Chapter 4: Chemical Bonding and Molecular Structure Long Questions | chemistry Class 11 CBSE

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Chapter 4: Chemical Bonding and Molecular Structure Long Questions | chemistry Class 11 CBSE | ByteNotes.in

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Question 1

Long Form

Compare and contrast the formation of ionic and covalent bonds with suitable examples, and explain the role of lattice enthalpy in determining the stability of ionic compounds.

Ionic bonds form by complete transfer of electrons from a less electronegative atom (usually a metal) to a more electronegative atom (usually a non-metal), resulting in oppositely charged ions held together by electrostatic attraction; e.g., Na transfers one electron to Cl, forming Na+ and Cl- which combine to give NaCl.
Covalent bonds form by the sharing of one or more electron pairs between atoms, with each contributing at least one electron; e.g., two Cl atoms each contribute one electron to form a shared pair in Cl2, with both achieving the argon octet.
Ionic bond formation is favoured by low ionization enthalpy in the cation-forming element and high (negative) electron gain enthalpy in the anion-forming element; covalent bonds form between atoms of similar electronegativity.
Ionic compounds in the crystalline state consist of ordered three-dimensional arrangements of cations and anions; the lattice enthalpy is defined as the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions.
Even if the sum of ionization enthalpy and electron gain enthalpy is positive (as in NaCl, where the sum is +147.1 kJ mol-1), the compound is still stable because the lattice enthalpy of formation (-788 kJ mol-1) more than compensates for this energy requirement.
Therefore, the qualitative measure of stability of an ionic compound is its lattice enthalpy of formation and not merely the attainment of octets around the ionic species; a higher lattice enthalpy indicates a more stable ionic compound.

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Question 2

Long Form

Discuss the limitations of the octet rule and identify the three types of exceptions with at least one example for each.

The octet rule, though very useful for understanding bonding in organic compounds and second-period elements, is not universally applicable; there are three major types of exceptions.
Exception 1 — Incomplete octet: In some compounds, the central atom has fewer than eight electrons. This occurs with elements having less than four valence electrons; examples include LiCl (Li, 2e-), BeH2 (Be, 4e-), BCl3 and BF3 (B, 6e-), and AlCl3 (Al, 6e-).
Exception 2 — Odd-electron molecules: Molecules with an odd number of electrons cannot satisfy the octet rule for all atoms; examples include nitric oxide (NO) and nitrogen dioxide (NO2), both of which have unpaired electrons.
Exception 3 — Expanded octet: Elements in the third period and beyond possess 3d orbitals of comparable energy to 3s and 3p orbitals, allowing accommodation of more than eight electrons; examples include PF5 (10e- around P), SF6 (12e- around S), and H2SO4 (12e- around S).
Additional drawbacks include: the octet rule is based on the chemical inertness of noble gases, yet xenon and krypton form stable compounds like XeF2 and KrF2, contradicting this premise.
The theory also fails completely to account for molecular shapes and says nothing about the relative stability or energy of a molecule, making it silent on crucial aspects of molecular behaviour.

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Question 3

Long Form

Describe the VSEPR theory and use it to predict and explain the shapes of BF3, CH4, NH3, H2O, and PCl5.

VSEPR theory proposes that the shape of a molecule is determined by the repulsion between electron pairs (bonding and non-bonding) in the valence shell of the central atom; electron pairs arrange themselves to minimize repulsion and maximize distance.
BF3 (AB3, 0 lone pairs): Three bond pairs around B are arranged in a trigonal planar geometry with bond angles of 120°; no lone pairs distort the ideal arrangement.
CH4 (AB4, 0 lone pairs): Four bond pairs around C adopt tetrahedral geometry with bond angles of 109.5°; all four pairs are equivalent, giving a perfectly symmetric structure.
NH3 (AB3E, 1 lone pair): Three bond pairs and one lone pair around N; the lone pair exerts greater repulsion on bond pairs, compressing H-N-H angles from 109.5° to 107°; shape is trigonal pyramidal.
H2O (AB2E2, 2 lone pairs): Two bond pairs and two lone pairs around O; lp-lp and lp-bp repulsions reduce H-O-H angle from 109.5° to 104.5°; shape is bent (V-shaped).
PCl5 (AB5, sp3d, 0 lone pairs): Five bond pairs around P give trigonal bipyramidal geometry; three equatorial P-Cl bonds at 120° and two axial bonds at 90° to the equatorial plane; axial bonds are slightly longer due to greater repulsion from three equatorial bond pairs.

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Question 4

Long Form

Explain the concept of resonance with reference to O3, CO3 2-, and CO2, and clarify common misconceptions about resonance structures.

Resonance is necessary when a single Lewis structure cannot accurately describe the bonding in a molecule; according to the resonance concept, a molecule is described as a hybrid of two or more canonical (resonance) structures of similar energy, nuclei positions, and electron arrangements.
For O3: Two canonical forms can be drawn, each with one O-O single bond (148 pm) and one O=O double bond (121 pm); but experimental O-O bond lengths in O3 are both 128 pm — intermediate; the resonance hybrid (with a double-headed arrow between canonical forms) represents this accurately.
For CO3 2-: Three equivalent canonical forms can be drawn, each with two C-O single bonds and one C=O double bond; experimentally all C-O bonds in carbonate are equivalent, so the hybrid of all three forms best represents the ion.
For CO2: The experimental C-O bond length (115 pm) lies between a C=O double bond (121 pm) and a C triple bond O triple bond (110 pm); three canonical forms including structures with C-O single and C triple bond O triple bond bonds are required.
Key misconceptions to avoid: canonical forms have no real, independent existence; the molecule does not flip between them at any fraction of time; there is no equilibrium between canonical forms (unlike tautomers).
The resonance hybrid stabilizes the molecule — its energy is lower than any individual canonical form; resonance also averages bond characteristics so all equivalent bonds have the same experimentally observed length.

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Question 5

Long Form

Explain sp, sp2, and sp3 hybridisation with one example each, including orbital diagrams and geometry.

sp hybridisation: One s and one p orbital combine to form two equivalent sp hybrid orbitals, each with 50% s-character and 50% p-character; these are oriented at 180° to each other (linear geometry); example is BeCl2 where Be undergoes sp hybridisation to form two sp orbitals overlapping axially with Cl's p orbitals.
sp2 hybridisation: One s and two p orbitals mix to form three equivalent sp2 hybrid orbitals arranged in a trigonal planar geometry at 120° to each other; example is BCl3 where the central B atom forms three B-Cl sigma bonds using sp2 hybrids; the unhybridised p orbital remains perpendicular to the plane.
sp3 hybridisation: One s and three p orbitals hybridise to form four equivalent sp3 orbitals with 25% s-character and 75% p-character, directed towards four corners of a tetrahedron at 109.5°; example is CH4 where C forms four C-H sigma bonds using sp3 hybrids.
In NH3, N is also sp3 hybridised; three sp3 orbitals form N-H bonds and the fourth contains the lone pair; lp-bp repulsion reduces the H-N-H angle to 107°, giving a pyramidal shape.
In H2O, O is sp3 hybridised; two sp3 orbitals form O-H bonds and two hold lone pairs; lp-lp repulsion reduces H-O-H to 104.5°, giving bent (V-shape) geometry.
The type of hybridisation directly determines molecular geometry: sp is linear, sp2 is trigonal planar, sp3 is tetrahedral (or distorted tetrahedral with lone pairs).

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Question 6

Long Form

Describe the formation of ethane (C2H6), ethene (C2H4), and ethyne (C2H2) in terms of hybridisation, and compare their bond lengths and bond angles.

Ethane (C2H6) — sp3 hybridisation: Both carbon atoms are sp3 hybridised; one sp3 orbital of each C overlaps axially to form a C-C sigma bond (sp3-sp3), and the three remaining sp3 orbitals of each C overlap with 1s orbitals of H to form six C-H sigma bonds; all bonds are sigma bonds.
In C2H6: C-C bond length is 154 pm, C-H bond length is 109 pm, and the H-C-H bond angle is approximately 109.5°; no pi bonds are present.
Ethene (C2H4) — sp2 hybridisation: Both C atoms are sp2 hybridised; sp2-sp2 sigma bond forms the C-C bond; each C uses two remaining sp2 orbitals for C-H sigma bonds; the unhybridised 2px (or 2py) orbitals of both C atoms overlap sidewise to form one pi bond.
In C2H4: C-C bond consists of one sigma and one pi bond (bond length 134 pm); C-H bond length is 108 pm; H-C-H angle is 117.6° and H-C-C angle is 121°.
Ethyne (C2H2) — sp hybridisation: Both C atoms are sp hybridised; sp-sp overlap forms the C-C sigma bond; each C uses the other sp orbital to bond with H; the two unhybridised p orbitals on each C (2px and 2py) overlap sidewise to form two pi bonds — hence a triple bond (1 sigma + 2 pi).
In C2H2: C-C triple bond length is 120 pm and C-H bond length is 106 pm; bond angle is 180° (linear); as bond order increases from C2H6 to C2H2, bond length decreases and bond enthalpy increases.

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Question 7

Long Form

Discuss the key features of Molecular Orbital Theory and explain how it accounts for the stability and magnetic nature of H2, He2, Li2, and O2 molecules.

MO theory (Hund and Mulliken, 1932) treats electrons as belonging to the whole molecule; molecular orbitals are formed by LCAO — addition gives a bonding MO (lower energy, sigma = psiA + psiB) and subtraction gives an antibonding MO (higher energy, sigma* = psiA - psiB); the number of MOs equals the number of AOs combined.
MOs are filled using the aufbau principle, Pauli exclusion principle, and Hund's rule; bond order = 1/2(Nb - Na); positive bond order = stable molecule; zero or negative = unstable.
H2: Electronic configuration (sigma1s)2; Nb=2, Na=0, bond order=1; no unpaired electrons, so H2 is diamagnetic and stable with bond length 74 pm and bond dissociation energy 438 kJ mol-1.
He2: Configuration (sigma1s)2(sigma*1s)2; Nb=Na=2, bond order=0; He2 is unstable and does not exist.
Li2: Configuration KK(sigma2s)2; Nb=4, Na=2, bond order=1; no unpaired electrons, Li2 is stable and diamagnetic; it exists in the vapour phase.
O2: Configuration has 10 bonding and 6 antibonding electrons, bond order=2 (double bond); two electrons are singly placed in degenerate pi*2px and pi*2py antibonding MOs (by Hund's rule), giving O2 two unpaired electrons and making it paramagnetic — a result confirmed experimentally and one that Lewis theory fails to predict.

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Question 8

Long Form

Describe the valence bond theory explanation for the formation of H2 molecule and discuss the orbital overlap concept including sigma and pi bonds.

Valence Bond theory (Heitler and London, 1927) explains bonding in terms of overlap of atomic orbitals; a covalent bond forms when two half-filled atomic orbitals containing electrons of opposite spins overlap — the greater the overlap, the stronger the bond.
For H2: As two H atoms (A and B with nuclei NA, NB and electrons eA, eB) approach, new attractive forces (NA-eB and NB-eA) arise; the magnitude of attraction exceeds repulsion (eA-eB and NA-NB) and potential energy decreases.
A minimum energy (most stable state) is reached at the equilibrium bond length of 74 pm; the energy released upon bond formation is the bond enthalpy (435.8 kJ mol-1); moving nuclei closer or farther from this distance increases energy.
Sigma (sigma) bond: Formed by end-to-end (axial) overlap of atomic orbitals along the internuclear axis; types include s-s, s-p, and p-p axial overlap; sigma bonds are cylindrically symmetrical around the bond axis and are stronger due to more extensive overlap.
Pi (pi) bond: Formed by sidewise (lateral) overlap of p orbitals whose axes are parallel and perpendicular to the internuclear axis; pi bonds consist of two electron cloud lobes above and below the internuclear plane; pi bonds are weaker than sigma bonds.
In multiple bonds, a sigma bond always forms first; a double bond has one sigma and one pi bond; a triple bond has one sigma and two pi bonds; this is seen in N2 (triple bond: 1 sigma + 2 pi) and O2 (double bond: 1 sigma + 1 pi).

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Question 9

Long Form

Explain the polarity of bonds and the dipole moment concept. Discuss how the geometry of a molecule affects its overall dipole moment using the examples of H2O, CO2, BF3, and NH3.

A polar covalent bond forms when atoms of different electronegativities share electrons unequally; the shared pair shifts toward the more electronegative atom, creating partial charges (delta+ and delta-); dipole moment mu = Q x r, expressed in Debye units.
In polyatomic molecules, the net dipole moment is the vector sum of all individual bond dipoles; molecular geometry is therefore as important as individual bond polarity in determining the overall dipole moment.
CO2 (linear): Despite having two polar C=O bonds, the two bond dipoles are equal and point in exactly opposite directions and cancel each other completely; net dipole moment = 0 D.
BF3 (trigonal planar): All three B-F bonds are polar, but the three bond moments are arranged symmetrically at 120° to each other; the resultant of any two is equal and opposite to the third; net dipole moment = 0 D.
H2O (bent, 104.5°): The two O-H bond dipoles do not cancel due to the angular geometry; they add vectorially to give a net resultant dipole moment of 1.85 D, pointing from O toward the bisector of the H-O-H angle.
NH3 (trigonal pyramidal): The three N-H bond dipoles and the orbital dipole of the lone pair all act in the same direction, giving NH3 a large dipole moment (1.47 D); this illustrates how lone pair orbital dipoles contribute significantly to the overall molecular dipole moment.

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Question 10

Long Form

Describe sp3d and sp3d2 hybridisation with reference to PCl5 and SF6 respectively, explaining why equatorial and axial bonds in PCl5 are not equivalent.

sp3d hybridisation (PCl5): In the excited state, phosphorus (ground state: [Ne]3s23p3) promotes one 3s electron to a 3d orbital, giving five unpaired electrons; one 3s, three 3p, and one 3d orbital hybridise to form five sp3d hybrid orbitals.
The five sp3d orbitals are directed towards five corners of a trigonal bipyramidal geometry; three bonds lie in the equatorial plane at 120° to each other (equatorial bonds) and two bonds lie along the vertical axis at 90° to the equatorial plane (axial bonds).
Axial bonds are longer and weaker than equatorial bonds because axial bond pairs experience more repulsion from three equatorial bond pairs at 90° (stronger repulsion) compared to equatorial bond pairs which interact with only two axial bonds at 90° and two other equatorial bonds at 120°.
sp3d2 hybridisation (SF6): Sulphur (ground state: 3s23p4) promotes electrons to the excited state, making six unpaired electrons; one 3s, three 3p, and two 3d orbitals hybridise to form six equivalent sp3d2 hybrid orbitals.
The six sp3d2 orbitals are directed towards the six corners of a regular octahedron, with all S-F bond angles of 90°; SF6 has a regular octahedral geometry with all six bonds equivalent.
All six S-F bonds in SF6 are identical in length and strength (unlike PCl5); SF6 is a highly stable, symmetric molecule; this type of hybridisation also occurs in [CrF6]3- and other coordination compounds.

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Chapter 4: Chemical Bonding and Molecular Structure | Long Questions

Compare and contrast the formation of ionic and covalent bonds with suitable examples, and explain the role of lattice enthalpy in determining the stability of ionic compounds.

Discuss the limitations of the octet rule and identify the three types of exceptions with at least one example for each.

Describe the VSEPR theory and use it to predict and explain the shapes of BF3, CH4, NH3, H2O, and PCl5.

Explain the concept of resonance with reference to O3, CO3 2-, and CO2, and clarify common misconceptions about resonance structures.

Explain sp, sp2, and sp3 hybridisation with one example each, including orbital diagrams and geometry.

Describe the formation of ethane (C2H6), ethene (C2H4), and ethyne (C2H2) in terms of hybridisation, and compare their bond lengths and bond angles.

Discuss the key features of Molecular Orbital Theory and explain how it accounts for the stability and magnetic nature of H2, He2, Li2, and O2 molecules.

Describe the valence bond theory explanation for the formation of H2 molecule and discuss the orbital overlap concept including sigma and pi bonds.

Explain the polarity of bonds and the dipole moment concept. Discuss how the geometry of a molecule affects its overall dipole moment using the examples of H2O, CO2, BF3, and NH3.

Describe sp3d and sp3d2 hybridisation with reference to PCl5 and SF6 respectively, explaining why equatorial and axial bonds in PCl5 are not equivalent.

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