Ans: Due to greater lone pair-bond pair repulsion, the HNH angle is slightly less than the tetrahedral angle. The actual HNH angle in NH3 is 107°.
In H₂O, greater repulsion between lone-pair-bond pair leads to a HOH bond angle lower than the tetrahedral angle. The actual angle is 104.5°.
Thus,ammonia due to stronger lone pair-lone pair repulsion, which compresses the bond angle more in water than in ammonia.
9. How do you express the bond strength in terms of bond order?
Ans: With an increase in bond order, bond strength increases and bond length decreases.Bond order is associated with the strength of bond and bond length. The higher the bond order, the smaller will be the bond length and stronger will be the bond. Arrange O2,O+2,O−2 and O2−2 in order of increasing bond order.
10. Define the bond length.
Ans. The internuclear separation between two atoms when energy of the system in minimum is called bond length. Bond strength increases with increase in bond order.
11. Explain the important aspects of resonance with reference to the CO3 2− ion.
Ans. Carbonate ion (CO32−) is best described as a resonance hybrid of the three canonical forms.
(i) In these structures, the position of the nuclei are the same.
(ii) The number of bonding and nonbonding pairs of electrons are the same in all these structures.
(iii) The three structures have nearly the same energy.
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.
Ans: The formulae 1 and 2 do not describe the canonical forms of the resonance hybrid representing H3PO3 because:
(i) One of the hydrogen atoms occupies a different position.
(ii) The number of bonding and nonbonding electron pairs is different.
13. Write the resonance structures for SO3, NO2 and NO3 −.
Ans:
14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions:
(a) Na and CI.
(b) K and S.
(c) Ca and O.
(d) Al and N.
(e) Li and H.
Ans:
15. Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Ans: The net dipole moment of CO₂ is zero because the dipole moments of the two C = O bonds cancel each other out.
16. Write the significance/applications of dipole moment.
Ans: Dipole moment of a molecule can help us in:
(i) Determining the ionic character of a polar bond.
(ii) Deciding the geometry of the molecule.
17. Define electronegativity. How does it differ from electron gain enthalpy?
Ans: Electronegativity of an element refers to the tendency of its atom in a molecule to attract a shared pair of electrons towards itself. On the other hand, electron affinity (or electron gain enthalpy) is the change in enthalpy when an electron is added to an isolated gaseous atom in its ground state.
18. Explain with the help of suitable example polar covalent bond.
Ans: A covalent bond between two unlike atoms that differ in their affinities for electrons is referred to as a polar covalent bond.
In the hydrogen chloride (HCI) molecule, the hydrogen and chlorine atoms are bonded through electron sharing. The shared pair of electrons lies more towards Cl atom (because Cl is more electronegative). Therefore, Cl atom acquires a slight negative charge, and H atom a slight positive charge. This causes the covalent bond between H and Cl to have an appreciable ionic character.
19. Arrange the bonds in order of increasing ionic character in the molecules:
LiF, K2O, N2, SO2 and ClF3.
Ans: For the Given molecule LiF, K2O, N2, SO2 and ClF3.
Diff. in the electronegativities of the two atoms
3.0 2.7 0 1.0 1.0
Thus, the ionic character in the bonds follows the order,
N₂ < ClF< SO₂ < K₂O < LiF
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.
Ans: The correct Lewis structures for acetic acid is:
21. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?
Ans: In CH4, the carbon atom undergoes sp3 hybridization. The hybrid orbitals are arranged tetrahedrally to minimise the electronic repulsions, and not in a plane (say horizontal plane). This arrangement allows for the most stable configuration of the molecule.
22. Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
Ans: In BeH2, the molecule adopts a linear geometry with an H-Be-H bond angle of 180°. Because the two Be-H bond dipoles are oriented directly opposite each other, they cancel out. This results in a net dipole moment of zero for the molecule. So, the dipole moment BeH2 is zero.
23. Which out of NH3 and NF3 has higher dipole moment and why?
Ans: The dipole moment of NH3(= 4.9 × 10−30 Cm) is significantly higher than that of NF3(= 0.8 × 10−30 Cm). This is because, in case of NH3 the orbital dipole due to lone pair, and the resultant dipole moment of the N – H bonds act in the same direction. In the case of NF3 the orbital dipole acts in the direction opposite to the resultant dipole moment of the three N – F bonds. As a result, the net dipole of NH3 becomes much less than that of NH3 molecule. This difference is illustrated in the figure.
24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Ans: The process of mixing of the atomic orbitals to form new hybrid orbitals is called hybridization. All hybrid orbitals of a particular kind have equal energy, identical shapes and are symmetrically oriented in space.
(i) sp hybrid orbitals are oriented linearly at an angle of 180°, forming a straight line.
(ii) sp² hybrid orbitals lie in a plane and are directed towards the corners of an equilateral triangle, with a bond angle of 120°.
(iii) sp3 hybrid orbitals are directed towards the four corners of a tetrahedron. Improve.
25. Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + C– → AICI 4.
Ans: AlCl3 + Cl– → AICI 4.
tetrahedral geometry sp3 hybridisation
Aluminium chloride exists as a dimerAI2CI6. In this molecule, chlorine atoms are arranged tetrahedrally around the two Al atoms.
In Al2Cl6, aluminium atoms are sp2 hybridised, not sp3. The hybridization changes when AlCl3 reacts with a chloride ion. In AlCl3, aluminium has sp2 hybridization. In AlCl4 aluminium has sp3 hybridization.
26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B – NH3?
Ans. Here, B atom in BF3 is sp2 hybridised and one of its p orbital is empty. N in NH3 is sp3 hybridised and one of its hybrid orbitals is occupied by a lone- pair of electrons. During the reaction,
BF3 + NH3 → [F3B – : NH3] → [F3B ← NH3]
A coordinate bond (or dative covalent bond) is formed when one atom provides both electrons for a shared pair in the bond.
There is no change in the hybridisation of any of two atoms in this reaction.
27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
Ans: (i) C2H4
Formation of double bond between carbon atoms in C2H4.
The Lewis structure of ethylene is represented as
(ii) C2H2
he carbon-carbon triple bond is made up of one strong o-bond and two weak pi bond. The Lewis structure of acetylene (C₂H₂) is represented as:
H:C::C: H
28. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2
(b) C2H4
Ans:
| Sigma Bonds | Pi Bonds | |
| (a) | 3 | 3 |
| (b) | 5 | 1 |
29. Considering x-axis as the internuclear 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.
Ans: (a) 1s and 1s, (b) 1s and 2px is will from sigma bonds.
30. Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH3–CH3;
Ans: Both carbon atoms are sp³ hybridized.
(b) CH3–CH=CH2;
Ans: C1 is sp³ hybridized, whereas both C2 and C3 are sp² hybridized.
(c) CH3-CH2-OH;
Ans: Both carbon atoms are sp³ hybridized.
(d) CH3-CHO
Ans: C1 is sp³ hybridized, while C2 is sp² hybridized.
(e) CH3COOH
Ans: C1 is sp³ hybridization, whereas C2 displays sp² hybridization.
31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Ans: In chemical bonding, the electron pair shared between two atoms is called a shared pair or bond pair. In contrast, the electron pair not involved in bonding is called a lone pair. For instance, in BeH2, the central Be atom has two bond pairs. In NH3, the central N atom has three bond pairs and one lone pair.
For example, in BeH, molecule, the central Be atom is surrounded by two bond pairs. In the molecule of NH3, the central B atom has three shared pairs and one lone pair around it.
32. Distinguish between a sigma and a pi-bond.
Ans: Distinguish between a sigma and a pi bond are:
| Basic of difference | sigma bond | pi bond |
| Formation | It is formed due to axial overlap of the two orbitals. | This bond is formed by the lateral (sideways) overlap of two p-orbitals |
| Orbital Overlap | There can be only one sigma bond between atoms | There can be more than one π-bonds between the two atoms. |
| Bond Strength | The bonding is relatively strong. | The bonding due to a π-bond is weak. |
| Bond Rotation | Free rotation of atoms about sigma (s) bond is possible. | Free rotation about a π-bond is not possible. |
| Electron Density | The electron density is maximum and cylindrically symmetri-cal about the bond axis | The electron density is high along a direction at right angle to the bond axis. |
33. Explain the formation of H2 molecule on the basis of valence bond theory.
Ans: Consider two hydrogen atoms, A and B approaching each other with nuclei NA and NB and electrons eA and eB. When these two atoms are far apart, there is no interaction between them. However, as the atoms begin to approach each other, new attractive and repulsive forces come into play.
Attractive forces arise between:
(i) nucleus of one atom and its own electron, that is NA-eA and NB-eB.
(ii) nucleus of one atom and electron of other atom i.e.,NA-eB and NB-eA.
Similarly repulsive forces arise between:
(i) electrons of two atoms like eA – eB,
(ii) nuclei of two atoms NA – NB
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
Forces of attraction and repulsion during the formation of H2 molecule.
Experimentally, it has been observed that the magnitude of the new attractive forces between the atoms is greater than the magnitude of the new repulsive forces. As a result, the two hydrogen atoms approach each other, and the potential energy of the system decreases. Eventually, a point is reached where the net attractive force exactly balances the repulsive force, and the system achieves its minimum energy state. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm.
34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Ans: Conditions for the combination of atomic orbitals:
(i) The combining atomic orbitals must have the same or nearly the same energy. This means that Is orbital can combine with another is orbital but not with 2s orbital because the energy of 2s orbital is appreciably higher than that of 1s orbital. This is not true if the atoms are very different.
(ii) The combining atomic orbitals must have the same symmetry about the molecular axis. By convention, the Z-axis is considered as the molecular axis.
(iii) The combining atomic orbitals must overlap to the maximum extent. The greater the extent of overlap, the higher the electron density between the nuclei in a molecular orbital.
35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
Ans: Molecular orbital configuration of Be2 is K (𝞂2 × s)2 (𝞂 × 2s)2
Bond order = ½ [2-2 ] = 0
Therefore Be2 does not exist.
36. Compare the relative stability of the following species and indicate their magnetic properties O2, O2+,O2− (superoxide), O22− (peroxide.)
Ans: Bond order of the different species are: O2 (2, 0) O2+ (2.5), O2– (1.5), O22- (1.0)
Relative stability: O2+ .> O2 > O2–> O22-
(i) O2 is paramagnetic.
(ii) O2+ is paramagnetic.
(iii) O2– paramagnetic.
(iv) O22- is diamagnetic.
37. Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans: Bonding molecular orbital is formed when the lobes of the combining atomic orbitals have same sign (+ and + or – and).
An antibonding molecular orbital is formed when the lobes of the combining atomic orbitals have opposite signs (one positive and one negative).
38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Ans: The ground state and the excited state outer electronic configuration of phosphorus (Z = 15) are represented below.
sp³d hybrid orbitals filled by electron pairs denoted by five Cl atoms.
Now the five orbitals (i.e., one s, three p and one d orbitals) are available for hybridisation to yield a set of five sp³d hybrid orbitals which are directed towards the five corners of a trigonal bipyramid.
Due to increased repulsive interactions between the axial bond pairs and the equatorial bond pairs, the axial bonds are slightly longer than the equatorial bonds.
39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Ans: The hydrogen bond can be defined as the attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. A hydrogen bond is shown by a dotted line (….)
However, hydrogen bonds are actually stronger than Van der Waals forces.
40. What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+ and O2 –.
Ans: It may be defined as half of the difference between the number of electrons present in bonding and antibonding molecular orbitals.
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