Important Questions for CBSE Class 11 Chemistry Chapter 4 – Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Class 11 Important Questions for Chemistry Chapter 4

Chemistry is a subject requiring deep, analytical and critical thinking. It has three categories namely Physical, Organic and Inorganic chemistry. 

Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure is the study of the structure of monoatomic and diatomic molecules and associated chemical bonding with them. The important concepts included in the chapter are:

• Kössel-Lewis Approach To Chemical Bonding
• Ionic Or Electrovalent Bond
• Bond Parameters
• The Valence Shell Electron Pair Repulsion (VSEPR) Theory
• Valence Bond Theory
• Hybridisation
• Molecular Orbital Theory 
• Bonding In Some Homonuclear Diatomic Molecules
• Hydrogen Bonding. 

Extramarks has been providing great online study resources for Class 11 and Class 12 students. Along with NCERT solutions and chapter-specific study notes, we have created the Important Questions Class 11 Chemistry Chapter 4 booklet that will help students to revise all the topics from the chapter while solving these questions. The questions are taken by our Chemistry expert teachers from various sources including NCERT textbooks, NCERT Exemplar books, past years’ question papers, CBSE sample papers, etc. 

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Class 11 Chemistry Chapter 4 Important Questions with Answers

Below are a few of the important questions and their solutions included in our Class 11 Chemistry Chapter 4 Important Questions:

Question 1. Isostructural species are those that have the same shape and hybridisation.

Among the given species, clarify the isostructural pairs.

(i) [NF3 and BF3]

(ii) [BF4– and NH4+]

(iii) [BCl3 and BrCl3]

(iv) [NH3 and NO3–]

Answer 1. Option (ii) is the answer.

Question 2. Polarity in a molecule and hence the dipole moment depends primarily on

the constituent atoms’ electronegativity and the molecule’s shape. Which of

the following has the highest dipole moment?

(i) CO2

(ii) HI

(iii) H2O

(iv) SO2

Answer 2. Option (iii) is the answer.

Question 3. The types of hybrid orbitals of nitrogen in NO2+, NO3– and NH4+respectively expected to be

(i) sp, sp3 and sp2

(ii) sp, sp2 and sp3

(iii) sp2, sp and sp3

(iv) sp2, sp3 and sp

Answer 3. Option (ii) is the answer.

Question 4. Hydrogen bonds are formed in many compounds, e.g., H2O, HF, and NH3. The boiling point of such compounds depends largely on the strength of the hydrogen bond and the number of the hydrogen bonds. The right decreasing order of the boiling points of the above compounds is :

(i) HF > H2O > NH3

(ii) H2O > HF > NH3

(iii) NH3 > HF > H2O

(iv) NH3 > H2O > HF

Answer 4. Option (ii) is the answer.

Question 5. In the PO43- ion, formal charge on the oxygen atom of the P–O bond is

(i) + 1

(ii) – 1

(iii) – 0.75

(iv) + 0.75

Answer 5. Option (ii) is the answer.

Question 6. In NO3–ion, the number of the bond pairs and lone pairs of electrons on the nitrogen

atom is

(i) 2, 2

(ii) 3, 1

(iii) 1, 3

(iv) 4, 0

Answer 6. Option (iv) is the answer.

Question 7. Which of the following species will have tetrahedral geometry?

(i) BH4–

(ii) NH2–

(iii) CO32-

(iv) H3O+

Answer 7. Option (i) is the answer.

Question 8. The number of π bonds and σ bonds in the following structure is–

(i) 6, 19

(ii) 4, 20

(iii) 5, 19

(iv) 5, 20

Answer 8. Option (iii) is the answer.

Question 9. Which molecule/ ion out of the following does not contain unpaired electrons?

(i) N2+

(ii) O2

(iii) O22-

(iv) B2

Answer 09. Option (iii) is the answer.

Question 10. In which of the following molecules and ions all the bonds are not equal?

(i) XeF4

(ii) BF4–

(iii) C2H4

(iv) SiF4

Answer 10. Option (iii) is the answer.

Question 11. In which of the following substances will the hydrogen bond be strongest?

(i) HCl

(ii) H2O

(iii) HI

(iv) H2S

Answer 11. Option (ii) is the answer.

Question 12. If the electronic configuration of the element is 1s2 2s2 2p6 3s2 3p6 3d2 4s2, the

four electrons involved in the chemical bond formation will be_____.

(i) 3p6

(ii) 3p6, 4s2

(iii) 3p6, 3d2

(iv) 3d2, 4s2

Answer 12. Option (iv) is the answer.

Question 13. Which of the following angles corresponds to sp2 hybridisation?

(i) 90°

(ii) 120°

(iii) 180°

(iv) 109°

Answer 13. Option (ii) is the answer

Question 14. Which of the given formulas may represent the stable form of A:

(i) A

(ii) A2

(iii) A3

(iv) A4

Answer 14. Option (i) is the answer.

Question 15. Which of the given formulas may represent the stable form of C:

(i) C

(ii) C2

(iii) C3

(iv) C4

Answer 15. Option (ii) is the answer.

Question 16. The molecular formula of the compound formed from the B and C will be

(i) BC

(ii) B2C

(iii) BC2

(iv) BC3

Answer 16. Option (iv) is the answer.

Question 17. The bond in between B and C will be

(i) Ionic

(ii) Covalent

(iii) Hydrogen

(iv) Coordinate

Answer 17. Option (ii) is the answer.

Question 18. Which of the following order of the energies of molecular orbitals of N2 is correct?

(i) (π2py ) < (σ2pz) < (π*2px) ≈ (π*2py)

(ii) (π2py ) > (σ2pz) > (π*2px) ≈ (π*2py )

(iii) (π2py ) < (σ2pz) > (π*2px) ≈ (π*2py )

(iv) (π2py ) > (σ2pz) < (π*2px) ≈ (π*2py )

Answer 18. Option (i) is the answer.

Question 19. Which of the following statement is not right from the viewpoint of the molecular

orbital theory?

(i) Be2 is not a stable molecule.

(ii) He2 is not stable but He2+

is expected to exist.

(iii) Bond strength of N2 is the maximum amongst the homonuclear diatomic

molecules belonging to the second period.

(iv) The order of energies of the molecular orbitals in the N2

molecule is

σ2s < σ*2s < σ2pz< (π2px = π2py ) < (π*2px= π*2py) < σ*2pz

Answer 19. Option (iv) is the answer.

Question 20. Which of the following options represents the correct bond order :

(i) O2–> O2 > O2+

(ii) O2–< O2 < O2+

(iii) O2–> O2 < O2+

(iv) O2–< O2 > O2

Answer 20. Option (ii) is the answer.

Question 21. Explain the non-linear shape of H2 S and the non-planar shape of PCl3 using valence shell electron pair repulsion theory.

Answer 21.

In H2S, the Sulphur atom is surrounded by four electron pairs (two bond pairs as well as two lone pairs).

These four electron pairs adopt tetrahedral geometry.

The repulsion in between lone pair electrons brings distortion in the shape of the H2S.

So, H2S is not linear in shape.

Question 22. Using the molecular orbital theory, compare the bond energy and magnetic character of O2+ and O2– species.

Answer 22.

The Molecular Orbital configuration of O2+ as well as O-2 has given below:

O2+ (15): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px1

O2– (17): σ1s2 σ *1s2 σ 2s2 σ *2s2 σ 2pz2 π2px2 = π 2py2π *2px2= π*2py1

Bond order for the O2+ = 10-5/2 = 2.5

Bond order for the O-2 = 10-7/2 = 1.5

According to the Molecular Orbital Theory, the greater the bond order greater the bond energy.

Thus, O2+ is more stable than O2–

Question 23. Explain the shape of BrF5.

Answer 23.

BrF5‘s central atom is bromine, which has the hybridisation sp3d2.

Br atom has seven valence electrons, out of which five uses to make the pair with the F atoms, as well as two uses to make lone pairs of the electrons.

The lone pair and the bond pair repel each other. So, the shape is square Pyramidal.

Question 24. Structures of molecules of the two compounds are given below :

(a) Which of the following two compounds will have intermolecular hydrogen bonding, and which compound expects to show the intramolecular hydrogen bonding?

(b) The compound’s melting point depends on, among other things, the extent of hydrogen bonding. Based on this, explain which of the above two compounds will show the higher melting point.

(c) The solubility of compounds in the water depends on the power to form hydrogen bonds with water. Which of the above compounds will easily form a hydrogen bond with water and be more soluble?

Answer 24.

(a) Compound 1 will be having intramolecular hydrogen bonding in o-nitrophenol

Compound (II) will have the intermolecular hydrogen bonding in p-nitrophenol.

(b) The compound (II) has a higher melting point because of the intermolecular bonding, a large number of molecules that will get attached.

(c) The compound (II) would be more soluble in water because it will easily form hydrogen bonding with the water molecules.

Question 25. Why does the type of overlap given in the following figure does not result in formation of bond?

Answer 25.

In figure (i), the area of the contact of ++ overlap is equal to the area of the +- overlap. The so-net overlap is zero.

In figure (ii), there is no overlap of the orbitals due to different symmetry.

Question 26. Explain why PCl5 is trigonal bipyramidal, whereas IF5 is square Pyramidal.

Answer 26.

In PCl5, P having the five valence electrons in the orbitals makes five bonds with 5 Cl atoms. It would share one of its electrons from the 3s to the 3d orbital. Therefore, the hybridisation will be sp3d, and the geometry will be trigonal bipyramidal.

IF5, the Iodine atom, has seven valence electrons in the molecular orbitals. It will form 5 bonds with the 5 Cl atoms using 5 electrons from its molecular orbital, and two electrons will form one lone pair on the Iodine atom, which gives the square pyramidal geometry.

Question 27. In both water and the dimethyl ether (CH3 —Ο — CH3 ), the oxygen atom is the central atom and having the same hybridisation, yet they have different bond angles. Which one has a greater bond angle? Elaborate with a reason.

Answer 27.

Dimethyl ether will have a greater bond angle. There will be more repulsion in between bond pairs of CH3 groups attached in ether than between bond pairs of the hydrogen atoms attached to oxygen in the water.

Question 28. Write the Lewis structure of the following compounds and shows a formal charge

on each atom.

HNO3, NO2, H2SO4

Answer 28.

The formal charge has been calculated by

Formal charge = ½ [total no: of bonding as well as shared electrons]

The formal charge on oxygen with the single bond =6-6-2/2 = -1

The formal charge on oxygen with the double bond 6-4-4/2 = 0

The formal charge on the nitrogen=5-2-6/2 = 0

The formal charge on oxygen 1 and 4 = 6-4-4/2= 0

The formal charge on oxygen 2 and 3 = 6-4-4/2=0

The formal charge on hydrogen 1 and 2 = 1-0-2/2=0

The formal charge on sulfur =6-0-12/2 = 0

Question 29. The energy of the σ2pz molecular orbital is greater than π2px as well as π2py molecular orbitals in the nitrogen molecule. Write the complete sequence of the energy levels in the increasing order of the energy in the molecule. Compares the relative stability and the magnetic behaviour of the following species :

N2, N2+ , N2– , N22+

Answer 29.

General sequence of the energy level of the molecular orbital has

σ1s < σ*1s < σ2s< σ*2s < π2px = π2py < σ2pz

N2 σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p2z

N2+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p1z

N2– σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y σ2p2z σ2p2x

N22+ σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x = π2p2y

thus, Bond order = [(½)electrons in BMO – (½)electrons in the ABMO]

For N2 = 10-4/2 = 3

Bond order for N2+= 9-4/2 = 2.5

Bond order for N2–= 10-5/2 = 2.5

Bond order for N22+= 8-4/2 = 2

so, the order of stability is:

N2> N2– > N2+> N22+

Question 30. What is the effect of the following process on the bond order in N2 and O2?

(i) N2→ N2+ + e-

(ii) O2→ O2+ + e-

Answer 30.

(i) N2 will have 14 electrons when it donates one electron. These electrons remove from Bonding molecular orbital. as BO for N2 = 3

(ii) O2 has 16 electrons, 8 electrons in the molecular orbitals and 4 in the antibonding molecular orbitals.

BO for O2 = 2

Question 31. Give reasons for the following :

(i) Covalent bonds are directional bonds, while ionic bonds are nondirectional.

(ii) The water molecule has a bent structure, whereas the carbon dioxide molecule is linear.

(iii) Ethyne molecule is linear.

Answer 31.

(i) A covalent bond is formed by overlapping atomic orbitals. The direction of the overlapping gives the direction of the bond.

(ii) In the water molecule, the oxygen atom is sp3 hybridised and has two lone pairs of electrons.

(iii) In the ethyne molecule, both of the carbon atoms are sp hybridised. The two sp hybrid orbitals of both of the carbon atoms orient in the opposite direction as, forming an angle of 180°.

Question 32. What is the ionic bond? With two suitable examples, explain the difference in between an ionic and the covalent bond?

Answer 32.

When the positively charged ion forms a bond with a negatively charged ion, one atom transfers electrons to another. An example of the ionic bond is the chemical compound Sodium Chloride (NaCl).

The difference between an ionic bond and a covalent bond is that an ionic bond essentially donates the electron to the other atom participating in the bond. In contrast, electrons in a covalent bond are shared equally between the atoms.

Question 33. Arrange the following of the bonds in order to increase the ionic character, giving a reason

N—H, F—H, C—H and O—H

Answer 33.

The ionic character is greater in the molecules with the highest electronegativity difference because the electron pair shifts toward the more electronegative atom, increasing the ionic character.

Thus, the ionic character order will be:

C-H<N-H<O-H< F-H

Question 34. Explain why CO32- the single Lewis structure cannot represent an ion. How can it be best represented?

Answer 34.

The carbonate ion is represented in the form of a resonating hybrid structure. These structures are equivalent. In resonance, all three C-O bonds get a double character in one of the resonating structures.

So, all the bonds are equivalent and have equal lengths; hence carbonate ions cannot be represented by a single Lewis structure.

Question 35. Predict the hybridisation of each carbon in the molecule of the organic compound given below. Also, indicate this molecule’s total number of sigma and pi bonds.

Answer 35.

The hybridisation of Carbon 1 is sp, carbon 2 is sp, carbon 3 sp2, carbon 4 is sp3, and carbon 5 has sp2.

The triple bond has two pi bonds as well as one sigma bond.

Each double bond has one sigma as well as one pie bond.

Every single bond is a sigma bond.

So, the total number of sigma bonds is 11 and pi bonds are 4 in the molecule.

Question 36. Group the following as linear as well as non-linear molecules :

H2O, HOCl, BeCl2, Cl2O

Answer 36.

BeCl2 has a linear structure

HOC1 is also non-linear in structure.

H2O has a V-shaped structure.

Cl2O has a V-shaped structure.

Question 37. Elements X, Y and Z have 4, 5 and 7 valence electrons, respectively.

(i) Write the molecular formula for the compounds formed by these elements individually with the Hydrogen.

(ii) Which of these compounds would have the highest dipole moment?

Answer 37.

(i); XH4, H3Y, and HZ

Hydrogen has only one electron in its outermost shell. It shares one electron to form the covalent bond or accepts or donates one electron to form an ionic bond.

(ii) The compound HZ has a linear shape, and the difference in the electronegativity of Hydrogen and element Z is maximum.

Question 38. Draw the resonating structure of

(i) Ozone molecule

(ii) Nitrate ion

Answer 38.

Question 39. Predict the shape of the following molecules based on the hybridisation.

BCl3, CH4, CO2, NH3

Answer 39.

In compound BCl3, Boron has sp2-hybridisation, and the shape is the Triangular Planar.

In methane CH4, Carbon has sp3 -hybridisation and its shape are Tetrahedral.

In carbon dioxide CO2, carbon has sp-hybridisation, and its shape is Linear.

In ammonia NH3, nitrogen has sp3-hybridisation, and its shape is Pyramidal.

Question 40. All the C-O bonds in the carbonate ion (CO2–3 ) are the same length. Explain

Answer 40.

The carbonate ion is represented in the form of a resonating structure. These structures are equivalent to nature. In resonance, all 3 C-O bonds get a double character in one of the resonating structures.

Question 41. What is meant by the term average bond enthalpy? Why is the O—H bond enthalpy difference between ethanol (C2H5OH) and water?

Answer 41.

Similar bonds in the molecule do not have the same bond enthalpies. Mainly in the term average bond enthalpy is used in the polyatomic molecules. It is obtained by dividing the bond dissociation enthalpy by the number of bonds broken. The bond enthalpy of the OH bond is different in ethanol and water because of the difference in electronegativity of Hydrogen and carbon. As electronegativity differs in the Hydrogen and oxygen is higher than that in carbon and oxygen, so the O-H bond in the water has more bond enthalpy than in ethanol.

Question 42. Explain the formation of a chemical bond

Answer 42.

“Chemical bonds are an attractive force that binds the constituents of the chemical species together.”

Thus, many theories have been suggested for chemical bond formation, such as the valence shell electron pair repulsion theory, electronic theory, molecular orbital theory as well as valence bond theory.

The formation of the chemical bond credits the system’s tendency to achieve stability. It noticed that the inertness of the noble gases directly resulted from their filled outermost orbitals. Consequently, that proposed that the elements having a deficiency of the electrons in outermost shells are unstable. Thus, atoms combine and finish their separate octets or duplets to achieve the stable configuration of the closest inert gases. Thus, this combination may occur either by sharing the electrons. Thus, The formed chemical bond results from sharing the electrons among atoms and is known as a covalent bond. Also, a formed ionic bond results from sharing of electrons among atoms.

Question 43. Write the Lewis dot symbols for the atoms of the following elements :

1. a) Mg
2. b) Na
3. c) B
4. d) O
5. e) N
6. f) Br

Answer 43.

a) Mg

The magnesium atom contains only two valence electrons. so, the lewis dot symbol for Mg is

b) Na

The sodium atom contains only one valence electron. Thus, the lewis dot symbol for Na is

c) B

Boron atom contains only three valence electrons. Thus, the lewis dot symbol for B is

d) O

The oxygen atom contains only six valence electrons. Thus, the lewis dot symbol for O is

e) N

The nitrogen atom contains only five valence electrons. Thus, the lewis dot symbol for N is

f) Br

The bromine atom contains only seven valence electrons. Thus, the lewis dot symbol for Br is

Question 44. Explain why the BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.

Answer 44.

Lewis structure of BeH2 is:

The central atom has no lone pair but two bond pairs. Thus, its shape is AB2. i.e. Linear shape.

so, the dipole moment of Be- H bond is equal and opposite in direction, nullifying one another. So, the dipole moment of BeH2 is 0.

Question 45. Which out of NH3 and NF3 has the higher dipole moment and why?

Answer 45.

N- atom is the central atom of the NF3 and NH3.

The central atom has one lone pair and three bond pairs. So, for both, the shape is AB3E. i.e. Pyramidal.

As, F-atom has more electronegativity than the H– atom, NF3 should have a higher dipole moment than the NH3. But the dipole moment of the NH3 is 1.46D which is higher than the dipole moment of NF3, which is 0.24D.

It gets clear from the directions of the dipole moments of individual bonds in NF3 and NH3.

As both the N-H bond is in the same direction, it adds to the bond moment of the lone pair, while N-F bonds are in the opposite direction, so they partly cut the bond moment of the lone pair.

Thus, the dipole moment of NH3 is higher than that of NF3.

Question 46. What is meant by the hybridisation of atomic orbitals? Describe the shapes of sp, sp2, and sp3 hybrid orbitals.

Answer 46.

“Hybridisation is the intermixing of a set of atomic orbitals of slightly different energies, forming a new set of the orbitals with equivalent energies and shapes”.

E.g. 1 s- orbital hybridises with the 3 p- orbitals to form the 4 sp3 hybrid orbitals.

(a) sp hybrid orbital

1 s- orbital hybridises with 1 p- orbitals to form 2 sp hybrid orbitals. Sp hybrid orbital has a linear shape. The formation of sp orbital is:

(b) sp2 hybrid orbital

here the 1 s- orbital hybridises with 2 p- orbitals in order to form 3 sp2 hybrid orbitals. The shape of the sp2 orbital is trigonal planar.

(c) sp3 hybrid orbital

here the 1 s- orbital hybridises with 3 p- orbitals in order to form 4 sp3 hybrid orbitals. The shape of the sp3 orbital is the tetrahedron.

Question 47. Is there any change in the B and N atoms hybridisation due to the following reaction?

BF3 + NH3 —> F3B.NH3

Answer 47.

N- atom in NH3 has sp3 hybridisation. The orbital picture of N- the atom shows below:

B- atom in NF3 has sp2 hybridisation. The orbital picture of B- the atom shows below:

On the reaction between NH3 and BF3, F3B.NH3 is obtained as a product, as hybridisation of B-atom changed to sp3. Although, the hybridisation of N- atoms remains unchanged.

Question 48. Which hybrid orbitals are used by the carbon atoms in the following molecules?

(a) CH3-CH3; (b) CH3-CH=CH2; (c) CH3CH2-OH; (d) CH3-CHO; (e) CH3COOH.

Answer 48.

(a) CH3-CH3

Here, C1 and C2 have sp3 hybridisation. 

(b) CH3-CH=CH2

Here, C3 and C2 have sp2 hybridisation and C1 has sp3 hybridisation. 

(c) CH3-CH2-OH

Here, C1 and C2 have sp3 hybridisation.

 (d) CH3-CHO

Here, C1 has sp3 hybridisation, and C2 has sp2 hybridisation. 

(e) CH3COOH

Here, C1 has sp3 hybridisation, and C2 has sp2 hybridisation.

Question 49. What do you understand by the bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Answer 49.

A covalent bond forms when two atoms combine by sharing their valence electrons.

The shared pairs of electrons between the bonded atoms are called bond pairs. Every electron cannot participate in bonding. So the pairs of electrons which do not participate in bonding are called lone pairs.

for example, ethane has seven bond pairs but zero lone pairs.

b) Water has two bond pairs and two lone pairs on the O- atom.

Question 50. Distinguish between sigma and a pi bond.

Answer 50.

 Question 51. Explain the formation of the H2 molecule based on the valence bond theory.

Answer 51.

Let us assume 2 H- atoms X and Y with the nuclei NX and NY and electrons eX and eY, respectively.

When X and Y are far from each other, they have no interaction. Thus, as they come closer, the attractive force, as well as a repulsive force, become active.

The repulsive forces will be acting:

(a) Between electrons of both the atoms, eX and eY.

(b) Between nuclei of both the atoms, NX and NY.

The attractive forces will be acting:

(a) Between the electron and nucleus of the same atom, NX – eX and NY -eY.

(b) Between the electron of one atom and the nucleus of other atoms, NX–eY and NY–eX.

The repulsive force pushes the two atoms apart, whereas the attractive force tends to bring them together.

Repulsive forces:

Attractive forces:

The values of repulsive forces are lesser than the attractive forces. Thus, two atoms will approach each other. Therefore, there is a decrease in potential energy. In the end, a stage will be reached when the repulsive forces will balance the attractive forces, and the overall system achieves minimum energy, which leads to the formation of the H2 molecule.

Question 52. Write the important conditions that are required for the linear combination of the atomic orbitals to form the molecular orbitals.

Answer 52.

The condition that is required for the linear combination of atomic orbitals to form the molecular orbitals are as follows:

(i) The joining of atomic orbitals might have approximately the same energy. This implies in the homo-nuclear molecule, the 1s-orbital of one atom can join with the 1s- orbital of another atom but can’t join with the 2s-orbital.

(ii) The joining atomic orbitals might have legitimate orientations to ensure the maximum overlap.

(iii) The overlapping might be to a large extent.

Question 53. Write the significance of a plus as well as a minus sign shown in representing the orbitals

Answer 53.

Generally, the molecular orbital is represented by the ‘wave function’.

A positive (+) sign representing a molecular orbital indicates a positive wave function.

A negative (-) sign representing a molecular orbital indicates a negative wave function.

Question 54. Define the octet rule. Write its significance and limitations

Answer 54.

The octet rule says, “atoms can combine either by transfer of valence electrons from one atom to another or by sharing the valence electrons to achieve the nearest inert gas configuration by having an octet in their valence shell.”

The octet rule explains chemical bond formation depending upon the nature of the element.

Limitations:

(a) Octet rules fail to predict the relative stability as well as the shape of the molecules.

(b) It is based on the inert nature of the noble gases. But, some inert gases, say, krypton(Kr) as well as xenon(Xe), form compounds like KrF2, XeF2 etc.

(c) The octet rule can’t apply to elements beyond the 3rd period. Elements present beyond the 3rd period have more than eight valence electrons surrounding the central atom. E.g. SF6, PF6 etc.

(d) The octet rule is not applied to atoms in a molecule with an odd number of electrons. E.g. For NO2 as well as NO octet rule is not applicable.

(e) If a compound has less than 8 electrons surrounding the central atom, then the octet rule can’t be applied to that compound. E.g. BeH2, AlCl3, LiCl etc., have not to obey the octet rule.

Question 55. Write favourable factors for the formation of the ionic bond.

Answer 55.

An ionic bond forms by transferring one or more electrons from one atom to another. So, ionic bond formation depends on the flexibility of the neutral atoms to lose or gain electrons. The formation of an ionic bond also depends on the compound’s lattice energy.

The factors that are favourable for the ionic bond formation:

(a) High electron affinity of atoms of non-metal.

(b) The high lattice energy of the compound which is formed.

(c) Low ionisation enthalpy of an atom of metal.

Question 56. Discuss the shape of the following molecule using the VSEPR model:

BeCl2, BCl3, SiCl4, AsF5, H2S, PH3

Answer 56.

BeCl2

The central atom has no lone pair but two bond pairs. Thus, its shape is AB2. i.e. Linear shape.

BCl3

The central atom has no lone pair but three bond pairs. Thus, its shape is AB3. i.e. Trigonal planar.

SiCl4

The central atom has no lone pair but 4 bond pairs. Thus, its shape is AB4. i.e. Tetrahedral.

AsF5

The central atom has no lone pair but 5 bond pairs. Thus, its shape is AB5. i.e. Trigonal bipyramidal.

 H2S

The central atom has one lone pair and two bond pairs. Thus, its shape is AB2E. i.e. Bent shape.

PH3

The central atom has one lone pair and three bond pairs. Thus, its shape is AB3E. i.e. Trigonal bipyramidal.

Question 57. Although geometries of NH3 and H2O molecules have distorted tetrahedral, the bond angle in water is less than that of  Ammonia. Discuss.

Answer 57.

The geometry of H2O and NH3:

The central atom(N) in ammonia has one lone pair and three bond pairs.

The water’s central atom(O) has two lone pairs and two bond pairs.

So, these two lone pairs on O- atom in the water molecule repels the two bond pairs. And this repulsion is between the lone pair and bond pair on O- the atom of H2O is stronger than the repulsion between the lone pair and bond pair on the N-atom of NH3.

So, the bond angle in H2O is less than NH3, even though they have a distorted tetrahedral structure.

Question 58. How do you express the bond strength in terms of the bond order?

Answer 58.

The bond strength presents the extent of bonding between two atoms while forming a molecule. As the bond strength increases, thus, the bond becomes stronger, and the bond order increases.

Question 59. Define Bond length.

Answer 59.

Bond length is defined as an equilibrium distance in between the nuclei of 2 bonded atoms in a molecule.

Question 60. H3PO3 could be represented by structures 1 and 2 shown below. Can these two structures be a canonical form of the resonance hybrid representing H3PO3? If not, give reasons for the same. 

Answer 60.

In given structures, the position of atoms is changed, so we can’t take the two given structures as a canonical form of resonance hybrid representing H3PO3.

Question 61. Use Lewis symbols to show the electron transfer between the following atoms to form cations and anions :

(i) K and S 

(ii) Ca and O

(iii) Al and N.

Answer 61.

(i) K and S

Electronic configurations of S and K are:

S: 2, 8, 6

K: 2, 8, 8, 1

Here, it is clear that K has one more electron than the nearest inert gas. i.e. Ne, whereas S need 2 electrons to complete its octet. So, the transfer of electrons takes place in the following way,

(ii) Ca and  O

Electronic configurations of O and Ca are:

O: 2, 6

Ca: 2, 8, 8, 2

Here, it is clear that Ca has two more electrons than the nearest inert gas. i.e. Ar, whereas O needs 2 electrons to complete its octet. So, the transfer of electrons takes place in the following way,

(iii) Al and N

Electronic configurations of N and Al are:

N: 2, 5

Al: 2, 8, 3

Here, it is clear that Al has three more electrons than the nearest inert gas. i.e. Ne, whereas N needs 3 electrons to complete its octet. So, the transfer of electrons takes place in the following way,

Question 62. Write the significance and applications of the dipole moment.

Answer 62.

Key significant points for the dipole moment are as below:

• The molecule’s shape can be determined by it. Symmetrical molecules, such as linear, will have zero dipole moments, whereas non-symmetrical molecules will take on varied shapes, such as angular, bent, etc.
• The polarity of molecules will be determined. The polarity will be lesser if the dipole moment is lesser and vice versa.
• One can conclude that if a molecule has zero dipole moment, then it is non-polar, and if it shows some polar character, it is non-polar.

Question 63. Define electronegativity. How does it differ from the electron gain enthalpy?

Answer 63.

Electronegativity can be defined as the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself.

Question 64. Explain with the help of the suitable example polar covalent bond.

Answer 64.

When two unique atoms with distinct electronegativities join to form a covalent bond, the bond pair of electrons are not shared equally. The nucleus of the atom which has greater electronegativity will attract the bond pair. So, in this case the electron distribution gets distorted, and an electronegativity atom will attract the electron cloud.

Therefore, the electronegative element will get slightly negatively charged, and on the other side, the other atom will get slightly positively charged. As a result of this, there are two opposite poles developed in a molecule, and this type of bond formed is termed as a ‘polar covalent bond’.

For e.g. HCl has a polar covalent bond. In HCl, Cl- atom has more electronegativity than the H- atom. So the bond pair shifts towards the Cl- atom, which acquires a positive charge.

Question 65. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2, and ClF3.

Answer 65.

The ionic character of a molecule depends on the difference in electronegativity between constituents atoms. So, the higher the difference, the higher the ionic character of a molecule.

So, the required sequence of ionic character of the above-given molecules is

N2< SO2< ClF3< K2O <LiF.

Question 66. As shown below, the skeletal structure of CH3COOH is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Answer 66.

Correct Lewis structure of CH3COOH is given below:

Question 67. Explain the shape of BrF5.

Answer 67.

In BrF5, the central atom Br is surrounded by the five bonded pairs and one lone pair. This forms the shape of a square Pyramidal.

Question 68. Structures of molecules of the two compounds have given below:

(a) Which of the following two compounds will have intermolecular hydrogen bonding, and which compound is expected to show the intramolecular hydrogen bonding?

(b) The compound’s melting point depends on, among other things, the extent of the hydrogen bonding. On the basis, explain which of the above two compounds would show the higher melting point.

(c) Solubility of the compounds in water depends on the power to form hydrogen bonds with water, which of the above compounds will easily form a hydrogen bond with water and be more soluble in it.

Answer 68.

(a) Hence the NO2 and OH groups in compound (I) are close together, and the intramolecular hydrogen bonding will form (II). Compound (II) will show the intermolecular hydrogen bonding.

(b) Hence it forms intramolecular hydrogen bonds, and compound (II) has a higher melting point. As a result, more and more molecules have linked together via hydrogen bond formation.

(c) Due to the intramolecular hydrogen bonding, compound (I) can’t form hydrogen bonds with the water and is less soluble in it, whereas compound (II) can form hydrogen bonds with the water more easily and is thus more soluble in water.

Question 69. Why does the type of overlap given in the following figure not result in bond formation?

Answer 69.

• In the first figure, the (++) overlap equals to the (+-) overlap; thus, these cancel out, and the net overlap is zero.
• Hence the two orbitals are perpendicular to each other in the second figure. No overlap is possible.

Question 70. Explain why PCl5 is trigonal bipyramidal, whereas IF5 is square Pyramidal.

Answer 70.

P is surrounded by the five bond pairs and no lone pairs in PCl5, whereas the iodine atom surrounds the five bond pairs and one lone pair in IF5, so the shape of PCl5 is trigonal bipyramidal, and IF5 is square Pyramidal.

Question 71. In both water and dimethyl ether (CH3–O–CH3), the oxygen atom is the central atom and has the same hybridisation, yet they have different bond angles. Which one has a greater bond angle? Specify with a reason.

Answer 71.

The bond angle of dimethyl ether would be greater. More repulsion will exist in between bond pairs of the CH3 groups attached in ether than in between bond pairs of hydrogen atoms attached to the oxygen in the water.

The carbon of the CH3 in ether is attached to three hydrogen atoms via bonds, and these bonds’ electron pairs contribute to the carbon atom’s electronic charge density. As a result, the repulsion in between two CH3 groups will be greater than between two hydrogen atoms.

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