NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure
4.1. Explain the formation of a chemical bond.
Ans. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called as a chemical bond. So many theories are suggested for chemical bond formation such as valence shell electron pair repulsion theory, electronic theory, molecular orbital theory and valence bond theory.
According to Kossel and Lewis, atoms unite to complete their individual octets in order to obtain the stable inert gas structure. Hence, it was postulated that the elements having outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called as covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.
4.2. Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
Ans. Mg : There are two valence electrons in Mg atom. Hence, the Lewis dot symbol for Mg is: Mg••
Na : There is only one valence electron in an atom of sodium. Hence, the Lewis dot structure is: Na•
B : There are 3 valence electrons in Boron atom. Hence, the Lewis dot structure is: •B••
O : There are six valence electrons in an atom of oxygen. Hence, the Lewis dot structure is: O••••••
N : There are five valence electrons in an atom of nitrogen. Hence, the Lewis dot structure is: N•••••
Br : There are seven valence electrons in bromine. Hence, the Lewis dot structure is: Br••••
4.3. Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H–.
Ans. S and S2–: The atomic number of Sulphur is 16 and its electronic configuration is 2, 8, 6. So, the outermost shell (valence shell) contains 6 electrons.
Therefore the lewis dot symbol of Sulphur is S••••••
In order to attain stable noble gas configuration, Sulphur requires two more electrons. Thus, the dinegative charge on S2– infers the additions of two more electrons in addition to the six valence electrons.
Therefore, the lewis dot symbol of sulphur ion is S••••••••2–.
Al and Al3+ : The atomic number of Aluminium is 13 and its electronic configuration is 2, 8, 3. So, the outermost shell (valence shell) contains 3 electrons.
Therefore, the lewis dot symbol of Aluminium is Al•••
In order to attain stable noble gas configuration, aluminium will lose three more electrons. Thus, the tripositive charge on Al3+ infers that loss of three valence shell electrons.
Therefore, the lewis dot symbol of Aluminium ion is [].Al3+
H and H– : The atomic number of Hydrogen is 1 and its electronic configuration is 1. So, the outermost shell (valence shell) contains 1 electrons.
Therefore the lewis dot symbol of Hydrogen is H•
In order to attain stable noble gas configuration and complete its duplet, hydrogen will gain one more electrons. This negative charge on H– infers that there will be addition of one more electron in addition to the one valence electron.
Therefore, the lewis dot symbol of Hydrogen ion is H••–
4.4. Draw the Lewis structures for the following molecules and ions :
H2S, SiCl4, BeF2, CO32–, HCOOH
Ans.
4.5. Define octet rule. Write its significance and limitations.
Ans. In 1916, Kössel and Lewis developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, 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 of Octet rule:
(i) The octet rule successfully explained the formation of chemical bonds depending upon the nature of the element.
(ii) It is useful in explaining why various atoms interact to generate ionic or covalent compounds.
Limitations of the Octet Rule:
(1) According to the Octet rule, atoms combine chemically to form the configuration of the nearest noble gas elements. However, some noble gas elements, such as Xenon and krypton form compounds with fluorine such as, XeF2, XeF4, KrF2 etc
(2) The form of molecules is not taken into account in this theory. In other words, octet rule failed to predict the shape and relative stability of molecules.
(3) If a compound is having less than 8 electrons surrounding the central atom than octet rule cannot be applied to that compound. For example, compounds such as BeH2, AlCl3, LiCl etc. do not obey the octet rule.
4.6. Write the favourable factors for the formation of ionic bond.
Ans. Ionic bonds are formed when one or more electrons are transferred from one atom to another. Hence, the ability of neutral atoms to form ionic bond depends upon the ease with which it lose or gain electrons.
The ionic bond formation is also influenced by the lattice energy of the compound being formed.
Factors that promote the formation of ionic bonds include :
(a) Non-metal atoms forming anions with high electrons gain enthalpy.
(b) A metal atoms forming cation with a low ionisation enthalpy.
(c) The high lattice energy of the ionic compound formed by the combination of cation and an anion.
4.7. Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
Ans. According to the VSEPR
theory, Type of bond formed
No. of Valence electrons
$$=\frac{+\text{No. of monovalent ions}}{2}$$
(i) BeCl2
4Be — 2, 2
Number of valence electrons = 2
Number of monovalent ions = 2
$$\text{Number of bonds =}\frac{2+2}{2}=2$$
The central atom has no lone pair and there are two bonds pairs. i.e., BeCl2 is of the type AB2. Hence, it has a linear shape.
(ii) BCl3
5B —2, 3
Number of valence electrons = 3
Number of monovalent ions = 3
$$\text{Number of bonds =}\frac{3+3}{2}=3$$
The central atom has no lone pair and there are three bonds pairs. Hence, it is of the type AB3. Hence, it is trigonal planar.
(iii) SiCl4
14Si—2, 8, 4
Number of valence electrons = 4
Number of monovalent ions = 4
$$\text{Number of bonds =}\frac{4+4}{2}=4$$
(iv) AsF5
33As—2, 8, 18, 5
Number of valence electrons = 5
Number of monovalent ions = 5
$$\text{Number of bonds =}\frac{5+5}{2}=5$$
The central atom has no lone pair and there are five bond pairs.. Hence, AsF5 is of the type AB5. Therefore, the shape is trigonal bipyramidal.
(v) H2S
16S — 2, 8, 6
Number of valence electrons = 6
Number of monovalent ions = 2
$$\text{Number of bonds =}\frac{6+2}{2}=4$$
The central atom has one lone pair and there are two bond pairs. Hence H2S is of the type AB2E. The shape is Bent.
(vi) PH3
15P — 2, 8, 5
Number of valence electrons = 5
Number of monovalent ions = 3
$$\text{Number of bonds =}\frac{5+3}{2}=4$$
The central atom has one lone pair and there are three bond pairs. Hence, PH3 is of the AB3E type. Therefore the shape is trigonal pyramidal.
4.8. Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Ans. According to VSEPR theory,
| NH3 | H2O |
| 7N—2, 5 Number of valence electrons = 5 | 8O—2, 6 Number of valence electrons = 6 |
| Number of monovalent ions = 3 | Number of monovalent ions = 2 |
| Number of bonds =$$\frac{5+3}{2}=4$$ | Number of bonds =$$\frac{6+2}{2}=4$$ |
| Bond type AB3L, shape-distorted tetrahedral or Trigonal pyramidal | Bond type AB2L2, shape- distorted tetrahedral or bent |
The central atom (N) in NH3 has one lone pair and there are three bond pairs. In H2O, there are two lone pairs and two bond pairs.
The two lone pairs present in the oxygen atom of H2O molecule repels the two bond pairs. This repulsion is stronger than the repulsion between the lone pair and the three bond pairs on the nitrogen atom.
Since the repulsion on the bond pairs in H2O molecule are greater than that in NH3, the bond angle in water is less than that of ammonia.
4.9. How do you express the bond strength in terms of bond order ?
Ans. Bond strength refers to the amount of bonding that happens between two atoms during the formation of a molecule.
Bond order is defined the number of chemical bonds present between a pair of atoms.
The larger the bond energy, the stronger is the bond and the greater is the bond order.
4.10. Define the bond length.
Ans. “Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Bond length are expressed in terms of Angstrom (10–10 m) or picometer (10–12 m) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques.
In an ionic compound, the bond length is the sum of ionic radii of the constituting atoms (d = r+ + r–. In a covalent compound, it is the sum of their covalent radii (d = rA + rB).
4.11. Explain the important aspects of resonance with reference to the CO32– ion.
Ans. According to experimental findings, all carbon to oxygen bonds in CO32– are equivalent. The three standard forms of resonance in CO32– are represented in structure I, II, and III which are as follows:
The nuclei in these formations are all in the same place.
All three forms have nearly equal amounts of energy.
The sole difference between paired and impaired electrons is that they differ only in their position.
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.
4.13. Write the resonance structures for SO3, NO2 and NO3–.
Ans.
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.
Ans. (a) The electronic configuration of K and S are as follows:
19K – 2, 8, 8, 1 and 16S – 2, 8, 6
Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:
Lewis symbols to show electron transfer between Potassium and Sulphur
(b) The electronic configuration of Ca and O are as follows:
20Ca – 2, 8, 8, 2 and 8O – 2, 6
Oxygen requires two electrons more to complex its octet, whereas calcium has two electron i.e., Argon. Hence, the electron transfer takes place as:
Lewis symbols to show electron transfer between Calcium and Oxygen
(c) The electronic configuration of Al and N are as follows:
13Al – 2, 8, 3 and 7N – 2, 5
Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons. Hence, the electron transference can be shown as:
Lewis symbols to show electron transfer between Aluminium and Nitrogen.
4.15. 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. 6C – 2, 4
Number of valence electrons = 4
Number of monovalent ions = 0
$$\text{Number of bonds =}\frac{4+0}{2}=2$$
Bond type AB2, shape - Linear
Despite the fact that C and O have distinct electronegativities and each C = O link is polar and has the same dipole moment, the CO2 molecule has a zero dipole moment. This means that the individual dipole moments are of identical amplitude and are pointing in opposing directions, cancelling each other out.
8O – 2, 6
Resultant μ = 0D
Number of valence electrons = 6
Number of monovalent ions = 2
$$\text{Number of bonds =} \frac{6+2}{2}=4$$
Bond type AB2L2, shape - distorted tetrahderal or bent
The H2O molecule, on the other hand, possesses a net dipole moment (1.84 D). Because the O—H bonds are oriented at an angle of 104.5° and do not cancel each other’s bond moments, the H2O molecule has a bent shape.
4.16. Write the significance/applications of dipole moment.
Ans. The significance of dipole moment are as follows :
- 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.
HF
2. In the determination of molecular shapes.
3. For determining the percentage of ionic character.
%(Ionic character) in A—B bond in AB
$$\text{molecule =}\bigg(\frac{100×\mu_\text{exp}}{\mu_{\text{ionic}}}\bigg)\%$$
µexp = Expotential value of dipole moment AB molecule
µionic = Calculated value of dipole moment AB molecule
4. It helps in distinguishing between cis and trans isomers.
4.17. Define electronegativity. How does it differ from electron gain enthalpy ?
Ans. Electronegativity is defined as the ability of an atom to attract a shared pair of electrons towards itself in a covalent bond.
Electronegativity of any given element is not constant. It varies according to the element to which it is bound. It is not a measurable quantity.
Electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gas anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.
4.18. Explain with the help of suitable example polar covalent bond.
Ans. When two dissimilar atoms having different electro-negativity combine to form a covalent bond, the bond pair of electron is not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, 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 the Hydrogen fluoride molecule, fluoride has a higher electronegativity than hydrogen. As a result, the shared electron pair is pushed closer to the fluorine atom, which gains a partial negative charge (δ–). The hydrogen atom will have a partial positive charge (δ+) at the same moment.
$$\delta^+\underset{(2.1)}{\text{H}}-\delta^+\underset{(4.0)}{\text{F}}$$
Polar covalent bonds or simply polar bonds, are examples of this type of covalent bond.
4.19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
Ans. The difference in electronegativities between constituent atoms determines the ionic nature of a molecule. As a result, the greater the difference, the greater will be the ionic character of a molecule.
The required ionic character order of the provided molecules is:
N2 < SO2 < ClF3 < K2O < LiF.
Ans. The correct lewis dot structure of acetic acid is as follows :
4.21. 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 ground state
$$^6\text{C}-1s^{2},2s^{2},2p_x^{1},2p_y^1\\\text{In excited state}\space 1s^{2},2s^{2},2p_x^{1},2p_y^{1},2p_z^{1}\\\text{sp}^{3}-\text{hybridised}$$
In CH4 molecule, carbon is sp3 hybridised, so it is tetrahedral in shape. For square planar, dsp2 hybridisation is required which is not possible in carbon due to the absence of d-orbital.
The tetrahedral and square planar structures of CH4 are shown as follows:
the force of repulsion in them is the minimum. Now, in a square planar geometry, the bond angle is 90° while it is 109°28′ in tetrahedral geometry. This clearly shows that the electron repulsions are less in the tetrahedral geometry as compared to the square planar geometry. Thus, methane cannot be represented by square planar structure.
4.22. Explain why BeH2 molecule has a zero dipole moment although the Be—H bonds are polar.
Ans. The molecule with symmetrical and linear geometries have zero dipole moment because they are vector in nature and the dipole of different bonds cancel with one another and in a unsymmetrical molecule and bent geometrices, the geometrices have specific dipole moment because the bond polarities do not cancel each other.
The Lewis structure of BeH2 is as follows:
$$\text{H}_\infty^\infty\text{Be}_\infty^\infty\text{H}$$
Since there is no lone pair at the central atom (Be) and there are two bond pairs, BeH2 is of the type AB2. It has a linear structure.
Dipole moments of each H–Be bond are equal and are in opposite directions.
Therefore, they nullify each other. Hence, BeH2 molecule has zero dipole moment.
While in Be – H, dipole moment of Be and H are not equal and they do not nullify each other. Hence, has a specific dipole moment.
4.23. Which out of NH3 and NF3 has higher dipole moment and why ?
Ans. In both molecules i.e., NH3 and NF3, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF3 is greater than NH3. However, the net dipole moment of NH3(1.46 D) is greater than that of NF3 (0.24 D).
This can be explained on the basis of the directions of the dipole moments of each individual bond in NF3 and NH3. These directions can be shown as:
In case of NH3, the orbital dipole due to lone pair of electron is in the same direction as the resultant dipole moment of three NH bonds. Where in NF3 the orbital dipole is in the opposite direction to the direction of resultant dipole moment of three NF bonds.
The orbital dipole because of lone pair decreases the effect of the resultant NF bond moments. This results in the low dipole moment of NF3 as compared to NH3. Therefore, the resultant dipole moment in NH3 is more than in NF3.
4.24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
Ans. Hybridisation: It is defined as the process of intermixing of atomic orbitals of slightly different energies to give rise to new hybridised orbitals having equivalent energy and identical shapes.
Orbitals’ Shapes:
(i) sp hybridisation: When one s-orbital and one p-orbital mix, this is referred to as sp-hybridisation. The Be atom, for example, undergoes sp-hybridisation in BeCl2. It has linear shape. The bond angle is 180°.
(ii) sp2 hybridisation: One s-orbital and two p-orbitals are hybridised to generate three equivalent hybrid orbitals. The three hybrid orbitals directed towards three corners of an equilateral triangle. It is, therefore, known as trigonal hybridisation.
The bond angle is 120°
(iii) sp3 hybridisation: One s-orbital and three p-orbitals are hybridised to generate four equivalent hybrid orbitals. These orbitals are directed towards the four corners of a regular tetrahedron. Thus, it has tetrahedral shape and bond angle is 109.5°.
4.25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
$$\textbf{AlCl}_\textbf{3}+\textbf{Cl}^{\textbf{-}}\xrightarrow{}\textbf{AlCl}_\textbf{4}^{\textbf{-}}$$
Ans. On the basis of the electronic configuration of 13Al atom, its valence shell electronic configuration is 3s23p1 in the ground state which changes to 3s13px13py1 in the excited state.
The valence orbital picture of aluminium in the ground state can be represented as:
The orbital picture of aluminium in the excited state can be represented as:
4.26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
$$\textbf{BF}_\textbf{3}+\textbf{NH}_\textbf{3}\xrightarrow{}\textbf{F}_\textbf{3}\textbf{B.NH}_\textbf{3}$$
Ans. Boron atom in BF3 is sp2 hybridzed. The orbital picture of boron in the excited state can be shown as:
C2H2 : In the formation of C2H2 molecule, each C-atom is sp hybridzed with two 2p-orbitals in an unhybridized state.
One sp orbital of each carbon atom overlaps with theother along the internuclear axis forming a C—C sigma bond. The second sp orbital of each C-atom overlaps a half-filled 1s-orbital to form a σ bond.
The two unhybridised 2p-orbitals of the first carbon undergo sidewise overlap with the 2p orbital of another carbon atom, thereby forming two pi (π) bonds between carbon atoms. Hence, the triple bond between two carbon atoms is made up of one sigma and two π-bonds.
4.28. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2, (b) C2H4.
Ans. A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A pi-bond is always present in it. A triple bond is a combination of two pi-bonds and one sigma bond.
(a) The structure of C2H2 can be represented as:
$$\text{H}\frac{\sigma}{}\text{C}\frac{\frac{\pi}{\sigma}}{\frac{\pi}{}}\frac{\sigma}{}\text{H}$$
Hence, there are three sigma and two pi-bonds in C2H2.
(b) The structure of C2H4 can be represented as:
Hence, there are five sigma bonds and one pi-bond in C2H4.
4.29. 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.
2py and 2py orbitals will not a form a sigma bond. Taking x-axis as the internuclear axis, 2py and 2py orbitals will undergo lateral overlapping, thereby formina pi (π) bond.
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
Ans. Hybrid orbital are formed by the overlapping of orbitals having almost same energy.
Single bond = one sigma bond = sp3 hybridization
Double bond = one sigma bond + one pie bond = sp2 hybridization
Triple bond = one sigma bond + two pie bond = sp hybridization
(a) According to structure, C1 and C2 are making four sigma bonds (single bond) each with the help of one s hybrid orbital and three p hybrid orbital, hence C1 and C2 are sp3 hybridised.
(d) From the structure it is clear that, C1 is making sigma bonds only, therefore, it is sp3 hybridised. Whereas, C2 is making a double bond therefore, it is sp2 hybridised Thus, C1 is sp3 hybridized and C2 is sp2 hybridized.
4.31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type.
Ans. A covalent bond is formed between two atoms when they mutually share one or more of their valence electrons.
Bond pairs are the shared pairs of electrons that exist between bonded atoms. All valence electrons may not participate in bonding. Thus,
lone pairs of electrons are electron pairs that do not participate in bonding.
For example, in C2H6(ethane), there are seven bond pairs but no lone pair present.
4.32. Distinguish between a sigma and a pi bond.
Ans.
| Sigma bond | π bond |
| It is formed by end-to-end (head-on)overlapping of atomic orbitals. | It is formed by sideways overlapping of atomic orbitals. |
| It involves over-lapping of s-s, s-p, and p-p orbitals. | It involves over-lapping of only p-p atomic orbitals. |
| It is a strong bond. | It is a weak bond. |
| Free rotation of atoms around s-bond is possible. | Free rotation of atoms around bond is not possible. |
| Electron cloud of sigma bond is symmetrical about internuclear axis. | Electron cloud of π bond is unsymmetrical. |
4.33. Explain the formation of H2 molecule on the basis of valence bond theory.
Ans. Consider two hydrogen atoms (A and B) with nuclei (NA and NB) and electrons (eA and eB) undergoing a process to form a hydrogen molecule. There is no interaction between A and B when they are separated by a considerable distance. The attracting and repulsive forces begin to work as they approach each other.
- An attractive force exists between:
(a) One atom’s nucleus and its own electron, i.e. NA—eA and NB—eB.
(b) One atom’s nucleus and another atom’s electron, i.e. NA—eB and NB—eA - A repulsive force exists between:
(a) Two atoms electrons, eA — eB, and
(b) Two atoms’ nuclei, NA — NB.
Since, the force of attraction brings the two atoms together, whereas the force of repulsion tends to push them apart.
The attracting forces have a greater magnitude than the repulsive forces. As a result, the two atoms are approaching one other. As a result, the potential energy of the system is reduced. Finally, the system reaches a condition where the attracting and repulsive forces are balanced and the system has the least amount of energy. A dihydrogen molecule is formed as a result of this reaction.
4.34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Ans. To generate molecular orbitals, atomic orbitals must satisfy the following conditions:
(a) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, an atom’s 1s-atomic orbital can join with another atom’s 1s-atomic orbital, but not with the 2s-orbital.
(b) To ensure maximum overlap, the combining atomic orbitals must be properly oriented.
(c) The amount of overlap should be substantial.
4.35. Use molecular orbital theory to explain why the Be2 molecule does not exist.
Ans. The electronic configuration of 4Be is 1s2 2s2
The molecular orbital electronic configuration for Be2 molecule can be written as :
(σ 1s)2, (σ *1s)2, (σ 2s)2, (σ *2s)2
The molecular orbital diagram of Be can be represented as:
$$\text{Since, bond order =}\frac{1}{2}(\text{N}_b-\text{N}_a)$$
As the bond order of Be2 molecule is zero, thus, Be2 molecule does not exist.
4.36. Compare the relative stability of the following species and indicate their magnetic properties;
O2, O2+, O2– (superoxide), O22– (peroxide)
Ans. The electronic configuration of 8O is 1s2 2s2 2p4
There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule be written as:
[σ (1s)2][σ*(1s)2][σ(2s)2][σ(1pz)2][π(2px)2]
[π(2py)2][π*(2px)1][p*(2py)1]
Since the 1s orbital of each oxygen atom is not involved in bonding, the number of bonding electrons = 8 = Nb and the number of anti-bonding orbitals = 4 = Na.
$$\text{Bond order =}\frac{1}{2}(\text{N}_b-\text{N}_a)\\=\frac{1}{2}(8-4)\\= 2$$
Due to presence of unpaired electrons in π*2px and π*2py orbitals., it is paramagnetic in nature.
Similarly,
The molecular orbital electronic configuration for O2+ molecule can be written as:
O2+ : ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2, (π2px)2 = (π2py)2, (π*2px)1
$$\text{Bond order of O}_2=\frac{1}{2}[10-5]=\frac{5}{2}=2.5$$
Due to presence of unpaired electrons in σ *2px orbital, it is paramagnetic in nature.
The molecular orbital electronic configuration for O2– molecule can be written as :
O2– : ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2, (π2px)2 = (π2py)2, (π*2px)2 = (π*2py)1
$$\text{Bond order of O}_2^{-}=\frac{1}{2}[10-7]=\frac{3}{2}=1.5$$
Due to presence of unpaired electrons in π*2py orbital., O2– is also paramagnetic in nature.
∴ O22– is also paramagnetic like O2+, but less paramagnetic than O2.
The molecular orbital electronic configuration for O22– molecule can be written as :
O22–: ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2,
(π2px)2 = (π2py)2, (π*2px)2 =(π*2py)2
$$\text{Bond order of O}_2^{2-}=\frac{1}{2}[10-8]=\frac{2}{2}=1$$
Since, there is no unpaired electrons, O22– is diamagnetic in nature.
Thus, O2 is more paramagnetic than O2+ or O2, whereas O22– is diamagnetic.
Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be stability. On this basis, the order of stability is : O2+ > O2 > O2– > O22–.
4.37. Write the significance of a plus and a minus sign shown in representing the orbitals.
Ans. Wave functions are used to illustrate molecular orbitals. A positive wave function is represented by a plus sign in an orbital, while a negative wave function is represented by a minus sign in an orbital.
4.38. Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
Ans. The ground state and excited state outer electronic configuration of phosphorus (Z = 15) is:
Thus, the hybridization of PCl5 is sp3d2 and its shape will be trigonal bipyramidal.
All the bond angles in this structure are not equivalent.
There are five P—Cl sigma bonds in PCl5. Three P—Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.
The remaining two P—Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds.
As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.
4.39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Ans. A hydrogen bond is defined as an attractive force between hydrogen attached to one molecule’s electronegative atom and hydrogen attached to another molecule’s electronegative atom (may be of the same kind).
The bond pair between hydrogen and the electronegative atom drifts far away from the hydrogen atom due to a difference in electronegativities. As a result, a hydrogen atom gains a positive charge and becomes electropositive with regard to the other atom.
Hδ+—Xδ– ......... Hδ+ — Xδ– ...........
In the solid state, the magnitude of H-bonding is maximum, while in the gaseous state, it is minimum.
H - bonds are again divided into two types:
A. Intermolecular H-bonds, such as HF and H2O.
B. Intramolecular H-bonds, such as o-nitrophenol
4.40. What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–.
Ans. Bond order is defined the number of chemical bonds present between a pair of atoms.
Or
It is defined as the one - half the difference between the number of electrons present in the bonding and antibonding orbitals of a molecule.
$$\text{Bond order =}\frac{1}{2}(\text{N}_b-\text{N}_a)$$
Where, Nb is the number of electrons in the bonding orbitals
Na is the number of electrons in the antibonding orbitals
N2 molecule : The molecular orbital electronic configuration for N2 molecule can be written as:
N2 : (s1s)2, (s*1s)2, (s2s)2, (s*2s)2, (p2px)2 = (p2py)2, (s2pz)2
Number of bonding electrons, Nb = 10
Number of anti-bonding electrons, Na = 4
$$\text{Bond order of nitrogen molecule =}\frac{1}{2}(10-4)=3$$
Bond order of O2 molecule : There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom.
The molecular orbital electronic configuration for O2 molecule can be written as :
O2 : ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2, (π2px)2 = (π2py)2, (π*2px)1 = (π*2py)1
$$\text{Bond order of O}_2 =\frac{\text{N}_b-\text{N}_a}{2}\\=\frac{10-6}{2}=\frac{4}{2}=2$$
Bond order of O2+ molecule :
The molecular orbital electronic configuration for O2+ molecule can be written as :
O2+ : ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2, (π2px)2 = (π2py)2, (π*2px)1
$$\text{Bond order of O}_2\space^+ =\frac{\text{N}_b-\text{N}_a}{2}\\=\frac{10-5}{2}=2\frac{1}{2}=2.5$$
Bond order of O2– molecule :
The molecular orbital electronic configuration for O2– molecule can be written as :
O2– : ( σ 1s)2, ( σ *1s)2, ( σ 2s)2, ( σ *2s)2, ( σ 2pz)2, (π2px)2 = (π2py)2, (p*2px)2 = (π*2py)1
$$\text{Bond order of O}_2\space^- =\frac{\text{N}_b-\text{N}_a}{2}\\=\frac{10-7}{2}=\frac{3}{2}=1\frac{1}{2}=1.5$$
Hence,
The bond order of nitrogen molecule is 3.
The bond order of oxygen (O2) molecule is 2.
The bond order of O2+ ion is 2.5.
The bond order of O2– ion is 1.5.
NCERT Solutions for Class 11 Chemistry Chapter 4 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
