Chemical Bonding and Molecular Structure — NCERT Solutions

Given/Concept: A chemical bond is the force of attraction that holds two atoms together in a molecule or compound.

Formation of a Chemical Bond:

Atoms combine to form chemical bonds in order to attain a state of minimum energy and maximum stability (usually by achieving the nearest noble gas configuration).

There are three main types of chemical bonds:

1. Ionic (Electrovalent) Bond: Formed by the complete transfer of one or more electrons from an electropositive atom to an electronegative atom. The resulting oppositely charged ions attract each other electrostatically. Example: NaCl — Na loses one electron to Cl, forming Na⁺ and Cl⁻.

2. Covalent Bond: Formed by the mutual sharing of electron pairs between two atoms, both of which are short of the noble gas configuration. Example: H₂ — each H atom shares its one electron with the other.

3. Coordinate (Dative) Bond: A special type of covalent bond where both electrons of the shared pair are donated by one atom (the donor) to another (the acceptor). Example: NH₄⁺ — the lone pair of NH₃ is donated to H⁺.

Driving Force: The formation of a bond lowers the potential energy of the system. At the equilibrium bond distance, the energy is at a minimum — the system is most stable.

Conclusion: Chemical bonds form because the bonded state is energetically more stable than the separated atoms.

Concept: Lewis dot symbols represent the valence electrons of an atom as dots placed around the chemical symbol.

The number of valence electrons for each element:

Detailed representation:

• Mg (2 valence electrons): $\overset{\cdot}{\underset{\cdot}{\text{Mg}}}$
• Na (1 valence electron): Na·
• B (3 valence electrons): ·B· with one dot on top
• O (6 valence electrons): has 2 lone pairs and 2 unpaired electrons — $:\overset{\cdot\cdot}{O}:$ with two dots on each side and two unpaired
• N (5 valence electrons): one lone pair and 3 unpaired electrons
• Br (7 valence electrons): 3 lone pairs and 1 unpaired electron

Summary in standard notation:
$$\text{Na}\cdot \quad \cdot\text{Mg}\cdot \quad \cdot\overset{\cdot}{\text{B}}\cdot \quad :\overset{\cdot\cdot}{\text{O}}: \quad :\overset{\cdot}{\text{N}}:\cdot \quad :\overset{\cdot\cdot}{\text{Br}}:$$

(Each pair of dots = lone pair; single dot = unpaired electron)

Concept: Lewis symbols show valence electrons as dots. For ions, electrons are added (for anions) or removed (for cations) accordingly.

(a) S and S²⁻:

• S: Group 16, 6 valence electrons → 2 lone pairs + 2 unpaired electrons

$$\text{S: } :\overset{\cdot\cdot}{\text{S}}:$$

• S²⁻: Gains 2 electrons → 8 valence electrons → 4 lone pairs

$$[:\overset{\cdot\cdot}{\text{S}}:]^{2-}$$

(b) Al and Al³⁺:

• Al: Group 13, 3 valence electrons → 1 lone pair + 1 unpaired (or 3 unpaired)

$$\cdot\overset{\cdot}{\text{Al}}\cdot$$

• Al³⁺: Loses 3 electrons → 0 valence electrons

$$[\text{Al}]^{3+}$$

(c) H and H⁻:

• H: 1 valence electron → 1 unpaired electron

$$\text{H}\cdot$$

• H⁻: Gains 1 electron → 2 electrons (like He)

$$[:\text{H}]^{-}$$

Summary:

Concept: Lewis structures show all bonding pairs and lone pairs. Total valence electrons are counted and distributed to satisfy the octet rule (duet for H).

(a) H₂S:

• Total valence electrons: 2(1) + 6 = 8
• S is the central atom; forms 2 bonds with H; 2 lone pairs on S

$$\text{H} - \overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{S}}} - \text{H}$$
(S has 2 bond pairs + 2 lone pairs)

(b) SiCl₄:

• Total valence electrons: 4 + 4(7) = 32
• Si is central atom; forms 4 single bonds with Cl; each Cl has 3 lone pairs

$$\text{Cl}_4\text{Si: Si surrounded by 4 Cl atoms, each Cl has 3 lone pairs}$$
$$\begin{array}{c} \text{Cl} \\ | \\ \text{Cl} - \text{Si} - \text{Cl} \\ | \\ \text{Cl} \end{array}$$
(Each Cl has 3 lone pairs; Si has no lone pairs)

(c) BeF₂:

• Total valence electrons: 2 + 2(7) = 16
• Be is central atom; forms 2 single bonds with F; each F has 3 lone pairs; Be has no lone pairs (incomplete octet)

$$:\overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{F}}} - \text{Be} - \overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{F}}}:$$

(d) CO₃²⁻:

• Total valence electrons: 4 + 3(6) + 2 = 24
• C is central atom; one C=O double bond and two C–O single bonds (resonance exists)
• One structure:

$$\left[\overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{O}}} - \text{C}\left(=\overset{\cdot\cdot}{\text{O}}\right) - \overset{\cdot\cdot}{\underset{\cdot\cdot}{\text{O}}}\right]^{2-}$$

• The actual structure is a resonance hybrid with all three C–O bonds equivalent (bond order = 4/3).

(e) HCOOH (Formic acid):

• Total valence electrons: 1 + 4 + 2(6) + 1 = 18
• Structure: H–C(=O)–O–H

$$\text{H} - \overset{\|}{\underset{O}{C}} - \overset{\cdot\cdot}{\text{O}} - \text{H}$$

• The carbonyl C has a double bond to one O and a single bond to the –OH group; H is bonded to C.

$$\text{H} - \text{C}\begin{pmatrix} =O \\ -O-H \end{pmatrix}$$

(C has one double bond to O and one single bond to O–H; H is also bonded to C)

Definition of Octet Rule:
The octet rule states that atoms tend to combine in such a way that each atom has eight electrons in its valence shell, thereby attaining the electronic configuration of the nearest noble gas.

Significance:

1. It explains why atoms form chemical bonds — to complete their octet.
2. It helps in predicting the formulae of ionic and covalent compounds. For example, NaCl: Na loses 1e⁻ and Cl gains 1e⁻, both achieving octets.
3. It helps in drawing Lewis dot structures of molecules.
4. It explains the valency of elements in many compounds.

Limitations of the Octet Rule:

1. Incomplete octet: Some elements have fewer than 8 electrons in their valence shell in stable compounds. Example: LiCl (Li has 2e⁻), BeCl₂ (Be has 4e⁻), BCl₃ (B has 6e⁻).

2. Expanded octet: Elements of the 3rd period and beyond can accommodate more than 8 electrons using d-orbitals. Example: PCl₅ (P has 10e⁻), SF₆ (S has 12e⁻).

3. Odd-electron molecules: Molecules with an odd number of electrons cannot satisfy the octet rule for all atoms. Example: NO (11 electrons), NO₂ (17 electrons).

4. It does not explain the shape of molecules or the relative stability of molecules.

5. It does not account for the energy of bond formation — it gives no information about bond enthalpies.

6. Transition metal compounds often do not obey the octet rule.

Concept: An ionic bond is formed by the complete transfer of electrons from a metal to a non-metal. The following factors favour its formation:

Favourable Factors for Ionic Bond Formation:

1. Low ionization enthalpy of the metal (cation-forming atom): The metal atom should have a low ionization enthalpy so that it can easily lose electrons to form a cation. Example: Alkali metals (Na, K) have low ionization enthalpies.

2. High electron gain enthalpy (high negative value) of the non-metal: The non-metal should have a high (large negative) electron gain enthalpy so that it readily accepts electrons to form an anion. Example: Halogens (F, Cl) have high electron gain enthalpies.

3. High electronegativity difference between the two atoms: A large difference in electronegativity (generally > 1.7 on Pauling scale) between the two atoms favours ionic bond formation.

4. High lattice enthalpy: The greater the lattice enthalpy (energy released when ions come together to form the crystal lattice), the more stable the ionic compound. High lattice enthalpy is favoured by:

• Smaller ionic radii
• Higher ionic charges

Summary (Born-Haber cycle perspective):
The overall process is thermodynamically favourable when the energy released in lattice formation exceeds the energy required for ionization and electron transfer, making $\Delta H_f$ (enthalpy of formation) negative.

Concept (VSEPR Model): The shape of a molecule is determined by the total number of electron pairs (bond pairs + lone pairs) around the central atom. Lone pairs cause more repulsion than bond pairs, distorting ideal geometry.

(a) BeCl₂:

• Valence electrons on Be = 2; forms 2 bond pairs with Cl; lone pairs = 0
• Total electron pairs = 2 (both bond pairs)
• Geometry: Linear; Bond angle = 180°

$$\text{Cl} - \text{Be} - \text{Cl}$$

(b) BCl₃:

• Valence electrons on B = 3; forms 3 bond pairs with Cl; lone pairs = 0
• Total electron pairs = 3
• Geometry: Trigonal planar; Bond angle = 120°
• All three Cl atoms are in the same plane.

(c) SiCl₄:

• Valence electrons on Si = 4; forms 4 bond pairs with Cl; lone pairs = 0
• Total electron pairs = 4
• Geometry: Tetrahedral; Bond angle = 109.5°

(d) AsF₅:

• Valence electrons on As = 5; forms 5 bond pairs with F; lone pairs = 0
• Total electron pairs = 5
• Geometry: Trigonal bipyramidal
• 3 equatorial F atoms (bond angle 120°) and 2 axial F atoms (bond angle 90° with equatorial)

(e) H₂S:

• Valence electrons on S = 6; forms 2 bond pairs with H; lone pairs = 2
• Total electron pairs = 4 (tetrahedral arrangement of electron pairs)
• Due to 2 lone pairs, geometry is Bent (V-shaped)
• Bond angle = ~92° (less than 109.5° due to lp–lp and lp–bp repulsions)

(f) PH₃:

• Valence electrons on P = 5; forms 3 bond pairs with H; lone pairs = 1
• Total electron pairs = 4 (tetrahedral arrangement)
• Due to 1 lone pair, geometry is Trigonal pyramidal
• Bond angle = ~93.6° (slightly less than 109.5°)

Given: Both NH₃ and H₂O have a distorted tetrahedral geometry (based on 4 electron pairs around the central atom).

Concept (VSEPR): Lone pair–lone pair (lp–lp) repulsion > lone pair–bond pair (lp–bp) repulsion > bond pair–bond pair (bp–bp) repulsion.

In NH₃:

• Central atom N has 3 bond pairs (N–H) and 1 lone pair
• 1 lone pair causes repulsion with the 3 bond pairs
• Bond angle = 107° (slightly less than 109.5° of a perfect tetrahedron)
• Shape: Trigonal pyramidal

In H₂O:

• Central atom O has 2 bond pairs (O–H) and 2 lone pairs
• 2 lone pairs cause greater repulsion on the bond pairs
• The two lone pairs repel the two bond pairs more strongly, compressing the H–O–H angle further
• Bond angle = 104.5° (less than NH₃)
• Shape: Bent (V-shaped)

Reason for smaller bond angle in H₂O:
H₂O has two lone pairs while NH₃ has only one lone pair. The greater lp–lp and lp–bp repulsions in H₂O compress the bond angle more than in NH₃.

$$\text{Bond angle: } \text{H}_2\text{O (104.5°)} < \text{NH}_3 \text{ (107°)} < \text{Ideal tetrahedral (109.5°)}$$

Bond Order and Bond Strength:

Bond order is defined as the number of bonds (covalent bonds) between two atoms in a molecule. In MO theory:
$$\text{Bond Order} = \frac{1}{2}(N_b - N_a)$$
where $N_b$ = number of electrons in bonding MOs and $N_a$ = number of electrons in antibonding MOs.

Relationship between bond order and bond strength:

1. Higher bond order → Greater bond strength (higher bond enthalpy):

• Single bond (bond order = 1): e.g., H–H, bond enthalpy ≈ 436 kJ/mol
• Double bond (bond order = 2): e.g., O=O, bond enthalpy ≈ 498 kJ/mol
• Triple bond (bond order = 3): e.g., N≡N, bond enthalpy ≈ 946 kJ/mol

2. Higher bond order → Shorter bond length:

• As bond order increases, the bond length decreases.

3. Higher bond order → More stable the molecule.

Summary:
$$\text{Bond strength} \propto \text{Bond order} \propto \frac{1}{\text{Bond length}}$$

Thus, bond order is a direct measure of bond strength.

Definition of Bond Length:

Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is the distance at which the attractive and repulsive forces between the two atoms are balanced and the potential energy of the system is minimum.

Key Points:

• It is measured in picometres (pm) or Ångströms (Å); 1 Å = 100 pm.
• Bond length depends on:

1. Size of the atoms: Larger atoms → longer bond length.
2. Bond order: Higher bond order → shorter bond length.

• C–C (154 pm) > C=C (134 pm) > C≡C (120 pm)

3. Hybridisation: More s-character → shorter bond.

• $sp^3$ C–H > $sp^2$ C–H > $sp$ C–H
• In polyatomic molecules, bond length is the average value determined experimentally by X-ray diffraction, electron diffraction, or spectroscopic methods.

Example:

• H–H bond length = 74 pm
• O=O bond length = 121 pm
• N≡N bond length = 109 pm

Concept of Resonance:
When a single Lewis structure cannot accurately represent a molecule or ion, two or more Lewis structures (called canonical forms or resonance structures) are written. The actual structure is a resonance hybrid — an average of all canonical forms.

Resonance in CO₃²⁻:

Total valence electrons = 4 + 3(6) + 2 = 24

Three resonance structures can be drawn:

Structure I: C=O double bond to one oxygen, C–O single bonds to the other two (with negative charges on the two singly bonded oxygens).

Structure II: C=O double bond to the second oxygen.

Structure III: C=O double bond to the third oxygen.

$$\left[O=C\begin{pmatrix}-O^- \\ -O^-\end{pmatrix}\right] \longleftrightarrow \left[^-O-C\begin{pmatrix}=O \\ -O^-\end{pmatrix}\right] \longleftrightarrow \left[^-O-C\begin{pmatrix}-O^- \\ =O\end{pmatrix}\right]$$

Important Aspects:

1. The actual CO₃²⁻ ion is not represented by any single structure — it is the resonance hybrid of all three.

2. All three C–O bonds are equivalent with bond length intermediate between a C–O single bond (143 pm) and a C=O double bond (122 pm). Experimentally, all C–O bonds in CO₃²⁻ are equal at ~129 pm.

3. Bond order of each C–O bond = $\frac{4}{3}$ (one double bond shared over 3 bonds).

4. The resonance hybrid is more stable than any individual canonical form — this extra stability is called resonance energy (or delocalization energy).

5. The negative charge is equally distributed over all three oxygen atoms.

6. The geometry of CO₃²⁻ is trigonal planar with bond angles of 120°.

Given: Two structures (1) and (2) for H₃PO₃ are provided (figure not visible, but based on standard chemistry knowledge).

Standard structures of H₃PO₃:

• Structure 1: P is bonded to 3 OH groups and 1 H directly (P–H bond), with one P=O double bond. This is the correct structure (phosphorous acid is diprotic).
• Structure 2: P is bonded to 3 OH groups and has a P=O bond but no direct P–H bond (all H are on O).

Answer: No, these two structures cannot be taken as canonical forms (resonance structures) of H₃PO₃.

Reasons:

1. Canonical forms (resonance structures) must have the same arrangement of atoms (same skeletal structure). They differ only in the position of electrons (bonds and lone pairs), not in the position of atoms.

2. In structures 1 and 2 of H₃PO₃, the positions of hydrogen atoms are different — in one structure, H is directly bonded to P (P–H bond), while in the other, all H atoms are bonded to O (O–H bonds). This means the atomic connectivity is different.

3. Since the two structures differ in the arrangement of atoms (not just electrons), they represent different compounds (structural isomers or tautomers), not resonance structures.

4. Resonance structures must have the same number of bond pairs and lone pairs distributed differently; here the bonding framework itself changes.

Conclusion: Structures 1 and 2 of H₃PO₃ are not canonical forms. They are different structural representations (tautomers) because the positions of hydrogen atoms differ between the two structures.

Concept: Resonance structures are drawn when a single Lewis structure is insufficient to represent the actual bonding. The double-headed arrow (↔) is used between resonance structures.

(a) SO₂:

• Total valence electrons = 6 + 2(6) = 18
• S is the central atom

Structure I: S=O double bond on left, S–O single bond on right (lone pairs adjusted)
Structure II: S–O single bond on left, S=O double bond on right

$$\overset{\cdot\cdot}{O} = S - \overset{\cdot\cdot}{\underset{\cdot\cdot}{O}}: \longleftrightarrow :\overset{\cdot\cdot}{\underset{\cdot\cdot}{O}} - S = \overset{\cdot\cdot}{O}$$

Actual S–O bond order = 1.5 (intermediate between single and double bond)

(b) NO₂:

• Total valence electrons = 5 + 2(6) = 17 (odd electron molecule)
• N is the central atom

Structure I: N=O on left, N–O on right, unpaired electron on N
Structure II: N–O on left, N=O on right, unpaired electron on N

$$\overset{\cdot\cdot}{O} = N\cdot - \overset{\cdot\cdot}{\underset{\cdot\cdot}{O}}: \longleftrightarrow :\overset{\cdot\cdot}{\underset{\cdot\cdot}{O}} - \cdot N = \overset{\cdot\cdot}{O}$$

Actual N–O bond order = 1.5

(c) NO₃⁻:

• Total valence electrons = 5 + 3(6) + 1 = 24
• N is the central atom

Three resonance structures:

Structure I: N=O double bond to one O, N–O single bonds to the other two (negative charges on singly bonded O atoms)

Structure II: N=O double bond to the second O

Structure III: N=O double bond to the third O

$$\left[O=N\begin{pmatrix}-O^- \\ -O^-\end{pmatrix}\right] \longleftrightarrow \left[^-O-N\begin{pmatrix}=O \\ -O^-\end{pmatrix}\right] \longleftrightarrow \left[^-O-N\begin{pmatrix}-O^- \\ =O\end{pmatrix}\right]$$

Actual N–O bond order = 4/3 ≈ 1.33; all three N–O bonds are equivalent.

Concept: In ionic bond formation, electrons are transferred from the metal (low ionization enthalpy) to the non-metal (high electron gain enthalpy). Lewis symbols track this transfer.

(a) K and S:

• K has 1 valence electron; S needs 2 electrons to complete its octet.
• 2 K atoms each donate 1 electron to S.

$$2\text{K}\cdot + :\overset{\cdot\cdot}{\underset{\cdot\cdot}{S}}: \rightarrow 2\text{K}^+ + [:\overset{\cdot\cdot}{\underset{\cdot\cdot}{S}}:]^{2-}$$

• Product: K₂S (2 K⁺ ions and 1 S²⁻ ion)

(b) Ca and O:

• Ca has 2 valence electrons; O needs 2 electrons to complete its octet.
• 1 Ca atom donates 2 electrons to 1 O atom.

$$\cdot\text{Ca}\cdot + :\overset{\cdot\cdot}{O}: \rightarrow \text{Ca}^{2+} + [:\overset{\cdot\cdot}{\underset{\cdot\cdot}{O}}:]^{2-}$$

• Product: CaO (1 Ca²⁺ ion and 1 O²⁻ ion)

(c) Al and N:

• Al has 3 valence electrons; N needs 3 electrons to complete its octet.
• 1 Al atom donates 3 electrons to 1 N atom.

$$\cdot\overset{\cdot}{\text{Al}}\cdot + :\overset{\cdot}{\underset{\cdot}{N}}\cdot \rightarrow \text{Al}^{3+} + [:\overset{\cdot\cdot}{\underset{\cdot\cdot}{N}}:]^{3-}$$

• Product: AlN (1 Al³⁺ ion and 1 N³⁻ ion)

Given: Both CO₂ and H₂O are triatomic molecules.

Concept: The shape of a molecule is related to its dipole moment. Individual bond dipoles may cancel (giving zero net dipole) or add up (giving a net dipole).

CO₂ (Linear, μ = 0):

• Structure: O=C=O
• Each C=O bond is polar (C is slightly positive, O is slightly negative) due to electronegativity difference.
• However, CO₂ is linear (bond angle = 180°), so the two C=O bond dipoles are equal in magnitude but point in opposite directions.
• They cancel each other vectorially:

$$\vec{\mu}_{C=O} + \vec{\mu}_{O=C} = 0$$

• Net dipole moment of CO₂ = 0 (non-polar molecule despite polar bonds)

H₂O (Bent, μ ≠ 0):

• Structure: H–O–H with bond angle ≈ 104.5°
• Each O–H bond is polar (O is δ⁻, H is δ⁺).
• Because H₂O is bent, the two O–H bond dipoles do not cancel — they add up to give a net dipole moment.
• The resultant dipole points from the midpoint of H–H towards O.
• Net dipole moment of H₂O = 1.85 D (polar molecule)

Conclusion:

• CO₂ is linear → bond dipoles cancel → μ = 0 → non-polar
• H₂O is bent → bond dipoles add up → μ ≠ 0 → polar

The bent shape of H₂O is due to the 2 lone pairs on O which cause lp–bp repulsion, while CO₂ has no lone pairs on C and adopts a linear geometry.

Dipole Moment (μ): It is the product of the magnitude of charge (q) and the distance (d) between the charges:
$$\mu = q \times d$$
Unit: Debye (D); 1 D = 3.336 × 10⁻³⁰ C·m

Significance and Applications of Dipole Moment:

1. Determining polarity of bonds: A non-zero dipole moment indicates a polar bond. Greater the dipole moment, greater the polarity. Example: HF (μ = 1.91 D) is more polar than HCl (μ = 1.07 D).

2. Predicting molecular geometry/shape:

• If μ = 0 for a molecule with polar bonds → symmetric (linear, tetrahedral, etc.) geometry. Example: CO₂ (μ = 0) is linear; BF₃ (μ = 0) is trigonal planar.
• If μ ≠ 0 → asymmetric geometry. Example: H₂O (μ = 1.85 D) is bent.

3. Distinguishing between isomers: Dipole moment helps distinguish between cis and trans isomers. Example: cis-1,2-dichloroethene has μ ≠ 0; trans-1,2-dichloroethene has μ = 0.

4. Determining percentage ionic character: The ratio of observed dipole moment to the theoretical dipole moment (assuming complete charge transfer) gives the percentage ionic character of a bond.

$$\% \text{ ionic character} = \frac{\mu_{\text{observed}}}{\mu_{\text{theoretical}}} \times 100$$

5. Predicting solubility: Polar molecules (high μ) dissolve in polar solvents (like water); non-polar molecules (μ = 0) dissolve in non-polar solvents.

6. Understanding intermolecular forces: Molecules with high dipole moments have stronger dipole–dipole interactions, leading to higher boiling points.

Electronegativity:
Electronegativity is defined as the tendency of an atom in a covalent molecule to attract the shared pair of electrons towards itself. It is a relative property and has no units. The Pauling scale is most commonly used (F = 4.0, most electronegative).

Electron Gain Enthalpy (Electron Affinity):
Electron gain enthalpy is defined as the enthalpy change when an electron is added to an isolated gaseous atom to form a gaseous anion:
$$X(g) + e^- \rightarrow X^-(g) \quad \Delta_{eg}H$$
It is measured in kJ/mol and is a measurable thermodynamic quantity.

Differences between Electronegativity and Electron Gain Enthalpy:

Key distinction: Electronegativity refers to the power of an atom to attract electrons within a bond, while electron gain enthalpy refers to the energy change when a free atom accepts an electron.

Polar Covalent Bond:
A polar covalent bond is formed when two atoms with different electronegativities share a pair of electrons. The more electronegative atom attracts the shared electron pair more strongly towards itself, acquiring a partial negative charge (δ⁻), while the less electronegative atom acquires a partial positive charge (δ⁺).

Example: HCl (Hydrogen Chloride)

• H has electronegativity = 2.1 (Pauling scale)
• Cl has electronegativity = 3.0 (Pauling scale)
• Electronegativity difference = 3.0 – 2.1 = 0.9

Since Cl is more electronegative, it pulls the shared electron pair towards itself:
$$\text{H}^{\delta+} - \text{Cl}^{\delta-}$$

• Cl acquires a partial negative charge (δ⁻)
• H acquires a partial positive charge (δ⁺)
• The molecule has a net dipole moment (μ = 1.07 D)

Other examples:

• H–F: F is more electronegative → H^{δ+}–F^{δ−} (μ = 1.91 D)
• H–O in water: O is more electronegative → H^{δ+}–O^{δ−}

Key Points:

1. The greater the electronegativity difference, the more polar the bond.
2. A purely covalent (non-polar) bond exists when both atoms have the same electronegativity (e.g., H–H, Cl–Cl).
3. A polar covalent bond is intermediate between a pure covalent bond and an ionic bond.

Concept: Ionic character of a bond depends on the electronegativity difference between the bonded atoms. Greater the electronegativity difference, greater the ionic character.

Electronegativity values (Pauling scale):

• Li = 1.0, F = 4.0 → ΔEN = 3.0 (LiF)
• K = 0.8, O = 3.5 → ΔEN = 2.7 (K₂O)
• N = 3.0, N = 3.0 → ΔEN = 0 (N₂)
• S = 2.5, O = 3.5 → ΔEN = 1.0 (SO₂)
• Cl = 3.0, F = 4.0 → ΔEN = 1.0 (ClF₃)

Ranking by electronegativity difference (increasing order):

$$N_2 (0) < SO_2 \approx ClF_3 (1.0) < K_2O (2.7) < LiF (3.0)$$

Order of increasing ionic character:
$$\boxed{N_2 < SO_2 \approx ClF_3 < K_2O < LiF}$$

Note: N₂ has zero ionic character (purely covalent, identical atoms). LiF has the highest ionic character due to the largest electronegativity difference.

Given: Skeletal structure of CH₃COOH (acetic acid) with some incorrect bonds (figure not fully visible from OCR).

Correct Lewis Structure of CH₃COOH (Acetic Acid):

Step 1: Count valence electrons

• C: 4 × 2 = 8 (two C atoms)
• H: 1 × 4 = 4 (four H atoms)
• O: 6 × 2 = 12 (two O atoms)
• Total = 24 valence electrons

Step 2: Connectivity

• CH₃ group: C bonded to 3 H atoms (all single bonds) and to the carbonyl C
• Carbonyl C: double bond to one O (C=O), single bond to O–H, single bond to CH₃

Correct Lewis Structure:

$$\text{H} - \overset{|}{\underset{H}{C}} - \overset{||}{\underset{O}{C}} - \overset{\cdot\cdot}{O} - H$$

More explicitly:

$$\begin{array}{ccccc} & H & & & \\ | & & & & \\ H-C-C(=O)-O-H & & & \\ | & & & & \\ & H & & & \end{array}$$

Key features of the correct structure:

1. The CH₃ carbon is sp³ hybridised — bonded to 3 H atoms and 1 C atom, all by single bonds.
2. The carboxyl carbon (COOH) is sp² hybridised — has one C=O double bond (carbonyl) and one C–O single bond (to –OH).
3. The –OH oxygen has 2 lone pairs and forms a single bond to H.
4. The carbonyl oxygen has 2 lone pairs and forms a double bond to C.
5. No bonds should be shown as double bonds in the CH₃ group — a common error in incorrect structures.