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

ByteNotes

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student draws a Lewis structure for SO3 with a double bond to one oxygen and single bonds to two others. Another student argues that all S-O bonds should be equal. Who is correct, and how would you resolve this using resonance?

The second student is correct in pointing out that experimentally, all three S-O bonds in SO3 are equivalent in length, which cannot be represented by a single Lewis structure showing one double bond and two single bonds.
The concept of resonance resolves this: SO3 has three equivalent canonical forms, each with one S=O double bond and two S-O single bonds at different positions; the three forms are in equilibrium as a resonance hybrid.
The resonance hybrid represents SO3 more accurately with all three S-O bonds having intermediate bond order (between 1 and 2), equal bond lengths, and equivalent character — consistent with experimental data.
This is analogous to the CO3 2- ion and O3, where single Lewis structures are inadequate; resonance energy makes the hybrid more stable than any single canonical form.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

With reference to the VSEPR theory, predict the shapes of the following molecules and explain any deviations from ideal geometry: H2O, NH3, SF4 (AB4E type).

H2O (AB2E2): Oxygen has 2 bond pairs and 2 lone pairs; the electron arrangement is tetrahedral but the molecular shape is bent/V-shaped; two lone pairs cause greater lp-lp and lp-bp repulsion, compressing the H-O-H bond angle to 104.5° from the ideal 109.5°.
NH3 (AB3E): Nitrogen has 3 bond pairs and 1 lone pair; electron arrangement is tetrahedral but molecular shape is trigonal pyramidal; one lone pair exerts more repulsion on bond pairs, reducing H-N-H angle to 107° from 109.5°.
SF4 (AB4E): Sulfur has 4 bond pairs and 1 lone pair; electron arrangement is trigonal bipyramidal; the lone pair preferentially occupies an equatorial position (fewer 90° lp-bp repulsions); this gives SF4 a see-saw (distorted tetrahedral) shape, with bond angles of approximately 101.6° (axial) and 187° (equatorial).
In all three cases, lone pairs cause deviation from ideal geometry because lone pair electrons occupy more space on the central atom and repel bond pairs more strongly than bond pairs repel each other.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

An industrial chemist wants to determine if a newly synthesised triatomic molecule AB2 is linear or bent. Describe how measuring the dipole moment can help make this determination.

If the triatomic molecule AB2 has a zero dipole moment, the two A-B bond dipoles must cancel completely, which is only possible if the molecule is linear (bond dipoles are equal and directly opposite, as in CO2).
If the molecule has a non-zero dipole moment, the bond dipoles do not cancel, indicating that the molecule is bent (V-shaped); the angle between the bonds results in a net vector sum, as seen in H2O (mu = 1.85 D) and H2S (mu = 0.95 D).
This method is particularly powerful because it is an experimentally measurable quantity; dipole moments are determined in the laboratory using dielectric constant measurements and can distinguish geometric isomers.
In practice, a linear symmetric AB2 molecule like CO2 has mu = 0, while a bent molecule like H2O or SO2 has mu > 0; this measurement directly reveals the spatial arrangement of bonds.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

With reference to the chapter, explain why the bond dissociation enthalpy of F2 (155 kJ mol-1) is much lower than that of Cl2 (243 kJ mol-1), even though F is more electronegative than Cl.

Bond dissociation enthalpy represents the energy required to break one mole of F-F or Cl-Cl bonds in the gaseous state; a lower value indicates a weaker bond.
Although F is more electronegative, the F-F bond is weaker than Cl-Cl because F atoms are very small; the two F atoms are much closer together in F2, causing strong repulsion between the lone pairs on adjacent F atoms at close range.
Cl is larger; in Cl2, the lone pairs are farther apart and this lone pair-lone pair repulsion is much less significant, making the Cl-Cl bond relatively stronger despite lower electronegativity.
This seemingly anomalous result highlights that bond strength is not solely determined by electronegativity but also by atomic size and lone pair repulsion, which are bond parameters discussed in section 4.3 of the chapter.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

Using the concept of formal charge, determine the best Lewis structure for CO from the following options: (a) :C=O: with a double bond, or (b) :C triple bond O: with a triple bond.

CO has 4 + 6 = 10 valence electrons. For structure (a) with C=O (double bond): C has 2 lone pair electrons and shares 4 in the double bond; FC of C = 4 - 2 - (4/2) = 0. O has 4 lone pair electrons and shares 4; FC of O = 6 - 4 - (4/2) = 0. But this leaves only 8 electrons and does not complete the octet on C with a lone pair.
For structure (b) with :C triple bond O: (triple bond, one lone pair on each): FC of C = 4 - 2 - (6/2) = -1. FC of O = 6 - 2 - (6/2) = +1. Total formal charge = 0, matching the neutral molecule.
Structure (b) with a triple bond and formal charges of -1 on C and +1 on O is the correct Lewis structure because it satisfies the octet rule on both atoms and accounts for all 10 valence electrons.
The lowest formal charge magnitude does not always guarantee the best structure; the triple bond in CO is consistent with its high bond dissociation enthalpy and short bond length.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

With reference to the chapter, explain why BF3 readily accepts a lone pair from NH3 to form F3B-NH3. Relate this to hybridisation and the octet rule.

In BF3, the central boron atom is sp2 hybridised and has only 6 electrons in its valence shell (an incomplete octet) despite satisfying valence requirements; the unhybridised empty p orbital makes BF3 an electron-deficient species.
NH3 has a lone pair of electrons on the sp3 hybridised nitrogen atom that is available for donation to the empty p orbital of B in BF3.
When F3B-NH3 (Lewis acid-base adduct) forms, the boron atom accepts the lone pair from N and changes its hybridisation from sp2 to sp3; this completes the octet around B and the resultant adduct has tetrahedral geometry around B.
This reaction illustrates a key limitation of the octet rule (BF3 violates it) and also demonstrates how hybridisation changes upon bond formation; it is an important concept linking hybridisation, octet limitations, and Lewis acid-base theory.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

Predict whether the following pairs of molecules would form hydrogen bonds with water: (a) HF, (b) CH4, (c) NH3, (d) CCl4. Justify your predictions.

HF and NH3 can form hydrogen bonds with water. HF contains H bonded to the highly electronegative F, satisfying both conditions for H-bonding; it can act as H-bond donor (H-F...O) and acceptor (O-H...F). NH3 has N bonded to H and has a lone pair on N that can accept H-bonds from water (O-H...N) and donate H-bonds (N-H...O).
CH4 cannot form hydrogen bonds with water. Although it contains H, carbon is not sufficiently electronegative to create a significant partial positive charge on H; also, C has no lone pairs to act as a hydrogen bond acceptor. CH4 is nonpolar and interacts with water only through weak van der Waals forces.
CCl4 similarly cannot form hydrogen bonds with water. Although Cl is electronegative, it is attached to C (not H); there are no O-H, N-H, or F-H bonds in CCl4, so the criterion for H-bond formation (H attached to F, O, or N) is not met.
The fundamental criterion: hydrogen bonding requires H to be directly bonded to F, O, or N; only then is the H polarised enough (delta+) to be attracted to the lone pairs of another electronegative atom.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A molecule has the general formula AB3 with the central atom A having one lone pair. Predict the shape, bond angles, and hybridisation of A. Give an example from the chapter.

An AB3E molecule type has the central atom A with 3 bond pairs and 1 lone pair; the total electron pairs around A = 4, giving a basic tetrahedral electron arrangement.
However, the molecular shape (considering only bond pairs and atoms) is trigonal pyramidal; the lone pair is not counted in describing the molecular shape but occupies one tetrahedral position.
The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, so the bond angle is reduced below the ideal 109.5° to approximately 107°; the central atom is sp3 hybridised with one hybrid orbital holding the lone pair.
Example from the chapter: NH3 (ammonia) — nitrogen is the central atom with 3 N-H bond pairs and 1 lone pair; it is sp3 hybridised; the molecule is trigonal pyramidal with H-N-H bond angle of 107°.

ByteNotes

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

An experiment shows that the O-O bond length in O3 is 128 pm, intermediate between O-O single bond (148 pm) and O=O double bond (121 pm). How does the resonance concept explain this observation?

A single Lewis structure for O3 would predict one O-O single bond (148 pm) and one O=O double bond (121 pm), giving two unequal bonds; but experiment shows both bonds are equal at 128 pm — a value intermediate between the two.
The resonance concept explains this by proposing two canonical forms: in form I, the first O-O is a double bond and the second is single; in form II, the reverse is true; neither form alone accurately describes the molecule.
The actual molecule exists as a resonance hybrid of both canonical forms simultaneously, not alternating between them; this hybrid has O-O bonds of bond order 1.5 (intermediate between 1 and 2), consistent with the observed 128 pm bond length.
The resonance hybrid is also more stable (lower energy) than either canonical form alone due to resonance stabilisation; canonical forms have no independent existence and the molecule has a single structure described only by the hybrid.

ByteNotes

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

Using molecular orbital theory, explain why He2 does not exist but He2+ is known to exist.

He2: Each He atom has 2 electrons (configuration 1s2); in He2, all 4 electrons fill the sigma1s (2 electrons) and sigma*1s (2 electrons) MOs. Bond order = 1/2(2-2) = 0; zero bond order means no net bonding and the molecule is unstable — He2 does not exist.
He2+: This species has 3 electrons total (one electron removed from He2); MO configuration is (sigma1s)2(sigma*1s)1; Nb=2, Na=1. Bond order = 1/2(2-1) = 0.5.
A positive bond order of 0.5 means He2+ has weak but measurable net bonding; the species is stable (though weakly so) and has indeed been detected experimentally.
This demonstrates the power of MO theory: removing one electron from an unstable He2 generates a weakly stable He2+, because the removed electron comes from the antibonding sigma*1s orbital, reducing the antibonding effect.

ByteNotes

Premium Access

Unlock all Application Questions

You are viewing 10 free application questions from this chapter. Upgrade to Premium to unlock the remaining 12 questions.

Upgrade to Premium12 questions locked

Chapter sections

Chapter 4: Chemical Bonding and Molecular Structure | Application Questions

A student draws a Lewis structure for SO3 with a double bond to one oxygen and single bonds to two others. Another student argues that all S-O bonds should be equal. Who is correct, and how would you resolve this using resonance?

With reference to the VSEPR theory, predict the shapes of the following molecules and explain any deviations from ideal geometry: H2O, NH3, SF4 (AB4E type).

An industrial chemist wants to determine if a newly synthesised triatomic molecule AB2 is linear or bent. Describe how measuring the dipole moment can help make this determination.

With reference to the chapter, explain why the bond dissociation enthalpy of F2 (155 kJ mol-1) is much lower than that of Cl2 (243 kJ mol-1), even though F is more electronegative than Cl.

Using the concept of formal charge, determine the best Lewis structure for CO from the following options: (a) :C=O: with a double bond, or (b) :C triple bond O: with a triple bond.

With reference to the chapter, explain why BF3 readily accepts a lone pair from NH3 to form F3B-NH3. Relate this to hybridisation and the octet rule.

Predict whether the following pairs of molecules would form hydrogen bonds with water: (a) HF, (b) CH4, (c) NH3, (d) CCl4. Justify your predictions.

A molecule has the general formula AB3 with the central atom A having one lone pair. Predict the shape, bond angles, and hybridisation of A. Give an example from the chapter.

An experiment shows that the O-O bond length in O3 is 128 pm, intermediate between O-O single bond (148 pm) and O=O double bond (121 pm). How does the resonance concept explain this observation?

Using molecular orbital theory, explain why He2 does not exist but He2+ is known to exist.

Unlock all Application Questions