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

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A chemistry teacher showed students two white crystalline solids — sodium chloride (NaCl) and naphthalene (C10H8). She dissolved both in water and tested their electrical conductivity. NaCl solution conducted electricity well, while naphthalene barely dissolved and its solution did not conduct. She then explained that NaCl forms when sodium transfers one electron to chlorine, with Na losing its outermost electron to achieve the neon configuration and Cl gaining one to achieve the argon configuration. The resulting ions are held together by strong electrostatic forces. Naphthalene, being a covalent compound, does not form ions. She also noted that the lattice enthalpy of NaCl is 788 kJ mol-1, a large value that accounts for its high melting point.

Question 1: Write the electron transfer equations for the formation of NaCl from sodium and chlorine atoms, showing the electronic configurations at each step.

Na (configuration [Ne]3s1) loses one electron to form Na+ (configuration [Ne]): Na → Na+ + e-
Cl (configuration [Ne]3s23p5) gains one electron to form Cl- (configuration [Ne]3s23p6 or [Ar]): Cl + e- → Cl-
The oppositely charged ions attract each other electrostatically to form NaCl: Na+ + Cl- → NaCl; the bond formed is called an electrovalent or ionic bond.

Question 2: Define lattice enthalpy and explain why NaCl is stable as a solid even though the sum of ionization enthalpy of Na (+495.8 kJ mol-1) and electron gain enthalpy of Cl (-348.7 kJ mol-1) is positive.

Lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions; for NaCl it is 788 kJ mol-1.
The sum of ionization enthalpy and electron gain enthalpy for NaCl = 495.8 + (-348.7) = +147.1 kJ mol-1, which is endothermic; energy must be supplied to form gaseous ions.
However, the lattice formation enthalpy of NaCl is -788 kJ mol-1 (highly exothermic), which more than compensates for the 147.1 kJ mol-1 required; the large energy released on forming the crystal lattice stabilises NaCl as a solid compound.

Question 3: Using Fajans' rules, compare the covalent character of NaCl and AgCl. Predict which compound would have a lower melting point and explain why.

Fajans' rules state that covalent character increases with smaller cation size, higher cation charge, and larger anion size; also, cations with (n-1)d^n ns0 configuration are more polarising than those with noble gas configuration.
Na+ has a noble gas (Ne) configuration and is a relatively large cation; Ag+ has a pseudo-noble gas (4d10) configuration and similar size but d electrons shield nuclear charge poorly, giving Ag+ much greater polarising power.
AgCl therefore has significantly more covalent character than NaCl; greater covalent character means weaker overall ionic lattice forces and a lower melting point; NaCl melts at 801°C while AgCl melts at 455°C, consistent with this prediction.

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In a laboratory practical, students were asked to draw Lewis dot structures for several molecules and ions: H2O, CO2, NH4+, and CO32-. The teacher observed that most students drew a single Lewis structure for CO32- with one C=O double bond and two C-O single bonds. She pointed out that experimental data shows all three C-O bonds in carbonate are identical in length (approximately 136 pm), which lies between typical C-O single bond (143 pm) and C=O double bond (121 pm). She introduced the concept of resonance to resolve this discrepancy and asked students to re-examine all structures. She also asked them to calculate formal charges on the atoms in ozone.

Question 1: Define resonance. Draw all canonical forms of CO32- and explain why none of them alone represents the true structure.

Resonance is the concept that when a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, and electron arrangements (canonical forms) are taken together as a resonance hybrid to describe the molecule accurately.
For CO32-, three canonical forms exist: each has carbon at the centre with one C=O double bond and two C-O single bonds, with the double bond rotating between each of the three oxygen positions in the three forms.
No single canonical form accurately represents CO32- because each would predict two longer C-O single bonds and one shorter C=O double bond; but experimentally all three C-O bonds are equivalent at 136 pm, which only the resonance hybrid (bond order 4/3) correctly represents.

Question 2: Calculate the total valence electrons in NH4+ and CO32- ions and describe the steps used to account for the ionic charge in Lewis structures.

For NH4+: N has 5 valence electrons, each of 4 H atoms contributes 1 electron (total 4), and the +1 charge means subtraction of 1 electron; total = 5 + 4 - 1 = 8 valence electrons for the Lewis structure.
For CO32-: C has 4 valence electrons, each of 3 O atoms contributes 6 electrons (total 18), and the 2- charge means addition of 2 electrons; total = 4 + 18 + 2 = 24 valence electrons.
The rule is: for each negative charge, add one electron to the count; for each positive charge, subtract one electron; this ensures the Lewis structure correctly reflects the total electron count of the ion as a whole.

Question 3: Calculate the formal charge on each atom in the ozone (O3) molecule. What does the pattern of formal charges tell us about the structure? Why do formal charges not represent real charge separation?

For O3 (Lewis structure with central O doubly bonded to one terminal O and singly bonded to the other): Central O (atom 1): FC = 6 - 2 - (1/2)(6) = +1. Terminal O with double bond (atom 2): FC = 6 - 4 - (1/2)(4) = 0. Terminal O with single bond (atom 3): FC = 6 - 6 - (1/2)(2) = -1.
The formal charge pattern (+1, 0, -1) indicates that the structure shown is not perfectly representative; the two resonance forms of O3 together give the hybrid where the charges are distributed symmetrically, explaining the equal bond lengths of 128 pm (intermediate between O-O 148 pm and O=O 121 pm).
Formal charges do not represent real charge separation because they are calculated assuming equal sharing of bonding electrons (pure covalent view); in reality, electronegative differences cause unequal sharing; formal charges simply help track valence electrons and select the most stable Lewis structure (smallest formal charges preferred).

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A research chemist was studying the shapes of several molecules: BeCl2, BF3, CH4, and PCl5. She noted that all these molecules have no lone pairs on the central atom, yet their shapes are entirely different. She used the VSEPR theory to predict their geometries, noting that electron pairs repel each other and arrange themselves in space to minimise repulsion. She also observed that while all bond angles within BF3 are equal, in PCl5 the five P-Cl bonds fall into two non-equivalent sets. She asked her students to explain this non-equivalence and to predict what would happen to the geometry if one bond pair were replaced by a lone pair in each case.

Question 1: Using VSEPR theory, predict and explain the shapes and bond angles of BeCl2, BF3, and CH4.

BeCl2 (AB2, 2 bond pairs, 0 lone pairs): The two bond pairs repel each other maximally at 180°; shape is linear with Cl-Be-Cl bond angle = 180°.
BF3 (AB3, 3 bond pairs, 0 lone pairs): Three bond pairs arrange in a trigonal planar geometry to minimise repulsion; all F-B-F bond angles = 120°; shape is trigonal planar.
CH4 (AB4, 4 bond pairs, 0 lone pairs): Four bond pairs adopt a tetrahedral arrangement to maximise distance; all H-C-H bond angles = 109.5°; shape is tetrahedral.

Question 2: Explain why the five P-Cl bonds in PCl5 are not equivalent, distinguishing between equatorial and axial bonds.

In PCl5, the five bond pairs adopt 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 bond pairs interact with three equatorial bond pairs at 90° (stronger repulsion); equatorial bond pairs interact with only two axial bond pairs at 90° and two equatorial pairs at 120° (weaker overall repulsion).
Because axial bond pairs experience greater repulsion, they are pushed slightly further from P, making axial bonds longer (~219 pm) and weaker than equatorial bonds (~204 pm); not all bond angles in trigonal bipyramidal geometry are equivalent, unlike linear, trigonal planar, tetrahedral, and octahedral geometries.

Question 3: If one bond pair in CH4 is replaced by a lone pair (giving NH3) and then another bond pair is replaced by a lone pair (giving H2O), trace the systematic change in shape and bond angle and explain the reasons for each change.

CH4 (4 bp, 0 lp): All four sp3 hybrid orbitals hold bond pairs; no lone pair distortion; tetrahedral shape with H-C-H = 109.5°; all electron pairs are equivalent.
NH3 (3 bp, 1 lp): One hybrid orbital now holds a lone pair; lp-bp repulsion > bp-bp repulsion; the lone pair compresses the three N-H bond pairs closer together; shape becomes trigonal pyramidal with H-N-H = 107° (reduced from 109.5°).
H2O (2 bp, 2 lp): Two hybrid orbitals hold lone pairs; both lp-lp repulsion (strongest) and lp-bp repulsion further compress the H-O-H angle; shape is bent (V-shaped) with H-O-H = 104.5° (further reduced); the systematic reduction in bond angle (109.5° → 107° → 104.5°) directly reflects the increasing number of lone pairs and their greater repulsive effect compared to bond pairs.

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A student was studying the dipole moments of various molecules. She found that water has a dipole moment of 1.85 D, ammonia (NH3) has 1.47 D, and nitrogen trifluoride (NF3) has only 0.23 D, even though fluorine is more electronegative than nitrogen or hydrogen. She was also puzzled that CO2 and BF3 have zero dipole moments despite having polar bonds. Her teacher explained that dipole moment is a vector quantity and the molecular geometry determines whether individual bond dipoles cancel or add. She further noted that the orbital dipole due to lone pairs plays a crucial role in NH3 and NF3.

Question 1: Define dipole moment. Explain why CO2 has zero dipole moment despite having two polar C=O bonds.

Dipole moment (mu) is defined as the product of the magnitude of the charge (Q) and the distance of separation (r) between the centres of positive and negative charge: mu = Q x r; it is expressed in Debye units (D) where 1 D = 3.33564 x 10-30 C m.
CO2 is a linear molecule (sp hybridised carbon) with two identical C=O bond dipoles pointing in exactly opposite directions (180° apart); being equal in magnitude and antiparallel, they cancel each other completely by vector addition.
The net dipole moment of CO2 = 0 D, making it a non-polar molecule despite having individual polar bonds; this illustrates that molecular symmetry and geometry are as important as bond polarity in determining the overall dipole moment.

Question 2: Explain why the dipole moment of NH3 (1.47 D) is much greater than that of NF3 (0.23 D), even though F is more electronegative than H.

Both NH3 and NF3 are pyramidal molecules with a lone pair on nitrogen; however, the direction of the orbital dipole due to the lone pair differs in the two molecules relative to the bond dipoles.
In NH3, the N-H bond dipoles (pointing from H to N, since N is more electronegative) and the orbital dipole of the lone pair on N all point in the same direction (toward N and beyond); they reinforce each other, giving a large net dipole moment (4.90 x 10-30 C m).
In NF3, the N-F bond dipoles point from N toward F (since F is more electronegative than N); the orbital dipole of the lone pair points in the opposite direction to these bond dipoles and partially cancels them, resulting in a much lower net dipole moment (0.8 x 10-30 C m) for NF3.

Question 3: A chemist has two unlabelled triatomic samples — one is SO2 (bent) and one is CS2 (linear). Describe how dipole moment measurement can identify which is which, and predict the dipole moments. Also comment on how bond polarity alone is insufficient.

SO2 is a bent molecule (similar to H2O) due to one lone pair on S; its two S=O bond dipoles do not cancel because of the non-linear geometry; SO2 therefore has a significant non-zero dipole moment (experimentally about 1.63 D); it is a polar molecule.
CS2 is linear (similar to CO2) with two identical C=S bond dipoles pointing in exactly opposite directions; these cancel completely; CS2 has zero dipole moment and is a non-polar molecule despite having polar C=S bonds.
Bond polarity alone is insufficient: CS2 has polar bonds but is non-polar overall due to symmetry; SO2 has the same number of identical bond types but is polar due to bent geometry; measuring the dipole moment directly identifies the linear molecule (mu = 0) as CS2 and the bent molecule (mu > 0) as SO2, demonstrating that geometry and not just electronegativity difference determines molecular polarity.

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Case Set 5

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Pauling introduced the concept of hybridisation to explain the characteristic geometrical shapes of molecules like CH4, NH3, H2O, BCl3, and BeCl2. In ethene (C2H4), each carbon atom is sp2 hybridised, leaving one unhybridised p orbital perpendicular to the molecular plane. In ethyne (C2H2), each carbon is sp hybridised with two unhybridised p orbitals. A chemistry student noticed that as carbon changes hybridisation from sp3 to sp2 to sp, the C-C bond becomes progressively shorter and stronger. She also noted that the H-C-H bond angle changes significantly across these molecules, and that the percentage of s-character in the hybrid orbital changes.

Question 1: Compare the percentage of s-character, bond angles, and C-C bond lengths in sp3, sp2, and sp hybridised carbon.

sp3 hybridisation: 25% s-character; H-C-H bond angle = 109.5°; C-C bond length in ethane = 154 pm (single bond, only sigma bonds present).
sp2 hybridisation: 33% s-character; H-C-H bond angle = 117.6° and H-C-C = 121° in ethene; C-C bond length = 134 pm (one sigma + one pi bond, double bond).
sp hybridisation: 50% s-character; bond angle = 180° (linear); C-C bond length in ethyne = 120 pm (one sigma + two pi bonds, triple bond); as s-character increases, the bond becomes shorter, stronger, and more directional.

Question 2: Describe the formation of the C-C double bond in ethene in terms of hybridisation and overlap, specifying which bonds are sigma and which are pi.

In ethene, each carbon atom is sp2 hybridised; one of the three sp2 orbitals of carbon atom 1 overlaps axially (end-to-end) with one sp2 orbital of carbon atom 2 to form the C-C sigma bond (sp2-sp2 overlap).
Each carbon uses its remaining two sp2 orbitals to form C-H sigma bonds (sp2-s overlap) with two hydrogen atoms; each carbon thus forms 3 sigma bonds in total.
The unhybridised 2p orbital (perpendicular to the molecular plane) of each carbon overlaps sidewise (laterally) with the corresponding 2p orbital of the other carbon to form one pi bond; the C=C double bond therefore consists of one sigma + one pi bond, with the pi bond consisting of two electron cloud lobes above and below the molecular plane.

Question 3: Explain why pi bonds are weaker than sigma bonds. In the context of ethyne (C2H2), describe the complete bonding and explain why free rotation around the C-C triple bond is not possible.

Sigma bonds form by axial (end-to-end) overlap of orbitals directly along the internuclear axis; this overlap is more extensive and concentrated between the nuclei, giving sigma bonds greater strength (higher bond enthalpy per bond component).
Pi bonds form by sidewise (lateral) overlap of parallel p orbitals; this overlap is less extensive and occurs above and below the internuclear axis rather than along it; consequently pi bonds have smaller overlap and are weaker than sigma bonds.
In ethyne (C2H2): Each sp carbon forms one C-C sigma bond (sp-sp) and one C-H sigma bond (sp-s); each carbon has two unhybridised p orbitals (2px and 2py); these overlap sidewise with corresponding orbitals on the other carbon to form two pi bonds; the triple bond (1 sigma + 2 pi) is very strong (bond enthalpy 839 kJ mol-1); free rotation around the C-C triple bond is not possible because rotation would require breaking the two pi bonds, as their sidewise overlap requires the p orbitals to remain parallel; rotation would misalign the p orbitals and destroy the pi bonds.

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Chapter 4: Chemical Bonding and Molecular Structure | Case Study Questions

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Question 1: Write the electron transfer equations for the formation of NaCl from sodium and chlorine atoms, showing the electronic configurations at each step.

Question 2: Define lattice enthalpy and explain why NaCl is stable as a solid even though the sum of ionization enthalpy of Na (+495.8 kJ mol-1) and electron gain enthalpy of Cl (-348.7 kJ mol-1) is positive.

Question 3: Using Fajans' rules, compare the covalent character of NaCl and AgCl. Predict which compound would have a lower melting point and explain why.

Passage

Question 1: Define resonance. Draw all canonical forms of CO32- and explain why none of them alone represents the true structure.

Question 2: Calculate the total valence electrons in NH4+ and CO32- ions and describe the steps used to account for the ionic charge in Lewis structures.

Question 3: Calculate the formal charge on each atom in the ozone (O3) molecule. What does the pattern of formal charges tell us about the structure? Why do formal charges not represent real charge separation?

Passage

Question 1: Using VSEPR theory, predict and explain the shapes and bond angles of BeCl2, BF3, and CH4.

Question 2: Explain why the five P-Cl bonds in PCl5 are not equivalent, distinguishing between equatorial and axial bonds.

Question 3: If one bond pair in CH4 is replaced by a lone pair (giving NH3) and then another bond pair is replaced by a lone pair (giving H2O), trace the systematic change in shape and bond angle and explain the reasons for each change.

Passage

Question 1: Define dipole moment. Explain why CO2 has zero dipole moment despite having two polar C=O bonds.

Question 2: Explain why the dipole moment of NH3 (1.47 D) is much greater than that of NF3 (0.23 D), even though F is more electronegative than H.

Question 3: A chemist has two unlabelled triatomic samples — one is SO2 (bent) and one is CS2 (linear). Describe how dipole moment measurement can identify which is which, and predict the dipole moments. Also comment on how bond polarity alone is insufficient.

Passage

Question 1: Compare the percentage of s-character, bond angles, and C-C bond lengths in sp3, sp2, and sp hybridised carbon.

Question 2: Describe the formation of the C-C double bond in ethene in terms of hybridisation and overlap, specifying which bonds are sigma and which are pi.

Question 3: Explain why pi bonds are weaker than sigma bonds. In the context of ethyne (C2H2), describe the complete bonding and explain why free rotation around the C-C triple bond is not possible.

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