Chemical Bonding and Molecular Structure

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:

| Element | Group | Valence electrons | Lewis dot symbol |
|---------|-------|-------------------|------------------|
| Mg | 2 | 2 | $\cdot$Mg$\cdot$ |
| Na | 1 | 1 | Na$\cdot$ |
| B | 13 | 3 | $\cdot$B$\cdot$ (with one extra dot) |
| O | 16 | 6 | $:\ddot{O}:$ |
| N | 15 | 5 | $\cdot\ddot{N}\cdot$ |
| Br | 17 | 7 | $:\ddot{Br}\cdot$ |

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:

| Species | Valence electrons | Lewis Symbol |
|---------|------------------|--------------|
| S | 6 | $:\overset{..}{S}:$ |
| S²⁻ | 8 | $[:\overset{..}{S}:]^{2-}$ |
| Al | 3 | $\cdot\overset{\cdot}{Al}\cdot$ |
| Al³⁺ | 0 | $[Al]^{3+}$ |
| H | 1 | H· |
| H⁻ | 2 | $[:H]^{-}$ |

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:

| Property | Electronegativity | Electron Gain Enthalpy |
|----------|-------------------|------------------------|
| Definition | Tendency to attract shared electrons in a bond | Energy change when an isolated atom gains an electron |
| State | Applies to atoms in a molecule (bonded state) | Applies to isolated gaseous atoms |
| Measurement | Relative, dimensionless (Pauling scale) | Absolute, measured in kJ/mol |
| Nature | Not directly measurable; calculated from bond energies | Experimentally measurable |
| Dependence | Depends on the bonding environment | Property of the free atom |

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.

Given: Two possible geometries for CH₄: tetrahedral and square planar.

Why CH₄ is NOT square planar:

1. VSEPR Argument:
- C in CH₄ has 4 bond pairs and 0 lone pairs.
- According to VSEPR theory, 4 electron pairs arrange themselves to be as far apart as possible to minimize repulsion.
- In a tetrahedral arrangement, the bond angle is 109.5° — maximum separation between 4 pairs.
- In a square planar arrangement, the bond angle would be 90° — much less separation, leading to greater repulsion.
- Therefore, tetrahedral geometry is more stable.

2. Hybridisation Argument:
- C undergoes sp³ hybridisation in CH₄, forming 4 equivalent sp³ hybrid orbitals directed towards the corners of a tetrahedron.
- Square planar geometry would require dsp² hybridisation (using d-orbitals), which is not available for carbon in its ground or excited state (carbon has no d-orbitals in the valence shell — it is in the 2nd period).

3. Energy Argument:
- The tetrahedral geometry minimises electron pair repulsions and is the lowest energy configuration for 4 bond pairs.
- Square planar geometry would have higher energy due to greater repulsions (90° angles vs 109.5°).

Conclusion: CH₄ is tetrahedral (bond angle = 109.5°) because this geometry minimises electron pair repulsions. Square planar geometry (bond angle = 90°) would be less stable and would require d-orbital participation, which is unavailable for carbon.

Given: BeH₂ has polar Be–H bonds but zero net dipole moment.

Explanation:

Step 1: Polarity of Be–H bond
- Electronegativity of Be = 1.5, H = 2.1
- Electronegativity difference = 0.6 → each Be–H bond is polar
- H is slightly more electronegative, so H carries δ⁻ and Be carries δ⁺ in each bond

Step 2: Geometry of BeH₂
- Be has 2 bond pairs and 0 lone pairs.
- VSEPR: 2 electron pairs → linear geometry (bond angle = 180°)
- BeH₂ undergoes sp hybridisation → linear molecule

Step 3: Cancellation of dipole moments
- The two Be–H bond dipoles are equal in magnitude but point in exactly opposite directions (180° apart).
- Vectorially:
$$\vec{\mu}_1 + \vec{\mu}_2 = 0$$
$$\text{H}^{\delta-} \leftarrow \text{Be} \rightarrow \text{H}^{\delta-}$$

- The two bond dipoles cancel each other completely.

Conclusion: Although each Be–H bond is polar, the linear geometry of BeH₂ causes the two bond dipoles to cancel vectorially, resulting in a net dipole moment of zero.

$$\mu_{\text{net}} = 0 \text{ D}$$

Given: NH₃ and NF₃ — both have trigonal pyramidal geometry with a lone pair on N.

Dipole Moment Analysis:

Both NH₃ and NF₃ have the same shape (trigonal pyramidal), so geometry alone does not differentiate them.

In NH₃:
- N is more electronegative than H (N = 3.0, H = 2.1)
- Each N–H bond dipole points from H towards N (towards N)
- The lone pair on N also contributes a dipole pointing away from N (in the same direction as the resultant of bond dipoles)
- The lone pair dipole and the resultant bond dipole add up (act in the same direction)
- μ(NH₃) = 1.46 D (relatively high)

In NF₃:
- F is more electronegative than N (F = 4.0, N = 3.0)
- Each N–F bond dipole points from N towards F (away from N)
- The lone pair on N contributes a dipole pointing away from N (opposite to the resultant of bond dipoles)
- The lone pair dipole and the resultant bond dipole oppose each other (partially cancel)
- μ(NF₃) = 0.24 D (much lower)

Conclusion:
\mu(\text{NH}_3) > \mu(\text{NF}_3)

NH₃ has a higher dipole moment than NF₃ because in NH₃, the lone pair dipole and bond dipoles reinforce each other, while in NF₃, they partially oppose each other.

Hybridisation:
Hybridisation is the process of intermixing of atomic orbitals of slightly different energies to form a new set of equivalent orbitals called hybrid orbitals. These hybrid orbitals have the same energy, shape, and are directed in space to minimise repulsion.

Key features:
- Number of hybrid orbitals formed = number of atomic orbitals mixed
- Hybrid orbitals are more effective in forming bonds than pure atomic orbitals
- Hybridisation explains the geometry of molecules

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(a) sp Hybridisation:
- Mixing: 1 s-orbital + 1 p-orbital → 2 sp hybrid orbitals
- Shape: Each sp hybrid orbital has one large lobe and one small lobe; the large lobes point in opposite directions.
- Geometry: Linear; bond angle = 180°
- Example: BeCl₂, C₂H₂ (acetylene)
- The two sp orbitals are oriented at 180° to each other.

(b) sp² Hybridisation:
- Mixing: 1 s-orbital + 2 p-orbitals → 3 sp² hybrid orbitals
- Shape: Each sp² orbital has one large lobe and one small lobe; the three large lobes lie in the same plane.
- Geometry: Trigonal planar; bond angle = 120°
- Example: BCl₃, C₂H₄ (ethylene), graphite
- One unhybridised p-orbital remains perpendicular to the plane (used for π bonding).

(c) sp³ Hybridisation:
- Mixing: 1 s-orbital + 3 p-orbitals → 4 sp³ hybrid orbitals
- Shape: Each sp³ orbital has one large lobe and one small lobe; the four large lobes point towards the corners of a tetrahedron.
- Geometry: Tetrahedral; bond angle = 109.5°
- Example: CH₄, NH₃, H₂O
- In NH₃ and H₂O, one or two sp³ orbitals are occupied by lone pairs, distorting the geometry.

Summary Table:

| Hybridisation | Orbitals mixed | No. of hybrid orbitals | Geometry | Bond angle |
|--------------|----------------|------------------------|----------|------------|
| sp | s + p | 2 | Linear | 180° |
| sp² | s + 2p | 3 | Trigonal planar | 120° |
| sp³ | s + 3p | 4 | Tetrahedral | 109.5° |

Given Reaction:
$$\text{AlCl}_3 + \text{Cl}^- \rightarrow \text{AlCl}_4^-$$

In AlCl₃:
- Al has 3 valence electrons; forms 3 bonds with Cl; no lone pairs on Al.
- Total electron pairs around Al = 3 (all bond pairs)
- Hybridisation of Al = sp²
- Geometry: Trigonal planar; bond angle = 120°
- Al has an empty p-orbital (incomplete octet — Lewis acid)

In AlCl₄⁻:
- Al accepts a Cl⁻ ion (Lewis base donates lone pair to Al's empty orbital)
- Now Al forms 4 bonds with 4 Cl atoms; no lone pairs on Al.
- Total electron pairs around Al = 4 (all bond pairs)
- Hybridisation of Al = sp³
- Geometry: Tetrahedral; bond angle = 109.5°

Change in Hybridisation:
$$\text{AlCl}_3: sp^2 \xrightarrow{+\text{Cl}^-} \text{AlCl}_4^-: sp^3$$

Conclusion: The hybridisation of Al changes from sp² (in AlCl₃) to sp³ (in AlCl₄⁻) as a result of the reaction. The empty p-orbital of Al in AlCl₃ participates in hybridisation upon accepting the lone pair from Cl⁻.

Given Reaction:
$$\text{BF}_3 + \text{NH}_3 \rightarrow \text{F}_3\text{B}\cdot\text{NH}_3$$

In BF₃ (before reaction):
- B has 3 valence electrons; forms 3 bonds with F; no lone pairs on B.
- Total electron pairs around B = 3
- Hybridisation of B = sp²
- Geometry: Trigonal planar; bond angle = 120°
- B has one empty p-orbital (Lewis acid)

In NH₃ (before reaction):
- N has 5 valence electrons; forms 3 bonds with H; 1 lone pair on N.
- Total electron pairs around N = 4
- Hybridisation of N = sp³
- Geometry: Trigonal pyramidal

In F₃B·NH₃ (after reaction):
- NH₃ donates its lone pair to the empty p-orbital of BF₃ (coordinate/dative bond forms: N→B)
- B: Now has 4 bond pairs (3 B–F + 1 B←N); hybridisation changes to sp³; geometry becomes tetrahedral
- N: Still has 4 electron pairs (3 N–H + 1 N→B lone pair now used in bond); hybridisation remains sp³; geometry becomes approximately tetrahedral (the lone pair is now used in bonding)

Conclusion:
- B: Hybridisation changes from sp² → sp³ ✓
- N: Hybridisation remains sp³ (no change), but the lone pair is now used to form the coordinate bond with B.

Yes, there is a change in hybridisation of B (sp² → sp³), but no change in hybridisation of N (remains sp³).

Concept:
- A double bond = 1 sigma (σ) bond + 1 pi (π) bond
- A triple bond = 1 sigma (σ) bond + 2 pi (π) bonds
- σ bond: head-on overlap of orbitals
- π bond: lateral (sideways) overlap of p-orbitals

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(a) C₂H₄ (Ethylene) — Double Bond:

Hybridisation: Each C is sp² hybridised.
- Each C: 3 sp² orbitals (used for σ bonds) + 1 unhybridised p-orbital (used for π bond)
- C–C σ bond: sp²–sp² head-on overlap
- C=C π bond: p–p lateral overlap (the two unhybridised p-orbitals, one on each C, overlap sideways)
- Each C also forms 2 C–H σ bonds (sp²–s overlap)

Structure:
$$\text{H}_2\text{C} = \text{CH}_2$$
- All 6 atoms (2C + 4H) lie in the same plane (planar molecule)
- The π bond is above and below the plane
- Bond angle ≈ 120°

Diagram description: Two C atoms connected by a σ bond (sp²–sp² overlap along the internuclear axis) and a π bond (p–p lateral overlap perpendicular to the molecular plane).

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(b) C₂H₂ (Acetylene) — Triple Bond:

Hybridisation: Each C is sp hybridised.
- Each C: 2 sp orbitals (used for σ bonds) + 2 unhybridised p-orbitals (used for 2 π bonds)
- C–C σ bond: sp–sp head-on overlap
- Two C≡C π bonds: two sets of p–p lateral overlaps (in two perpendicular planes)
- Each C also forms 1 C–H σ bond (sp–s overlap)

Structure:
$$\text{H} - \text{C} \equiv \text{C} - \text{H}$$
- Linear molecule; bond angle = 180°
- Two π bonds are in mutually perpendicular planes

Diagram description: Two C atoms connected by one σ bond (sp–sp overlap) and two π bonds (two sets of p–p lateral overlaps in perpendicular planes). The molecule is linear.

Note: The question lists (a) C₂H₄ and (b) C₂H₄ in the source, but based on context and standard NCERT questions, it should be (a) C₂H₂ and (b) C₂H₄.

Concept:
- Every single bond = 1 σ bond
- Every double bond = 1 σ bond + 1 π bond
- Every triple bond = 1 σ bond + 2 π bonds

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(a) C₂H₂ (Acetylene/Ethyne): H–C≡C–H

Bonds present:
- 2 C–H single bonds → 2 σ bonds
- 1 C≡C triple bond → 1 σ bond + 2 π bonds

Total σ bonds = 2 + 1 = 3
Total π bonds = 2

$$\text{C}_2\text{H}_2: \text{ 3 sigma bonds, 2 pi bonds}$$

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(b) C₂H₄ (Ethylene/Ethene): H₂C=CH₂

Bonds present:
- 4 C–H single bonds → 4 σ bonds
- 1 C=C double bond → 1 σ bond + 1 π bond

Total σ bonds = 4 + 1 = 5
Total π bonds = 1

$$\text{C}_2\text{H}_4: \text{ 5 sigma bonds, 1 pi bond}$$

Concept: A sigma (σ) bond is formed by head-on (axial) overlap of orbitals along the internuclear axis (x-axis here). For σ bond formation, the orbitals must overlap along the internuclear axis.

(a) 1s and 1s:
- Both are spherically symmetric; they overlap along the x-axis (internuclear axis).
- Forms a σ bond. ✓

(b) 1s and 2pₓ:
- The 2pₓ orbital is oriented along the x-axis (internuclear axis).
- The 1s orbital overlaps with the lobe of 2pₓ along the x-axis.
- Forms a σ bond. ✓

(c) 2p_y and 2p_z:
- The 2p_y orbital is oriented along the y-axis.
- The 2p_z orbital is oriented along the z-axis.
- Neither is oriented along the x-axis (internuclear axis).
- They cannot overlap along the internuclear axis (x-axis).
- They are perpendicular to each other and to the internuclear axis.
- Does NOT form a σ bond. ✗
- (They also cannot form a π bond with each other as they are perpendicular to each other.)

(d) 1s and 2s:
- Both s-orbitals are spherically symmetric.
- They can overlap along the x-axis (internuclear axis).
- Forms a σ bond. ✓

Conclusion: (c) 2p_y and 2p_z will NOT form a sigma bond because both orbitals are perpendicular to the internuclear axis (x-axis) and hence cannot undergo head-on overlap along the internuclear axis.

Concept:
- Carbon with 4 single bonds → sp³ hybridisation
- Carbon with 1 double bond → sp² hybridisation
- Carbon with 1 triple bond → sp hybridisation

(a) CH₃–CH₃ (Ethane):
- Both C atoms form only single bonds (4 single bonds each)
- Both C atoms: sp³ hybridised

(b) CH₃–CH=CH₂ (Propene):
- CH₃ carbon: 4 single bonds → sp³
- CH= carbon: part of C=C double bond → sp²
- =CH₂ carbon: part of C=C double bond → sp²

(c) CH₃–CH₂–OH (Ethanol):
- CH₃ carbon: 4 single bonds → sp³
- CH₂ carbon: 4 single bonds (to 2H, 1C, 1O) → sp³

(d) CH₃–CHO (Acetaldehyde/Ethanal):
- CH₃ carbon: 4 single bonds → sp³
- CHO carbon (carbonyl C): has C=O double bond → sp²

(e) CH₃COOH (Acetic acid):
- CH₃ carbon: 4 single bonds → sp³
- COOH carbon (carboxyl C): has C=O double bond → sp²

Summary Table:

| Molecule | C₁ | C₂ | C₃ |
|----------|----|----|----|
| CH₃–CH₃ | sp³ | sp³ | — |
| CH₃–CH=CH₂ | sp³ | sp² | sp² |
| CH₃–CH₂–OH | sp³ | sp³ | — |
| CH₃–CHO | sp³ | sp² | — |
| CH₃COOH | sp³ | sp² | — |

Bond Pairs:
Electron pairs that are shared between two atoms in a covalent bond are called bond pairs (or shared pairs). These electrons are involved in bonding and are located between the two bonded atoms.

Example: In H₂ molecule:
$$\text{H} : \text{H} \quad \text{or} \quad \text{H} - \text{H}$$
The two electrons shared between the two H atoms constitute a bond pair.

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Lone Pairs:
Electron pairs that are not involved in bonding and belong exclusively to one atom are called lone pairs (or non-bonding pairs). They are located on the atom and do not participate in bond formation.

Example: In H₂O molecule:
$$\text{H} - \overset{\cdot\cdot}{\underset{\cdot\cdot}{O}} - \text{H}$$
- O has 2 bond pairs (O–H bonds) and 2 lone pairs (shown as dots on O)
- The 2 lone pairs on O are not shared with any atom — they are lone pairs.

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Key Differences:

| Property | Bond Pairs | Lone Pairs |
|----------|-----------|------------|
| Location | Between two bonded atoms | On one atom only |
| Involvement | Participate in bonding | Do not participate in bonding |
| Repulsion | Less repulsive | More repulsive |
| Effect on geometry | Determine basic geometry | Distort geometry |

Sigma (σ) Bond vs Pi (π) Bond:

| Property | Sigma (σ) Bond | Pi (π) Bond |
|----------|---------------|-------------|
| Formation | Formed by head-on (axial) overlap of orbitals along the internuclear axis | Formed by lateral (sideways) overlap of p-orbitals perpendicular to the internuclear axis |
| Electron density | Electron density is concentrated along the internuclear axis (between the nuclei) | Electron density is concentrated above and below (or in front and behind) the internuclear axis |
| Rotation | Free rotation about the bond axis is possible | Free rotation is not possible (would break the π bond) |
| Strength | Generally stronger than π bond (greater overlap) | Generally weaker than σ bond |
| Occurrence | Present in all covalent bonds (single, double, triple) | Present only in double bonds (1 π) and triple bonds (2 π); always in addition to a σ bond |
| Orbitals involved | s-s, s-p, p-p (head-on), sp-sp, etc. | p-p (lateral), d-p, etc. |
| Bond order contribution | Forms the first bond between two atoms | Forms the second or third bond |

Example:
- In C₂H₄ (ethylene): C=C has 1 σ bond (sp²–sp² overlap) + 1 π bond (p–p lateral overlap)
- In C₂H₂ (acetylene): C≡C has 1 σ bond + 2 π bonds

Valence Bond (VB) Theory — Formation of H₂:

Given: Two hydrogen atoms H_A and H_B, each with one electron in its 1s orbital.

Step 1: Initial State (atoms far apart)
- When the two H atoms are far apart, there is no interaction between them.
- Each H atom has one electron in its 1s orbital.
- Potential energy of the system = 0 (reference)

Step 2: Approach of atoms
- As the two H atoms approach each other, the electron of H_A starts to feel the attraction of nucleus B, and vice versa.
- The electron density between the two nuclei increases.
- The potential energy of the system decreases (system becomes more stable).

Step 3: Orbital Overlap
- When the 1s orbitals of H_A and H_B overlap, the electron density between the nuclei increases significantly.
- This increased electron density between the nuclei attracts both nuclei, pulling them together.
- The potential energy continues to decrease.

Step 4: Equilibrium (Bond Formation)
- At the equilibrium internuclear distance (bond length = 74 pm for H₂), the potential energy reaches a minimum (= –436 kJ/mol, the bond enthalpy).
- At this point, the attractive forces (nucleus–electron) balance the repulsive forces (nucleus–nucleus, electron–electron).
- A stable H–H bond is formed.

Step 5: Further approach
- If the nuclei are brought closer than the equilibrium distance, nuclear–nuclear repulsion increases sharply, raising the potential energy rapidly → destabilisation.

Conclusion:
$$\text{H}_A(1s^1) + \text{H}_B(1s^1) \rightarrow \text{H}_A - \text{H}_B \quad (\Delta H = -436 \text{ kJ/mol})$$

The H–H bond is a σ bond formed by the head-on overlap of two 1s orbitals. The bond is stable because the energy of H₂ is lower than that of two separate H atoms by 436 kJ/mol (bond enthalpy).

Linear Combination of Atomic Orbitals (LCAO) — Conditions for MO Formation:

According to LCAO method, molecular orbitals are formed by the linear combination (addition or subtraction) of atomic orbitals. The following conditions must be satisfied:

1. Similar Energy of Combining Atomic Orbitals:
- The atomic orbitals combining to form MOs should have comparable (similar) energies.
- Example: 1s of H can combine with 1s of H, but not effectively with 2s of H (large energy difference).

2. Proper Symmetry:
- The combining atomic orbitals must have the same symmetry about the molecular (internuclear) axis.
- Example: 1s and 2pₓ (along x-axis) can combine; 1s and 2p_y cannot (different symmetry).

3. Significant Overlap:
- The combining atomic orbitals must have significant overlap (overlap integral S > 0).
- Greater the overlap, more stable the bonding MO formed.
- If overlap is negligible, effective MO formation does not occur.

4. Same Phase (for Bonding MO):
- For the formation of a bonding MO, the orbitals must combine with the same phase (constructive interference → increased electron density between nuclei).
- For the formation of an antibonding MO, orbitals combine with opposite phases (destructive interference → node between nuclei).

Mathematical Representation:
$$\psi_{\text{bonding}} = \psi_A + \psi_B \quad (\text{BMO: lower energy})$$
$$\psi_{\text{antibonding}} = \psi_A - \psi_B \quad (\text{ABMO: higher energy})$$

Note: The number of MOs formed always equals the number of AOs combined.

Given: Be₂ molecule — does it exist?

Electronic Configuration of Be: $1s^2 2s^2$ (4 electrons per Be atom)

Total electrons in Be₂: 4 + 4 = 8 electrons

Filling MOs in order of increasing energy:
$$\sigma_{1s},\ \sigma^*_{1s},\ \sigma_{2s},\ \sigma^*_{2s}$$

MO Electronic Configuration of Be₂:
\sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2

Bond Order Calculation:
$$\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{4 - 4}{2} = 0$$

where:
- $N_b$ = electrons in bonding MOs = 2 (in $\sigma_{1s}$) + 2 (in $\sigma_{2s}$) = 4
- $N_a$ = electrons in antibonding MOs = 2 (in $\sigma^*_{1s}$) + 2 (in $\sigma^*_{2s}$) = 4

Conclusion:
- Bond order of Be₂ = 0
- A bond order of zero means no net bonding between the two Be atoms.
- The stabilisation due to bonding MOs is exactly cancelled by the destabilisation due to antibonding MOs.
- Therefore, Be₂ molecule does not exist.

(Note: The $\sigma_{1s}$ and $\sigma^*_{1s}$ contributions also cancel each other, so effectively only the 2s electrons matter, and they too cancel.)

Concept: Stability is determined by bond order (higher bond order = more stable). Magnetic property depends on unpaired electrons (paramagnetic if unpaired electrons present; diamagnetic if all paired).

MO energy order for O₂ species:
\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}

(a) O₂ (16 electrons):
$$\text{MO config: } (\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1$$
- $N_b = 8,\ N_a = 4$ (ignoring inner shell)
- Bond Order $= \frac{8-4}{2} = 2$
- 2 unpaired electrons → Paramagnetic

(b) O₂⁺ (15 electrons — one electron removed from O₂):
- One electron removed from $\pi^*_{2p}$
$$\pi^*_{2p_x})^1(\pi^*_{2p_y})^0$$
- $N_b = 8,\ N_a = 3$
- Bond Order $= \frac{8-3}{2} = 2.5$
- 1 unpaired electron → Paramagnetic

(c) O₂⁻ (superoxide, 17 electrons — one electron added to O₂):
- One electron added to $\pi^*_{2p}$
$$\pi^*_{2p_x})^2(\pi^*_{2p_y})^1$$
- $N_b = 8,\ N_a = 5$
- Bond Order $= \frac{8-5}{2} = 1.5$
- 1 unpaired electron → Paramagnetic

(d) O₂²⁻ (peroxide, 18 electrons — two electrons added to O₂):
- Two electrons added to $\pi^*_{2p}$
$$\pi^*_{2p_x})^2(\pi^*_{2p_y})^2$$
- $N_b = 8,\ N_a = 6$
- Bond Order $= \frac{8-6}{2} = 1$
- 0 unpaired electrons → Diamagnetic

Summary Table:

| Species | Bond Order | Stability | Magnetic Property |
|---------|-----------|-----------|-------------------|
| O₂⁺ | 2.5 | Most stable | Paramagnetic |
| O₂ | 2.0 | Second | Paramagnetic |
| O₂⁻ | 1.5 | Third | Paramagnetic |
| O₂²⁻ | 1.0 | Least stable | Diamagnetic |

Order of stability: O_2^+ > O_2 > O_2^- > O_2^{2-}

Significance of Plus (+) and Minus (−) Signs in Orbital Representation:

The plus (+) and minus (−) signs used in representing atomic orbitals do not represent electric charges. Instead, they represent the phase (sign of the wave function, ψ) of the orbital.

Wave Function (ψ):
- Atomic orbitals are mathematical wave functions (ψ) that describe the probability of finding an electron.
- The wave function can have positive (+) or negative (−) values in different regions of space.
- The sign of ψ represents the phase of the orbital.

Significance:

1. Phase determines the type of overlap:
- When two orbitals of the same phase overlap (+ with + or − with −): constructive interference occurs → electron density between nuclei increases → Bonding MO (lower energy, more stable) is formed.
- When two orbitals of opposite phase overlap (+ with −): destructive interference occurs → electron density between nuclei decreases (node is created) → Antibonding MO (higher energy, less stable) is formed.

2. Node formation: A node (region of zero electron density) is formed where the wave function changes sign (from + to −).

3. Symmetry of orbitals: The signs help in determining whether two orbitals have the correct symmetry to combine and form MOs.

Example:
- 1s orbital: positive phase throughout (spherically symmetric, ψ > 0)
- 2p orbital: one lobe has + phase, the other has − phase
- When two 1s orbitals (both +) overlap: bonding σ MO forms
- When one + lobe and one − lobe overlap: antibonding σ* MO forms

Hybridisation in PCl₅:

Electronic configuration of P: [Ne] 3s² 3p³ 3d⁰

In PCl₅:
- P forms 5 bonds with 5 Cl atoms.
- P uses: 1 (3s) + 3 (3p) + 1 (3d) = 5 orbitals
- Hybridisation: sp³d
- 5 sp³d hybrid orbitals are formed, directed towards the 5 corners of a trigonal bipyramid.

Geometry: Trigonal Bipyramidal
- 3 equatorial bonds: In the equatorial plane, at 120° to each other (bond angle = 120°)
- 2 axial bonds: Above and below the equatorial plane, at 90° to the equatorial bonds (bond angle = 90° with equatorial)

Why are axial bonds longer than equatorial bonds?

1. Greater repulsion on axial bonds:
- Each axial bond is at 90° to the 3 equatorial bonds → experiences repulsion from 3 bond pairs at 90°.
- Each equatorial bond is at 90° to the 2 axial bonds and at 120° to the 2 other equatorial bonds → experiences repulsion from only 2 bond pairs at 90° (less repulsion).

2. Repulsion at 90° is much stronger than at 120° (closer proximity).

3. Due to greater repulsion, the axial bond pairs are pushed further away from the central P atom → axial bonds are longer (P–Cl axial ≈ 212 pm) than equatorial bonds (P–Cl equatorial ≈ 202 pm).

4. The equatorial bonds have more s-character (from sp² component) compared to axial bonds (pure p and d character), making equatorial bonds shorter and stronger.

Conclusion: Axial bonds in PCl₅ are longer because they experience greater repulsion from the three equatorial bond pairs (at 90°) compared to equatorial bonds.

Definition of Hydrogen Bond:
A hydrogen bond is the electrostatic force of attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) and another electronegative atom (F, O, or N) of the same or a different molecule.

It is represented as: X–H···Y
where X and Y are electronegative atoms (F, O, or N), '–' is the covalent bond, and '···' is the hydrogen bond.

Conditions for hydrogen bond formation:
1. The molecule must contain H bonded to a highly electronegative atom (F, O, or N).
2. The electronegative atom must be small in size.

Examples:
- H₂O: O–H···O (intermolecular H-bond)
- HF: F–H···F
- o-nitrophenol: intramolecular H-bond

Hydrogen Bond vs Van der Waals Forces:

Hydrogen bonds are STRONGER than van der Waals forces.

| Property | Hydrogen Bond | Van der Waals Forces |
|----------|--------------|---------------------|
| Strength | 10–40 kJ/mol | 0.1–10 kJ/mol |
| Nature | Electrostatic (dipole–dipole type) | Weak dispersion/induced dipole |
| Specificity | Specific (requires F, O, N) | Non-specific (all molecules) |

Conclusion: Hydrogen bonds are stronger than van der Waals forces but weaker than covalent or ionic bonds. They have a significant effect on the physical properties (boiling point, solubility, structure) of compounds.

Bond Order:
Bond order is defined as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and the number of electrons in antibonding molecular orbitals ($N_a$):
$$\text{Bond Order} = \frac{N_b - N_a}{2}$$

Higher bond order → stronger bond, shorter bond length, more stable molecule.

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(a) N₂ (14 electrons):

MO configuration:
\sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \pi_{2p_x}^2\ \pi_{2p_y}^2\ \sigma_{2p_z}^2

(For N₂, $\pi_{2p}$ is filled before $\sigma_{2p_z}$)

- $N_b = 2+2+2+2+2 = 10$ (counting $\sigma_{1s}, \sigma_{2s}, \pi_{2p_x}, \pi_{2p_y}, \sigma_{2p_z}$)
- $N_a = 2+2 = 4$ (counting $\sigma^*_{1s}, \sigma^*_{2s}$)

$$\text{Bond Order of N}_2 = \frac{10-4}{2} = \frac{6}{2} = \boxed{3}$$

N₂ has a triple bond (N≡N). Diamagnetic (no unpaired electrons).

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(b) O₂ (16 electrons):

MO configuration:
\sigma_{1s}^2\ \sigma^*_{1s}^2\ \sigma_{2s}^2\ \sigma^*_{2s}^2\ \sigma_{2p_z}^2\ \pi_{2p_x}^2\ \pi_{2p_y}^2\ \pi^*_{2p_x}^1\ \pi^*_{2p_y}^1

- $N_b = 2+2+2+2+2 = 10$
- $N_a = 2+2+1+1 = 6$

$$\text{Bond Order of O}_2 = \frac{10-6}{2} = \frac{4}{2} = \boxed{2}$$

O₂ has a double bond. Paramagnetic (2 unpaired electrons in $\pi^*_{2p}$).

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(c) O₂⁺ (15 electrons — one electron removed from O₂):

One electron removed from $\pi^*_{2p}$:
\pi^*_{2p_x}^1\ \pi^*_{2p_y}^0

- $N_b = 10$
- $N_a = 2+2+1 = 5$

$$\text{Bond Order of O}_2^+ = \frac{10-5}{2} = \frac{5}{2} = \boxed{2.5}$$

Paramagnetic (1 unpaired electron).

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(d) O₂⁻ (17 electrons — one electron added to O₂):

One electron added to $\pi^*_{2p}$:
\pi^*_{2p_x}^2\ \pi^*_{2p_y}^1

- $N_b = 10$
- $N_a = 2+2+2+1 = 7$

$$\text{Bond Order of O}_2^- = \frac{10-7}{2} = \frac{3}{2} = \boxed{1.5}$$

Paramagnetic (1 unpaired electron).

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Summary:

| Species | Bond Order | Magnetic Property |
|---------|-----------|-------------------|
| N₂ | 3 | Diamagnetic |
| O₂ | 2 | Paramagnetic |
| O₂⁺ | 2.5 | Paramagnetic |
| O₂⁻ | 1.5 | Paramagnetic |