RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Rajasthan Board RBSE Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

RBSE Class 11 Chemistry Chapter 4 Text Book Questions

RBSE Class 11 Chemistry Chapter 4 Multiple Choice Questions

Question 1.
PCl5 exists but NCl5 does not, why?
(a) Nitrogen do not have vacant d-orbitals
(b) Ionisation energy of nitrogen is very high.
(c) Nitrogen molecule is not similar to chlorine molecule.
(d) None of the above.
Answer:
(a) Nitrogen do not have vacant d-orbitals

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 2.
Which of the following has zero dipole moment?
(a) ClF
(b) PCl3
(c) SiF5
(d) CFCl3
Answer:
(c) SiF5

Question 3.
The bond order of N2 is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(a) 3

Question 4.
Which of the following has lone pair of electrons?
(a) NO
(b) CO
(c) NH3
(d) O2
Answer:
(a) NO

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 5.
Which of the following molecule/ion shows sp2 hybridisation?
(a) BF3 and \(\mathrm{NH}_{2}^{-}\)
(b) NO2 and NH3
(c) BF3 and \(\mathrm{NO}_{2}^{-}\)
(d) \(\mathrm{NH}_{2}^{-}\) and H2O
Answer:
(c) BF3 and \(\mathrm{NO}_{2}^{-}\)

RBSE Class 11 Chemistry Chapter 4 Very Short Answer Type Questions

Question 6.
Write Lewis dot structure for the following compounds: S, S2-, Al, Al3+
Answer:
(a) S and S2-: Atomic number of sulphur is 16 and electronic configuration is [Ne] 3s2 3p4, i.e., it has 6 electrons in its valence shell. The Lewis dot symbol of sulphur (S) is
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 1
For S2-, di-negative anion is formed by addition of two electrons. Now, the electronic configuration will be [Ne] 3s2 3p6, Hence, the Lewis dot symbol of S2-is
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 2
(b) Al and Al3+: Atomic number of aluminium is 13. Its electronic configuration will be [Ne] 3s23p1 i.e., the valence electron is 3. The Lewis dot symbol of aluminium (Al) is
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 3
The tri-positive ion of aluminium is formed by loss of 3 electrons. Thus, it has three less electrons than aluminium. Its possible electronic configuration will be [Ne]. Hence, the Lewis dot symbol is [Al]3+.

Question 7.
Write the electronic configuration of Li2 molecule.
Answer:
Electronic configuration of Li2 molecule is (σ 1 s)2 (σ*1s)2 (σ 2 s)2.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 8.
How do you express the bond strength in terms of bond order?
Answer:
Bond order is directly proportional to bond strength. More the bond order, more will be the bond strength. Bond strength increases with increase in bond order.

Question 9.
Arrange the following molecules in increasing order of energy: N2, O2, Cl2, F2.
Answer:
N2 > O2 > Cl2 > F2 .

Question 10.
Explain the important aspects of resonance with reference to the \(\mathrm{CO}_{3}^{2-}\) ion.
Answer:
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately. These contributing structures are known as canonical forms. Resonance is represented by a double headed arrow. For resonance, the contributing structures should have same number of unpaired electrons, should have same atomic positions but differ in position of electrons, should have nearly same energy. Resonance in carbonate ion can be represented as :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 4
The actual structure is an average of these three resonating structures.

Question 11.
Write the hybridization of central atom of the following molecules: CCl4, H2O, CO2, SO2. [Hint : sp3, sp2, sp] .
Answer:
Hybridization of carbon atom in CCl4 is sp3.
Hybridization of oxygen atom in H2O is sp3.
Hybridization of carbon atom in CO2 is sp.
Hybridization of sulphur atom in SO2 is sp2.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 12.
Water is liquid at room temperature. Why?
Answer:
It is the polarity of the water molecules that makes it a liquid at room temperature. The water molecule is dipolar because of the great difference in the polarities of oxygen and hydrogen. Thus, the hydrogen bond develops between the H and O of adjacent water molecules and it is because of this strong hydrogen bond that packs the water molecules closely together, water is a liquid at room temperature.

Question 13.
Melting and Boiling point of ionic compounds are high. Why?
Answer:
Ionic compounds have high melting and boiling points because the electrostatic interactions that hold the compounds together are very strong. Large amount of energy is needed to break strong electrostatic forces that hold oppositely charged ions together in solid lattice of ionic compounds like NaCl. So, ionic compounds have high melting, and boiling points.

Question 14.
AlCl3 is covalent where as AlF3 is ionic. Why ?
Answer:
According to Fajan’s rule, with increasing size of the anion and with decreasing size of the cation, ionic character of the compound decreases while the covalent character increases.

Here, since the cation is same in both the compounds, the size of the anion decides the nature of the compound. Size of chlorine being greater than fluorine, AlCl3 is covalent while AlF3 is ionic.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 15.
Write the resonance structures of SO3 and NO2.
Answer:
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 5

Question 16.
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) Is and Is (b) Is and 2px ; (c) 2py and 2py (d) 1s and 2s.
Answer:
A sigma bond is formed by axial overlapping of atomic orbital. It is always formed between two half filled atomic orbitals along their internuclear axis i.e the line joining the centers of the nuclei of two atoms. 2py, and 2py will not form sigma bond because they can only undergo sidewise overlapping of atomic orbitals and form pi bond.

Question 17.
Which of the following has maximum dipole moment: Na+, K+, Li+ and Cs+.
Answer:
Dipole moment is directly proportional to electronegativity. Electronegativity decreases with increasing atomic size. Thus, the ion having smallest size has maximum electronegativity and thus, maximum dipole moment. Also, atomic size increases down the group. Therefore, in the above ions, Li+ has smallest size. Hence, it has maximum dipole moment.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 18.
Arrange the halides of silver in increasing order of solubility in water.
Answer:
As we move down the group in periodic table, the bond’s covalent character increases, which invariably is related to the decrease in difference between the electronegativities of the bonding atoms, which means decrease in polarity of the molecule. Also as the size of anion increases, the tendency to polarise cation increases. Covalent compounds are insoluble in water. AgF will be most ionic compound. Therefore, the order of solubility of halides of silver will be:
Agl < AgBr < AgCl < AgF

Question 19.
NaCl is an ionic compound whereas CuCl is a covalent compound. Explain the reason.
Answer:
When a cation approaches an anion, the electron cloud of the anion is attracted towards the cation and hence gets distorted. This effect is called polarization. The greater the polarisation produced, more is the concentration of the electrons between the two atoms thereby decreasing the ionic character or increasing the covalent character. The covalent character of any compound in general depends upon size of anion, charge on ions and electronic configuration.

If two cations have the same charge and size, the one with pseudo noble gas configuration i.e. having 18 electrons in the outermost shell has greater polarizing power than a cation with noble gas configuration i.e having 8 electrons.
Electronic configuration of Cu+ : 2, 8, 18
Electronic configuration of Na+ : 2, 8
Therefore, CuCl is more covalent than NaCl although Cu+ ion (0.96Å) and Na+ ion (0.95Å) have same size and charge.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 20.
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl ➝ \(\mathrm{AlCl}_{4}^{-}\)
Answer:
The formation of AlCl3 can be explained as :
Electronic configuration of 13 Al = 1s2 2s2 2p6 3s2 3p1
Electronic configuration of 17 Cl = 1s2 2s2 2p6 3s2 3p5
In excited state, one electron in Al gets promoted to p-orbital. It can be shown as:
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 6
Thus, it has three unpaired electrons. Chlorine needs one electron to complete its octet. Three Cl atoms hybridise with Al to undergo sp2 hybridisation to give a trigonal planar arrangement in AlCl3. But in the formation of \(\mathrm{AlCl}_{4}^{-}\) the empty 3pz orbital will also be involved in the hybridisation and thus the hybridization changes from sp2 to sp3. As a result, the shape gets changed to tetrahedral.

RBSE Class 11 Chemistry Chapter 4 Short Answer Type Questions

Question 21.
Define octet rule. Write its significance and limitations.
Answer:
According to “Electronic theory of chemical bonding” of chemical combination, 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 and become stable. This is known as “Octet Rule”. Significance of Octet Rule : Octet rule is helpful in understanding the bonding and structures of most of the organic compounds.

Limitations of Octet Rule :
1. In some compounds, the number of electrons surrounding the central atom is less than eight especially with elements having less than four valence electrons. Examples are LiCl, BeH2 and BCl3.Li, Be and B have 1, 2 and 3 valence electrons only.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 7
2. The elements of third period and beyond that have 3d orbitals in addition to 3s and 3p. In these elements, there are more than eight valence electrons around central metal atom. Some of the examples of such compounds are: PF6, SF6, H2SO4 and a number of coordination compounds.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 8
3. In molecules with an odd number of electrons like nitric oxide and nitrogen dioxide, the octet rule is not satisfied for all the atoms. In NO, octet of nitrogen is not complete.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 9
4. This theory does not account for shape of molecules.
5. It does not explain the relative stability of the molecules being totally silent about the energy of a molecule.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 22.
Which of the following pairs are more covalent and why?
(a) CuO and CuS
(b) AgCl and Agl
(c) PbCl2 and PbCl4
(d) NaCl and CuCl
Answer:
Factors Governing Polarization and Polarisability (Fajan’s Rule)
Cation Size : Smaller is the cation, more is the value of charge density (φ) and hence more its polarising power.
Cationic Charge : More is the charge on the cation, the higher is the value of (φ) and higher is the polarising power.

Noble Gas vs Pseudo Noble Gas Cation : Greater departure from noble gas configuration, greater covalency.

Anion Size : Larger is the anion, more is the polarisability and hence more covalent character is expected.

Anionic Charge : Larger is the anionic charge, the more is the polarisability.
(a) CuO and CuS-Here cation is same and S anion is larger than Oxygen anion. So, there will be more covalent character in CuS than CuO.

(b) AgCl and Agl— Since Cl is smaller in size, its polarisability is less and therefore it is having more ionic

(c) PbCl2 and PbCl4 — Greater the charge on cation, more is the polarising power and hence more covalent character. As polarising power of Pb4+ > Pb2+. Thus PbC l4 is more covalent than PbCl2.

(d) NaCl and CuCl—If two cations have the same charge and size, the one with pseudo noble gas configuration i.e. having 18 electrons in the outermost shell has greater polarizing power than a cation with noble gas configuration i.e. having 8 electrons.
Electronic configuration of Cu+ : 2, 8, 18
Electronic configuration of Na+ : 2,8
Therefore, CuCl is more covalent than NaCl although Cu+ ion (0.96Å) and Na+ ion (0.95Å) have same size and charge.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 23.
Which out of NH3 and NF3 has higher dipole moment and why?
Answer:
NH3 and NF3 both the molecules have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH3 (4.90 × 10-30 C m) is greater than that of NF3 (0.8 × 10-30 Cm). This is because, in case of NH3 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N – H bonds, whereas in NF3 the orbital dipole is in the direction opposite to the resultant dipole moment of the three N-F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N- F bond moments,-which results in the low dipole moment of NF3 as represented below :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 10

Question 24.
Distinguish between a sigma and a pi bond.
Answer:
The difference between a sigma and pi bond is:
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 11

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 25.
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
The linear combination of atomic orbitals to form molecular orbitals takes place only if following conditions are fulfilled :
1. The combining atomic orbitals must have the same or nearly the same energy. This means that in diatomic homonuclear molecules, 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 like HF, HCl etc.
2. The combining atomic orbitals must overlap to the maximum extent. Greater the extent of overlap, the greater will be the electron density between the nuclei of a molecular orbital.
3. The combining atomic orbitals must have the same symmetry about the molecular axis. By convention, z-axis is taken as the molecular ahxis. It is important to note that atomic orbitals having same Or nearly the same energy will not combine if they do not have the same symmetry. For example, 2pz orbital of one atom can combine with 2pz obital of the other atom but not with the 2px or 2py orbitals because of their different symmetries.

Question 26.
\(\mathrm{He}_{2}^{+2}\) ion is more stable than He2 molecule. Explain.
Answer:
The \(\mathrm{He}_{2}^{+2}\) ion has only two valence electrons (two from each He atom minus two for the +2 charge). We can now fill the molecular orbital diagram :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 12
The two electrons occupy the lowest-energy molecular orbital, which is the bonding (σ 1s) orbital, giving a (σ 1s)2 electron configuration. So the bond order is
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 13
\(\mathrm{He}_{2}^{+2}\) is therefore predicted to contain a single He-He bond. Thus it should be a stable species.
Let us examine the He2 molecule, formed from two He atoms with 1s2 electronic configurations.
With a total of four valence electrons, both the σ 1s bonding and (σ * 1s) antibonding orbitals must contain two electrons. This gives a (σ1s)2 (σ1s)2 electronic configuration.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 14
which indicates that the He2 molecule has no net covalent bond and is not a stable species.

Question 27.
Among HCl, H2O and NH3, which has higher boiling point and why?
Answer:
The order of the boiling point is H2O > NH3 > HCl. The extent of Hydrogen bonding is H2O > NH3 > HCl. Each water molecule can potentially form “four” hydrogen bonds with surrounding water molecules. There are exactly the right numbers of hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. Ammonia could form only “three” hydrogen bond per molecule. So, due to strong hydrogen bonds, more energy (heat) is required to boil water. Due high electronegativity of nitrogen then chlorine, bond between N-H is stronger than bond between H-Cl. That is why boiling point of NH3 is more than HCl.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 28.
Explain the structure of ClF3 on the basis of VSEPR theory.
Answer:
The standard application of VSEPR theory to Chlorine trifluoride is as follows: In chlorine trifluoride, central atom is chlorine and valence electrons on central atom is 7. Also, contribution of three fluorine atoms is 1 electron each. Therefore, there are total 10 electrons or five electron pairs. As we know that the highest repulsion is between any two “lone electron pairs”, resulting in these moving apart as far as possible. The next highest repulsion is between one lone pair and a bond pair and the lowest is between two bond pairs.

As applied to chlorine trifluoride, it results in a trigonal bipyramidal geometry for the shape-determining five electron pairs. Three of these are bond pairs and two are lone pairs. These keep as far apart as possible, minimising repulsion between each of the negatively charged clouds by adopting a trigonal bipyramidal arrangement. The two lone pairs occupy equatorial position at an angle of 120° to each other, this gives the lowest energy arrangement of electron pairs in the molecule. Because repulsion involving lone pairs are stronger than bond pairs. Thus, F-Cl-F angle is a little less than 180°. Therefore, molecule has a T-shape geometry.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 15

Question 29.
Dipole moment of CO2 molecule is zero where as SO2 has some dipole moment. Explain the reason.
Answer:
The best way to understand geometry is by drawing the Lewis dot structure of the molecules.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 16
The geometry of CO2 is linear (electron geometry is linear too) because the two bonds are the farthest apart and the bond angle is 180°. This is what makes of CO2 a non polar molecule because the dipole moments of both bonds cancel out.

The geometry of SO2 is bent (electron geometry is trigonal planar) because the lone pair on the central sulphur atom pushes the two bonds down for a bond angle of 120° due to the electrons repulsion. This is what makes of SO2. a polar molecule because the dipole moments of both bonds do not cancel out, and the molecule will have a net dipole moment.

Question 30.
BaSO4 is ah ionic compound, yet it is insoluble in water. Why ?
Answer:
BaSO4 is an ionic compound, yet it is insoluble in water because both Ba2+ and \(\mathrm{SO}_{4}^{2-}\) are big and bigger cation stabilises bigger anion strongly hence BaSO4 has high lattice enthalpy. For a compound to be soluble, it’s lattice enthalpy should be low and it should decrease more rapidly than hydration enthalpy. Now since both so, it is insoluble in water.

Question 31.
Explain any four factors which affect the solubility of ionic compounds.
Answer:
The solubilities of ionic compounds are affected by solute-solvent interactions, the common ion effect, temperature and molecular size.
1. Solute-solvent attractions : Strong solute -solvent attractions increase solubility of ionic compounds. Ionic compounds are most soluble in polar solvents like water, because the ions of the solid are strongly attracted to the polar solvent molecules.

2. Common ion Effect : Ionic compounds are less soluble in solvents that contain a common ion. For example, CaSO4 is slightly soluble in water.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 17
If the water already Contains calcium ions or sulphate ions, the position of equilibrium moves to the left and the solubility decreases.

3. Temperature : Increasing the temperature usually increases the solubility of an ionic compound because the reaction is usually endothermic.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 18

4. Molecular size : The larger the molecules of the solute, the larger is their molecular weight and their size. It is more difficult for solvent molecules to surround “bigger molecules. If all of the above mentioned factors are excluded, a general rule can be found that larger particles are generally less soluble. If the pressure and temperature are the same then out of two solutes of the same polarity, the one with smaller particles is usually more soluble.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 32.
What do you understand by polar covalent bond. Explain with example.
Answer:
Covalent bond is formed by sharing of electrons. When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons will not be shared by both the atoms equally. As a result, the bond pair shifts towards the nucleus of the atom having greater electronegativity. Thus, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom. 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.

Example of polar covalent bond is HCl molecule. In HC1 molecule, Cl is more electronegative than H. As a result, electrons cloud gets displaced towards chlorine and it gets partial negative charge whereas hydrogen gets slightly positive charge.

Question 33.
Which of the following compound has weakest and strongest hydrogen bonding: NH3, PH3, H2O and H2S.
Answer:
There are two conditions for hydrogen bonding to occur :
1. The hydrogen in the molecule is bonded to a highly electronegative atom (usually, N, O, or F)
2. An electronegative atom (also usually N, O, or F) should possess small size that can be used to form a hydrogen bond.
The stability of hydrides decreases down the group due to decrease in bond dissociation energy down the group.
NH3 > PH3 > ASH3 > SbH3 > BiH3

The electronegativity of N (N- 3.0) is much higher that of H( H-2.1). Due to this, N-H bond is quite polar and hence NH3 undergoes intermolecular H-bonding whereas no hydrogen bonding occurs in PH3.
Hydrogen sulphide has sulphur; while it is one period below oxygen, it is not as electronegative as oxygen . This does not allow condition (1) to occur, although sulphur does have lone pairs which could possible be used to form a hydrogen bond. On the other hand, both P and H have an electronegativity of 2.1. Therefore, P-H bond is non polar and hence PH3 does not form H-bonds.

Question 34.
Explain the factors affecting bond length.
Answer:
Factors affecting bond length :
(i) The bond length increases with increase in the size of the atoms. For example, bond length of H – X is in the order, HI > HBr > HCl > HF.

(ii) The bond length decreases with the multiplicity of the bond. Thus, bond length of carbon – carbon bonds is in the order, C ≡ C < C = C < C-C.

(iii) As an s-orbital is smaller in size, greater the s-character, shorter is the hybrid orbital and hence shorter is the bond length.
For example, sp3 (C – H) > sp2 (C = H) > sp (C ≡ H)

(iv) Polar bond length is usually smaller than the theoretical non-polar bond length.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 35.
Draw diagrams showing the formation of a triple bond between carbon atoms in C2H2 molecule.
Answer:
C2H2
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 19

RBSE Class 11 Chemistry Chapter 4 Long Answer Type Questions

Question 36.
Explain the formation of H2 molecule on the basis of Valence Bond Theory.
Answer:
According to VBT, atoms combine so that their energy decreases and stability increases. Consider two hydrogen atoms A and B are approaching each other. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms approach each other, new attractive and repulsive forces begin to operate. These are: .

  • Attractive force between nucleus of HA and electron of HB.
  • Attractive force between nucleus of HB and electron of HA.
  • Repulsive force between nuclei of both the atoms.
  • Repulsive force between electrons of both the atoms.

Experimentally, it has been found that the magnitude of new attractive force is more than new repulsive force. As a result, two atoms approach each other and potential energy decreases. Ultimately a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage, two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm. At this stage, the minimum distance between the two nuclei which is necessary for minimum energy and maximum stability is called bond length and the energy released in gaseous state by combination of two isolated atoms to form a molecule is called bond energy.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 20

Question 37.
Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to
Answer:
The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are 1s2 2s2 2p6 3s2 3p3 and 1s2 2s2 2p6 3s1 3p3 3d1 respectively.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 21
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 bipyramidal. 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 bonds 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 axial bonds suffer more repulsion than equatorial bonds, these are found to be slightly longer and hence slightly weaker than equatorial bonds.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 22

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 38.
What is meant by hybridisation of atomic orbitals? Describe sp, sp2 and sp3 hybridization with examples.
Answer:
Hybridization is the phenomenon of intermixing of orbitals of slightly different energies so as to redistribute their energies and to give new set of orbitals of equivalent energy and shape. The atomic orbitals taking part in hybridisation are called hybrid orbitals. These hybrid orbitals have minimum repulsion between their electron pairs and thus are more stable. Only orbitals of almost similar energies and belonging to same atom or ion undergo hybridization. The number of hybrid orbitals produced is equal to the number of pure orbitals mixed during hybridization. The type of hybridisation indicates the geometry of the molecules. Only atomic orbitals participate in hybridisation. Electron’s present in them do not take part.

(a) sp hybridisation : This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two equivalent sp hybrid orbitals. Each sp hybrid orbital has 50% s-character and 50% p-character. Such a molecule in which the central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry.

This type of hybridisation is also known as diagonal hybridisation. The bond angle is 180°.

(b) sp2 hybridisation : In this hybridisation, there is involvement of one s and two p-orbitals in order to form three equivalent sp2 hybridised orbitals. The three hybrid orbitals so formed are oriented in a plane at maximum distance to form an angle of 120° between the hybrid orbitals. The hybrid orbitals are oriented in a trigonal planar arrangement

(c) sp3 hybridisation : In this type of hybridisation, there is mixing of one s-orbital and three p-orbitals of the valence shell to form four sp3 hybrid orbital of equivalent energies and shape. There is 25% s-character and 75% p-character in each sp3 hybrid orbital. This hybridisation has tetrahedral geometry.

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 39.
What is Molecular Orbital Theory. With the help of energy levels in homonuclear diatomic orbitals, arrange the following species in increasing order of stability
\(\mathrm{O}_{2}^{2-}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}\) Answer:
Molecular Orbital Theory (MOT) : Molecular Orbital Theory (MOT) was developed by F. Hund and R.S
Mulliken in 1932. Main postulates of this theory are :
1. Atomic orbitals of comparable energy and proper symmetry combine together to form molecular orbitals.
2. The movement of electrons in a molecular orbital is influenced by all the nuclei of combining atoms. (Molecular orbital is polycentric in nature)
3. The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals (AO’s) combine together two molecular orbitals (MO’s) are formed. One molecular orbital possesses higher energy than corresponding atomic orbitals and is called anti bonding molecular orbital (ABMO) and the other has lower energy and is called bonding molecular orbitals (BMO).
4. In molecules, electrons are present in molecular orbitals. The electron filling is in accordance with Pauli’s exclusion principle, Aufbau principle and Hund’s rule.
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 23
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 24
Bond order = (10 – 5)/2 = 5/2 = 2.5
The bond order decreases in the order is :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 25
As we know that stability is directly proportional to bond order.
Hence, increasing order of stability is :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 26

RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Question 40.
What do you mean by term bond order? Calculate the bond order of: N2, C2, H2, N2.
Answer:
Bond order is defined as one half the difference between the number of electrons present in the bonding and the anti-bonding orbitals.
Bond order (B.O.) = 1/2 (Nb – Na)
where, Nb and Na are number of electrons in bonding and anti bonding molecular orbitals respectively. Integral bond order values of 1, 2 or 3 correspond to single, double or triple bonds respectively.

A positive bond order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb < Na) or zero (i.e., Nb = Na) bond order means an unstable molecule.

(a) Bond order of N2
Molecular orbital electronic configuration of N2
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 27
Number of bonding electrons = 10
Number of anti-bonding electrons = 4
Bond order of nitrogen molecule = 1/2 (10 – 4) = 3

(b) Bond order of C2
Total number of electrons in C2 molecule is 6 + 6 = 12.
Molecular orbital electronic configuration of C2 :
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 28
The bond order is 2 and so carbon molecule will have a double bond.

(c) Bond order of \(\mathbf{H}_{2}^{-}\)
Total number of electrons =2 + 1 = 3
Molecular orbital electron is configuration
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 29

(d) Bond order of \(\mathbf{N}_{2}^{-}\)
Total number of electrons = 14 + 1 = 15
Molecular orbital electronic configuration
RBSE Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure img 30
Electrons in bonding orbitals : 10
Electrons in anti bonding orbitals : 5
Bond order of \(\mathbf{N}_{2}^{-}\) = 1/2(10 – 5) = 2.5

RBSE Solutions for Class 11 Chemistry