Grade 11 Chemistry – Unit 2: Chemical Bonding
Hello dear student! Welcome to Unit 2. In Unit 1, we studied atoms and their structure. Now we will answer a very important question: How do atoms join together to form compounds? This process is called chemical bonding, and it is one of the most important topics in all of chemistry. Let’s learn it step by step.
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2.1 Introduction to Chemical Bonding
Have you ever wondered why sodium (Na) reacts violently with chlorine (Cl) gas, but two chlorine atoms quietly join together? The answer lies in how atoms achieve stability by forming bonds.
2.1.1 The Octet Rule
Noble gases (He, Ne, Ar, Kr, Xe, Rn) are very stable and unreactive. Why? Because their outermost shell is completely filled with electrons. This observation led to the octet rule:
The Octet Rule: Atoms tend to gain, lose, or share electrons in order to have eight electrons in their valence shell (like the nearest noble gas). For hydrogen and helium, the goal is two electrons (duplet rule).
Think about it: sodium (Na) has 1 valence electron. It is easier to lose that one electron than to gain seven. Chlorine (Cl) has 7 valence electrons. It is easier to gain one electron than to lose seven. So Na loses one electron and Cl gains it — both achieve noble gas configurations!
2.1.2 Types of Chemical Bonding
There are three main types of chemical bonds:
- Ionic bonding — transfer of electrons from metal to non-metal
- Covalent bonding — sharing of electrons between non-metals
- Metallic bonding — delocalized electrons shared among metal atoms
Key Exam Notes — Introduction:
• Octet rule: atoms want 8 valence electrons (2 for H, He).
• Three types of bonding: ionic, covalent, metallic.
• The type of bond depends on the elements’ positions in the periodic table and their electronegativity difference.
• Octet rule: atoms want 8 valence electrons (2 for H, He).
• Three types of bonding: ionic, covalent, metallic.
• The type of bond depends on the elements’ positions in the periodic table and their electronegativity difference.
Practice Questions:
1. Why do noble gases rarely form chemical bonds?
2. How many valence electrons does an atom of oxygen need to gain to satisfy the octet rule?
1. Why do noble gases rarely form chemical bonds?
2. How many valence electrons does an atom of oxygen need to gain to satisfy the octet rule?
Answer 1: Noble gases already have a complete valence shell (8 electrons, or 2 for He). They are already in their most stable electron configuration, so they have no tendency to gain, lose, or share electrons.
Answer 2: Oxygen has 6 valence electrons. It needs to gain 2 electrons to reach 8 (octet). Alternatively, it can share 2 electrons by forming two covalent bonds.
Answer 2: Oxygen has 6 valence electrons. It needs to gain 2 electrons to reach 8 (octet). Alternatively, it can share 2 electrons by forming two covalent bonds.
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2.2 Ionic Bonds
An ionic bond is the electrostatic attraction between positively charged ions (cations) and negatively charged ions (anions) formed by the complete transfer of one or more electrons from a metal to a non-metal.
2.2.1 Lewis Electron-Dot Symbols
The Lewis symbol (electron-dot symbol) represents the valence electrons of an atom as dots placed around the element symbol. Each dot represents one valence electron.
Na Mg Al Si P S Cl Ar
. . . . . . ..
. . . . . . . ..
. . .. . . . . ..
.. .. .. .. ..
Notice that we place dots on all four sides (top, right, bottom, left) before pairing them. The number of unpaired dots tells you how many bonds the atom typically forms.
Worked Example 2.1: Write the Lewis symbols for Na, Ca, O, and N.
Solution:
• Na (Group 1, 1 valence e⁻): Na with 1 dot
• Ca (Group 2, 2 valence e⁻): Ca with 2 dots
• O (Group 16, 6 valence e⁻): O with 6 dots (2 pairs + 2 unpaired)
• N (Group 15, 5 valence e⁻): N with 5 dots (1 pair + 3 unpaired)
Solution:
• Na (Group 1, 1 valence e⁻): Na with 1 dot
• Ca (Group 2, 2 valence e⁻): Ca with 2 dots
• O (Group 16, 6 valence e⁻): O with 6 dots (2 pairs + 2 unpaired)
• N (Group 15, 5 valence e⁻): N with 5 dots (1 pair + 3 unpaired)
2.2.2 Formation of Ionic Bonds
Let’s trace the formation of an ionic bond step by step using sodium chloride (NaCl) as our example.
Step 1: Sodium atom loses one electron to form Na⁺:
$$\text{Na} \rightarrow \text{Na}^+ + e^-$$
Na: [Ne] 3s¹ → Na⁺: [Ne] — achieves neon configuration.
Step 2: Chlorine atom gains that electron to form Cl⁻:
$$\text{Cl} + e^- \rightarrow \text{Cl}^-$$
Cl: [Ne] 3s² 3p⁵ → Cl⁻: [Ne] 3s² 3p⁶ = [Ar] — achieves argon configuration.
Step 3: The oppositely charged ions attract each other electrostatically, forming the ionic bond:
$$\text{Na}^+ + \text{Cl}^- \rightarrow \text{NaCl}$$
The electrostatic force of attraction is described by Coulomb’s law:
$$F = k \frac{q_1 q_2}{r^2}$$
where $q_1$ and $q_2$ are the ion charges, $r$ is the distance between ion centers, and $k$ is a constant. The force is stronger for higher charges and smaller ions (smaller $r$).
Worked Example 2.2: Show the formation of magnesium oxide (MgO) using Lewis symbols and electron transfer.
Solution:
Mg has 2 valence electrons; O has 6 valence electrons.
$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$
$$\text{Mg}^{2+} + \text{O}^{2-} \rightarrow \text{MgO}$$
Note: Two electrons are transferred from Mg to O. Mg achieves [Ne] configuration; O⁻² achieves [Ne] configuration.
Solution:
Mg has 2 valence electrons; O has 6 valence electrons.
$$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$$
$$\text{O} + 2e^- \rightarrow \text{O}^{2-}$$
$$\text{Mg}^{2+} + \text{O}^{2-} \rightarrow \text{MgO}$$
Note: Two electrons are transferred from Mg to O. Mg achieves [Ne] configuration; O⁻² achieves [Ne] configuration.
Worked Example 2.3: Show the formation of calcium fluoride (CaF₂).
Solution:
Ca has 2 valence electrons; each F atom needs 1 electron.
$$\text{Ca} \rightarrow \text{Ca}^{2+} + 2e^-$$
$$2\text{F} + 2e^- \rightarrow 2\text{F}^-$$
$$\text{Ca}^{2+} + 2\text{F}^- \rightarrow \text{CaF}_2$$
One Ca atom transfers 1 electron to each of 2 F atoms.
Solution:
Ca has 2 valence electrons; each F atom needs 1 electron.
$$\text{Ca} \rightarrow \text{Ca}^{2+} + 2e^-$$
$$2\text{F} + 2e^- \rightarrow 2\text{F}^-$$
$$\text{Ca}^{2+} + 2\text{F}^- \rightarrow \text{CaF}_2$$
One Ca atom transfers 1 electron to each of 2 F atoms.
Lattice Energy
The lattice energy is the energy released when one mole of an ionic compound is formed from its gaseous ions. It is a measure of the strength of the ionic bond. Higher lattice energy means a stronger bond and higher melting point.
Lattice energy depends on:
- Ion charges: Higher charges → stronger attraction → higher lattice energy. MgO has much higher lattice energy than NaCl.
- Ion sizes: Smaller ions → shorter distance → higher lattice energy. NaF has higher lattice energy than NaCl.
Key Exam Notes — Ionic Bond Formation:
• Metals lose electrons (become cations); non-metals gain electrons (become anions).
• The formula of the ionic compound must be electrically neutral (total + charge = total − charge).
• Transition metals can form multiple ions (e.g., Fe²⁺ and Fe³⁺).
• Lattice energy ∝ (charge₁ × charge₂) / (r₁ + r₂).
• Ionic compounds do not exist as individual molecules — they form a crystal lattice.
• Metals lose electrons (become cations); non-metals gain electrons (become anions).
• The formula of the ionic compound must be electrically neutral (total + charge = total − charge).
• Transition metals can form multiple ions (e.g., Fe²⁺ and Fe³⁺).
• Lattice energy ∝ (charge₁ × charge₂) / (r₁ + r₂).
• Ionic compounds do not exist as individual molecules — they form a crystal lattice.
Practice Questions:
1. Write the Lewis symbol for Al. Show the formation of aluminium oxide (Al₂O₃) using electron transfer.
2. Which has higher lattice energy: NaCl or MgO? Explain why.
1. Write the Lewis symbol for Al. Show the formation of aluminium oxide (Al₂O₃) using electron transfer.
2. Which has higher lattice energy: NaCl or MgO? Explain why.
Answer 1:
Al has 3 valence electrons: Al with 3 dots.
$$2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^-$$
$$3\text{O} + 6e^- \rightarrow 3\text{O}^{2-}$$
$$2\text{Al}^{3+} + 3\text{O}^{2-} \rightarrow \text{Al}_2\text{O}_3$$
Total positive charge: 2 × (+3) = +6; Total negative charge: 3 × (−2) = −6. Balanced! ✓
Answer 2:
MgO has much higher lattice energy. Both NaCl and MgO have ions of similar size, but Mg²⁺ and O²⁻ have charges of +2 and −2 respectively, while Na⁺ and Cl⁻ have charges of +1 and −1. Since lattice energy is proportional to the product of charges (4 for MgO vs 1 for NaCl), MgO has roughly four times the lattice energy of NaCl.
Al has 3 valence electrons: Al with 3 dots.
$$2\text{Al} \rightarrow 2\text{Al}^{3+} + 6e^-$$
$$3\text{O} + 6e^- \rightarrow 3\text{O}^{2-}$$
$$2\text{Al}^{3+} + 3\text{O}^{2-} \rightarrow \text{Al}_2\text{O}_3$$
Total positive charge: 2 × (+3) = +6; Total negative charge: 3 × (−2) = −6. Balanced! ✓
Answer 2:
MgO has much higher lattice energy. Both NaCl and MgO have ions of similar size, but Mg²⁺ and O²⁻ have charges of +2 and −2 respectively, while Na⁺ and Cl⁻ have charges of +1 and −1. Since lattice energy is proportional to the product of charges (4 for MgO vs 1 for NaCl), MgO has roughly four times the lattice energy of NaCl.
2.2.3 Exceptions to the Octet Rule in Ionic Compounds
The octet rule works well for many ionic compounds, but there are important exceptions:
1. Incomplete Octet: Some ions do not achieve 8 electrons. For example, Li⁺ achieves the He configuration (2 electrons), and Be²⁺ also has only 2 electrons.
2. Expanded Octet: Elements in Period 3 and beyond can have more than 8 electrons in their valence shell because they have available d-orbitals. For example, in some compounds, sulfur can have 10 or 12 electrons.
3. Odd-Electron Species: Some molecules/ions have an odd number of electrons and cannot satisfy the octet rule for all atoms (e.g., NO).
Key Exam Notes — Exceptions:
• Li⁺ and Be²⁺ have duplet (2 electrons), not octet.
• Elements in Period 3+ can exceed the octet (expanded octet).
• The octet rule is a useful guideline, NOT a strict law.
• Li⁺ and Be²⁺ have duplet (2 electrons), not octet.
• Elements in Period 3+ can exceed the octet (expanded octet).
• The octet rule is a useful guideline, NOT a strict law.
2.2.4 Properties of Ionic Compounds
Now, why does knowing whether a compound is ionic matter? Because ionic compounds have characteristic properties that you must know for your exam:
- High melting and boiling points: Strong electrostatic forces between ions require a lot of energy to overcome. For example, NaCl melts at 801 °C.
- Hard and brittle: The ions are held in fixed positions. If you apply force, layers of ions shift so that like charges come adjacent → strong repulsion → crystal shatters.
- Conduct electricity when molten or dissolved, but NOT when solid: In solid state, ions are fixed. When melted or dissolved, ions are free to move and carry charge.
- Soluble in polar solvents (like water): Polar water molecules can surround and separate the ions (hydration).
- Form crystalline solids: Ions arrange in regular, repeating patterns (crystal lattice).
Practice Questions:
1. Why does solid NaCl NOT conduct electricity, but molten NaCl does?
2. Why are ionic crystals brittle rather than malleable?
1. Why does solid NaCl NOT conduct electricity, but molten NaCl does?
2. Why are ionic crystals brittle rather than malleable?
Answer 1: In solid NaCl, the Na⁺ and Cl⁻ ions are locked in fixed positions in the crystal lattice and cannot move. Without mobile charge carriers, electricity cannot be conducted. When NaCl is melted, the thermal energy breaks the lattice and the ions become free to move toward the electrodes, carrying electric current.
Answer 2: In an ionic crystal, positive and negative ions alternate. When a force is applied, one layer of ions shifts relative to another. This causes ions of the same charge to come next to each other. The resulting strong electrostatic repulsion forces the layers apart, causing the crystal to crack or shatter (brittleness). Metals, in contrast, can deform because their electrons are delocalized and can adjust.
Answer 2: In an ionic crystal, positive and negative ions alternate. When a force is applied, one layer of ions shifts relative to another. This causes ions of the same charge to come next to each other. The resulting strong electrostatic repulsion forces the layers apart, causing the crystal to crack or shatter (brittleness). Metals, in contrast, can deform because their electrons are delocalized and can adjust.
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2.3 Covalent Bonds and Molecular Geometry
A covalent bond is formed when two non-metal atoms share one or more pairs of electrons. Each shared pair counts toward the octet of both atoms.
Types of Covalent Bonds
- Single bond: Sharing of one electron pair (e.g., H—H in H₂).
- Double bond: Sharing of two electron pairs (e.g., O=O in O₂).
- Triple bond: Sharing of three electron pairs (e.g., N≡N in N₂).
Bond strength: triple bond > double bond > single bond.
Bond length: triple bond < double bond < single bond.
Drawing Lewis Structures for Covalent Molecules
Follow these steps carefully:
- Count total valence electrons: Add up all valence electrons from all atoms. For polyatomic ions, add electrons for negative charge, subtract for positive charge.
- Identify the central atom: Usually the least electronegative atom (but never H — it is always terminal). If there is only one of a certain atom, it is likely central.
- Connect atoms with single bonds: Each single bond uses 2 electrons.
- Complete octets of terminal atoms: Add lone pairs to outer atoms first.
- Complete octet of central atom: Put remaining electrons on the central atom.
- If central atom lacks an octet: Convert lone pairs from terminal atoms into bonding pairs (double or triple bonds).
Worked Example 2.4: Draw the Lewis structure of water (H₂O).
Solution:
Step 1: Total valence electrons = 2(1) + 6 = 8
Step 2: O is the central atom (least electronegative of the two types).
Step 3: Connect H—O—H with single bonds (uses 4 electrons).
Step 4: Complete octet of O: add 2 lone pairs (4 electrons).
Step 5: Total used = 4 + 4 = 8. O has 8 electrons (2 bonding + 4 lone pair = 6, plus 2 from bonds = 8). ✓
Result: O with 2 lone pairs and 2 single bonds to H atoms. The shape will be bent (not linear!).
Solution:
Step 1: Total valence electrons = 2(1) + 6 = 8
Step 2: O is the central atom (least electronegative of the two types).
Step 3: Connect H—O—H with single bonds (uses 4 electrons).
Step 4: Complete octet of O: add 2 lone pairs (4 electrons).
Step 5: Total used = 4 + 4 = 8. O has 8 electrons (2 bonding + 4 lone pair = 6, plus 2 from bonds = 8). ✓
Result: O with 2 lone pairs and 2 single bonds to H atoms. The shape will be bent (not linear!).
Worked Example 2.5: Draw the Lewis structure of carbon dioxide (CO₂).
Solution:
Step 1: Total valence electrons = 4 + 2(6) = 16
Step 2: C is central (least electronegative).
Step 3: O—C—O with single bonds (uses 4 electrons).
Step 4: Complete octets of terminal O atoms: each O needs 6 more electrons as lone pairs (uses 12 electrons). Total used = 16.
Step 5: C has only 4 electrons (2 single bonds). It needs 4 more. Convert one lone pair from each O into a bonding pair with C.
Result: O=C=O (two double bonds). Each atom has 8 electrons. ✓
Solution:
Step 1: Total valence electrons = 4 + 2(6) = 16
Step 2: C is central (least electronegative).
Step 3: O—C—O with single bonds (uses 4 electrons).
Step 4: Complete octets of terminal O atoms: each O needs 6 more electrons as lone pairs (uses 12 electrons). Total used = 16.
Step 5: C has only 4 electrons (2 single bonds). It needs 4 more. Convert one lone pair from each O into a bonding pair with C.
Result: O=C=O (two double bonds). Each atom has 8 electrons. ✓
Worked Example 2.6: Draw the Lewis structure of the carbonate ion (CO₃²⁻).
Solution:
Step 1: Total valence electrons = 4 + 3(6) + 2 (for 2− charge) = 24
Step 2: C is central.
Step 3: Connect three O atoms to C with single bonds (uses 6 electrons).
Step 4: Complete octets of terminal O atoms: each O needs 6 more (uses 18 electrons). Total = 24.
Step 5: C has only 6 electrons. Convert one lone pair from one O into a double bond with C.
Result: One C=O double bond and two C—O single bonds (with negative charges on those two O atoms). This is a resonance structure — the double bond can be with any of the three O atoms.
Solution:
Step 1: Total valence electrons = 4 + 3(6) + 2 (for 2− charge) = 24
Step 2: C is central.
Step 3: Connect three O atoms to C with single bonds (uses 6 electrons).
Step 4: Complete octets of terminal O atoms: each O needs 6 more (uses 18 electrons). Total = 24.
Step 5: C has only 6 electrons. Convert one lone pair from one O into a double bond with C.
Result: One C=O double bond and two C—O single bonds (with negative charges on those two O atoms). This is a resonance structure — the double bond can be with any of the three O atoms.
Resonance
When a molecule or ion can be represented by two or more Lewis structures that differ only in the position of electrons (not atoms), these are called resonance structures. The real structure is a resonance hybrid — an average of all the resonance forms.
Example: In CO₃²⁻, all three C—O bonds are actually identical (intermediate between single and double). We draw three resonance structures and connect them with double-headed arrows (↔).
Formal Charge
To decide which resonance structure is the best (most stable), we use formal charge:
$$\text{Formal charge} = \text{valence electrons} – \text{non-bonding electrons} – \frac{1}{2}(\text{bonding electrons})$$
Rules for best structure:
- Formal charges closest to zero are best.
- Negative formal charge should be on the most electronegative atom.
- Like charges should not be adjacent.
Worked Example 2.7: Calculate the formal charges on all atoms in the two possible Lewis structures of CO₂: (a) O=C=O and (b) O—C≡O (with a lone pair and negative charge on left O, positive on C).
Solution:
(a) O=C=O:
• Each O: FC = 6 − 4 − ½(4) = 6 − 4 − 2 = 0
• C: FC = 4 − 0 − ½(8) = 4 − 0 − 4 = 0
All formal charges = 0. This is the best structure! ✓
(b) O⁻—C⁺≡O:
• Left O: FC = 6 − 6 − ½(2) = 6 − 6 − 1 = −1
• C: FC = 4 − 0 − ½(6) = 4 − 0 − 3 = +1
• Right O: FC = 6 − 2 − ½(6) = 6 − 2 − 3 = +1… wait, let me recalculate: FC = 6 − 2 − ½(6) = 0
This structure has non-zero formal charges and is less stable than (a).
Solution:
(a) O=C=O:
• Each O: FC = 6 − 4 − ½(4) = 6 − 4 − 2 = 0
• C: FC = 4 − 0 − ½(8) = 4 − 0 − 4 = 0
All formal charges = 0. This is the best structure! ✓
(b) O⁻—C⁺≡O:
• Left O: FC = 6 − 6 − ½(2) = 6 − 6 − 1 = −1
• C: FC = 4 − 0 − ½(6) = 4 − 0 − 3 = +1
• Right O: FC = 6 − 2 − ½(6) = 6 − 2 − 3 = +1… wait, let me recalculate: FC = 6 − 2 − ½(6) = 0
This structure has non-zero formal charges and is less stable than (a).
Coordinate Covalent Bond (Dative Bond)
A coordinate covalent bond is a covalent bond in which both shared electrons come from the same atom (the donor). The acceptor atom provides an empty orbital.
Example: In the ammonium ion (NH₄⁺), the bond between NH₃ and H⁺ is a coordinate bond — NH₃ donates its lone pair to H⁺ (which has no electrons).
Once formed, a coordinate bond is indistinguishable from a regular covalent bond.
Polarity of Covalent Bonds
When two identical atoms share electrons (e.g., H—H, Cl—Cl), the electrons are shared equally → nonpolar covalent bond.
When two different atoms share electrons, the more electronegative atom pulls the shared electrons closer → polar covalent bond. This creates a dipole (partial charges: δ+ and δ−).
We can estimate bond polarity using electronegativity difference (ΔEN):
| ΔEN Range | Bond Type |
|---|---|
| 0 – 0.4 | Nonpolar covalent |
| 0.5 – 1.7 | Polar covalent |
| > 1.7 | Ionic |
Note: This is a guideline, not a strict rule. The ionic/covalent character is a continuum.
Key Exam Notes — Covalent Bonds:
• Lewis structure steps: count valence e⁻ → identify central atom → connect with single bonds → complete octets → adjust with multiple bonds if needed.
• For ions: add e⁻ for negative charge, subtract for positive charge.
• Resonance: when multiple valid Lewis structures exist, the real structure is a hybrid.
• Formal charge = valence e⁻ − non-bonding e⁻ − ½(bonding e⁻). Best structure has FC closest to 0.
• Coordinate bond: both electrons from one atom. Common in complex ions.
• Bond polarity depends on electronegativity difference.
• Lewis structure steps: count valence e⁻ → identify central atom → connect with single bonds → complete octets → adjust with multiple bonds if needed.
• For ions: add e⁻ for negative charge, subtract for positive charge.
• Resonance: when multiple valid Lewis structures exist, the real structure is a hybrid.
• Formal charge = valence e⁻ − non-bonding e⁻ − ½(bonding e⁻). Best structure has FC closest to 0.
• Coordinate bond: both electrons from one atom. Common in complex ions.
• Bond polarity depends on electronegativity difference.
Practice Questions:
1. Draw the Lewis structure of NH₃ and calculate the formal charge on N.
2. Draw TWO resonance structures for the nitrate ion (NO₃⁻).
3. Is the H—Cl bond polar or nonpolar? Calculate ΔEN. (EN: H = 2.1, Cl = 3.0)
1. Draw the Lewis structure of NH₃ and calculate the formal charge on N.
2. Draw TWO resonance structures for the nitrate ion (NO₃⁻).
3. Is the H—Cl bond polar or nonpolar? Calculate ΔEN. (EN: H = 2.1, Cl = 3.0)
Answer 1:
NH₃: N has 5 valence e⁻, each H has 1. Total = 5 + 3(1) = 8.
N is central. Connect 3 H atoms with single bonds (6 e⁻ used). Remaining 2 e⁻ go as a lone pair on N.
N has 3 bonds + 1 lone pair = 8 electrons. ✓
Formal charge on N: FC = 5 − 2 − ½(6) = 5 − 2 − 3 = 0.
Answer 2:
NO₃⁻: Total valence e⁻ = 5 + 3(6) + 1 = 24.
N is central. Three N—O single bonds (6 e⁻). Complete octets of O atoms (18 e⁻). Total = 24. But N has only 6 e⁻.
Convert one lone pair from one O into a double bond.
Structure 1: N=O (top), N—O⁻ (left), N—O⁻ (right)
Structure 2: N—O⁻ (top), N=O (left), N—O⁻ (right)
Structure 3: N—O⁻ (top), N—O⁻ (left), N=O (right)
All three are resonance forms of the hybrid. ↔ connects them.
Answer 3:
ΔEN = 3.0 − 2.1 = 0.9
Since 0.5 < 0.9 < 1.7, the H—Cl bond is polar covalent. Cl is δ− (more electronegative) and H is δ+.
NH₃: N has 5 valence e⁻, each H has 1. Total = 5 + 3(1) = 8.
N is central. Connect 3 H atoms with single bonds (6 e⁻ used). Remaining 2 e⁻ go as a lone pair on N.
N has 3 bonds + 1 lone pair = 8 electrons. ✓
Formal charge on N: FC = 5 − 2 − ½(6) = 5 − 2 − 3 = 0.
Answer 2:
NO₃⁻: Total valence e⁻ = 5 + 3(6) + 1 = 24.
N is central. Three N—O single bonds (6 e⁻). Complete octets of O atoms (18 e⁻). Total = 24. But N has only 6 e⁻.
Convert one lone pair from one O into a double bond.
Structure 1: N=O (top), N—O⁻ (left), N—O⁻ (right)
Structure 2: N—O⁻ (top), N=O (left), N—O⁻ (right)
Structure 3: N—O⁻ (top), N—O⁻ (left), N=O (right)
All three are resonance forms of the hybrid. ↔ connects them.
Answer 3:
ΔEN = 3.0 − 2.1 = 0.9
Since 0.5 < 0.9 < 1.7, the H—Cl bond is polar covalent. Cl is δ− (more electronegative) and H is δ+.
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2.3.1 Molecular Geometry (VSEPR Theory)
Knowing the Lewis structure tells us how atoms are connected, but not the shape of the molecule. For that, we use the Valence Shell Electron Pair Repulsion (VSEPR) Theory.
VSEPR Theory states that electron pairs (both bonding and lone pairs) around a central atom arrange themselves as far apart as possible to minimize repulsion.
Key principles:
- Lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion.
- Lone pairs occupy more space than bonding pairs (they are closer to the nucleus and spread out more).
- Multiple bonds are treated as one “region” of electron density (a double bond counts as one region, same as a single bond).
- The molecular geometry describes the arrangement of ATOMS (not lone pairs). The electron-pair geometry includes lone pairs.
Geometries for 2, 3, and 4 Electron Domains
| Electron Domains | Electron-Pair Geometry | Bonding Pairs | Lone Pairs | Molecular Shape | Example | Bond Angle |
|---|---|---|---|---|---|---|
| 2 | Linear | 2 | 0 | Linear | BeCl₂, CO₂ | 180° |
| 3 | Trigonal planar | 3 | 0 | Trigonal planar | BF₃ | 120° |
| 3 | Trigonal planar | 2 | 1 | Bent | SO₂ | <120° |
| 4 | Tetrahedral | 4 | 0 | Tetrahedral | CH₄ | 109.5° |
| 4 | Tetrahedral | 3 | 1 | Trigonal pyramidal | NH₃ | <109.5° |
| 4 | Tetrahedral | 2 | 2 | Bent | H₂O | <109.5° |
Worked Example 2.8: Predict the molecular geometry and bond angle of NH₃.
Solution:
Lewis structure of NH₃: N has 3 bonding pairs and 1 lone pair = 4 electron domains.
Electron-pair geometry: tetrahedral.
Molecular geometry (considering only atoms): trigonal pyramidal (the lone pair “pushes” the bonds down).
Bond angle: less than 109.5° (actual: about 107°) because lone pair repulsion compresses the bond angles.
Solution:
Lewis structure of NH₃: N has 3 bonding pairs and 1 lone pair = 4 electron domains.
Electron-pair geometry: tetrahedral.
Molecular geometry (considering only atoms): trigonal pyramidal (the lone pair “pushes” the bonds down).
Bond angle: less than 109.5° (actual: about 107°) because lone pair repulsion compresses the bond angles.
Worked Example 2.9: Predict the molecular geometry of H₂O.
Solution:
Lewis structure: O has 2 bonding pairs and 2 lone pairs = 4 electron domains.
Electron-pair geometry: tetrahedral.
Molecular geometry: bent (V-shaped).
Bond angle: less than 109.5° (actual: about 104.5°) — even smaller than NH₃ because two lone pairs create more repulsion.
Solution:
Lewis structure: O has 2 bonding pairs and 2 lone pairs = 4 electron domains.
Electron-pair geometry: tetrahedral.
Molecular geometry: bent (V-shaped).
Bond angle: less than 109.5° (actual: about 104.5°) — even smaller than NH₃ because two lone pairs create more repulsion.
Worked Example 2.10: Predict the geometry of XeF₄.
Solution:
Xe has 8 valence e⁻ + 4 from F atoms = 12 e⁻ total. Four Xe—F bonds use 8 e⁻. Remaining 4 e⁻ = 2 lone pairs on Xe.
Electron domains = 4 bonding + 2 lone = 6 domains.
Electron-pair geometry: octahedral.
Molecular geometry: square planar (the two lone pairs are opposite each other to minimize repulsion).
Bond angles: 90°.
Solution:
Xe has 8 valence e⁻ + 4 from F atoms = 12 e⁻ total. Four Xe—F bonds use 8 e⁻. Remaining 4 e⁻ = 2 lone pairs on Xe.
Electron domains = 4 bonding + 2 lone = 6 domains.
Electron-pair geometry: octahedral.
Molecular geometry: square planar (the two lone pairs are opposite each other to minimize repulsion).
Bond angles: 90°.
Key Exam Notes — VSEPR:
• Count electron DOMAINS (not individual electrons). A double bond = 1 domain.
• Lone pairs compress bond angles (lone pair repulsion > bonding pair repulsion).
• NH₃: trigonal pyramidal (~107°); H₂O: bent (~104.5°); both have tetrahedral electron-pair geometry.
• Linear = 180°; Trigonal planar = 120°; Tetrahedral = 109.5°; Octahedral = 90°.
• To determine polarity of a molecule: check both bond polarity AND molecular shape.
• Count electron DOMAINS (not individual electrons). A double bond = 1 domain.
• Lone pairs compress bond angles (lone pair repulsion > bonding pair repulsion).
• NH₃: trigonal pyramidal (~107°); H₂O: bent (~104.5°); both have tetrahedral electron-pair geometry.
• Linear = 180°; Trigonal planar = 120°; Tetrahedral = 109.5°; Octahedral = 90°.
• To determine polarity of a molecule: check both bond polarity AND molecular shape.
Practice Questions:
1. Predict the molecular geometry and bond angle of BF₃.
2. Why is the bond angle in H₂O (104.5°) smaller than in NH₃ (107°)?
3. Is CO₂ a polar molecule? Explain using both bond polarity and molecular shape.
1. Predict the molecular geometry and bond angle of BF₃.
2. Why is the bond angle in H₂O (104.5°) smaller than in NH₃ (107°)?
3. Is CO₂ a polar molecule? Explain using both bond polarity and molecular shape.
Answer 1:
BF₃: B has 3 bonding pairs and 0 lone pairs = 3 electron domains.
Electron-pair geometry: trigonal planar.
Molecular geometry: trigonal planar.
Bond angle: 120°.
Answer 2:
H₂O has 2 lone pairs while NH₃ has only 1 lone pair. Lone pair–lone pair repulsion is the strongest type of repulsion. In H₂O, the two lone pairs push the bonding pairs closer together more than the single lone pair does in NH₃. This is why H₂O’s bond angle (104.5°) is smaller than NH₃’s (107°), even though both have tetrahedral electron-pair geometry.
Answer 3:
CO₂ is NOT polar. Each C=O bond is polar (C: EN 2.5, O: EN 3.5, ΔEN = 1.0 → polar). However, CO₂ has a linear geometry (bond angle 180°). The two bond dipoles are equal in magnitude but point in opposite directions, so they cancel out. The net dipole moment = 0. A molecule can have polar bonds but be nonpolar overall if its shape causes the dipoles to cancel.
BF₃: B has 3 bonding pairs and 0 lone pairs = 3 electron domains.
Electron-pair geometry: trigonal planar.
Molecular geometry: trigonal planar.
Bond angle: 120°.
Answer 2:
H₂O has 2 lone pairs while NH₃ has only 1 lone pair. Lone pair–lone pair repulsion is the strongest type of repulsion. In H₂O, the two lone pairs push the bonding pairs closer together more than the single lone pair does in NH₃. This is why H₂O’s bond angle (104.5°) is smaller than NH₃’s (107°), even though both have tetrahedral electron-pair geometry.
Answer 3:
CO₂ is NOT polar. Each C=O bond is polar (C: EN 2.5, O: EN 3.5, ΔEN = 1.0 → polar). However, CO₂ has a linear geometry (bond angle 180°). The two bond dipoles are equal in magnitude but point in opposite directions, so they cancel out. The net dipole moment = 0. A molecule can have polar bonds but be nonpolar overall if its shape causes the dipoles to cancel.
2.3.2 Intermolecular Forces in Covalent Compounds
Chemical bonds (ionic, covalent, metallic) hold atoms together within a substance. Intermolecular forces (IMFs) are the forces between molecules. They are much weaker than chemical bonds but are very important for determining physical properties like boiling point, melting point, and solubility.
There are three main types of IMFs:
1. London Dispersion Forces (LDF)
- Present in ALL molecules (polar and nonpolar).
- Caused by temporary, instantaneous dipoles due to uneven electron distribution.
- Weakest IMF.
- Increases with molecular size/mass (more electrons → larger electron cloud → easier to distort → stronger LDF).
2. Dipole-Dipole Forces
- Present only in polar molecules.
- Attraction between the positive end of one polar molecule and the negative end of another.
- Stronger than LDF.
3. Hydrogen Bonding
- A special, strong type of dipole-dipole interaction.
- Occurs when H is bonded directly to F, O, or N (the most electronegative elements).
- The H atom is strongly δ+ and is attracted to a lone pair on F, O, or N of a neighboring molecule.
- Strongest IMF (but still much weaker than a chemical bond).
| IMF Type | Strength | Present In |
|---|---|---|
| London dispersion | Weakest | All molecules |
| Dipole-dipole | Moderate | Polar molecules only |
| Hydrogen bonding | Strongest IMF | Molecules with H—F, H—O, or H—N bonds |
Worked Example 2.11: For each pair, identify the stronger IMF and explain:
(a) H₂O vs H₂S (b) CH₄ vs CCl₄ (c) HF vs HCl
Solution:
(a) H₂O has hydrogen bonding (H bonded to O). H₂S has only dipole-dipole (H bonded to S, which is not electronegative enough for H-bonding). H₂O has stronger IMFs.
(b) Both are nonpolar, so both have only LDF. CCl₄ has more electrons and a larger electron cloud, so its LDF are stronger. CCl₄ has stronger IMFs.
(c) HF has hydrogen bonding (H bonded to F). HCl has only dipole-dipole (Cl is not electronegative enough for H-bonding with H). HF has stronger IMFs.
(a) H₂O vs H₂S (b) CH₄ vs CCl₄ (c) HF vs HCl
Solution:
(a) H₂O has hydrogen bonding (H bonded to O). H₂S has only dipole-dipole (H bonded to S, which is not electronegative enough for H-bonding). H₂O has stronger IMFs.
(b) Both are nonpolar, so both have only LDF. CCl₄ has more electrons and a larger electron cloud, so its LDF are stronger. CCl₄ has stronger IMFs.
(c) HF has hydrogen bonding (H bonded to F). HCl has only dipole-dipole (Cl is not electronegative enough for H-bonding with H). HF has stronger IMFs.
Key Exam Notes — Intermolecular Forces:
• H-bonding requires H bonded DIRECTLY to F, O, or N. H-bonding in HCl is WRONG!
• LDF increases with molecular mass/size (more electrons).
• Boiling point order reflects IMF strength: H-bonding > dipole-dipole > LDF.
• H₂O has an unusually high boiling point for its size because of hydrogen bonding.
• IMF strength: ionic bonding > hydrogen bonding > dipole-dipole > LDF.
• H-bonding requires H bonded DIRECTLY to F, O, or N. H-bonding in HCl is WRONG!
• LDF increases with molecular mass/size (more electrons).
• Boiling point order reflects IMF strength: H-bonding > dipole-dipole > LDF.
• H₂O has an unusually high boiling point for its size because of hydrogen bonding.
• IMF strength: ionic bonding > hydrogen bonding > dipole-dipole > LDF.
Practice Questions:
1. Explain why NH₃ has a higher boiling point (−33 °C) than PH₃ (−88 °C).
2. Which has a higher boiling point: CH₄ or C₂H₆? Explain.
1. Explain why NH₃ has a higher boiling point (−33 °C) than PH₃ (−88 °C).
2. Which has a higher boiling point: CH₄ or C₂H₆? Explain.
Answer 1:
NH₃ can form hydrogen bonds because it has H atoms bonded directly to N (highly electronegative). PH₃ cannot form hydrogen bonds because P is not electronegative enough (H-bonding only with F, O, N). PH₃ has only dipole-dipole forces and LDF. Since hydrogen bonding is much stronger than dipole-dipole, NH₃ has a higher boiling point.
Answer 2:
Both CH₄ and C₂H₆ are nonpolar, so both have only London dispersion forces. C₂H₆ has more electrons (18 vs 8 for CH₄) and a larger electron cloud, so its LDF are stronger. Therefore, C₂H₆ has a higher boiling point.
NH₃ can form hydrogen bonds because it has H atoms bonded directly to N (highly electronegative). PH₃ cannot form hydrogen bonds because P is not electronegative enough (H-bonding only with F, O, N). PH₃ has only dipole-dipole forces and LDF. Since hydrogen bonding is much stronger than dipole-dipole, NH₃ has a higher boiling point.
Answer 2:
Both CH₄ and C₂H₆ are nonpolar, so both have only London dispersion forces. C₂H₆ has more electrons (18 vs 8 for CH₄) and a larger electron cloud, so its LDF are stronger. Therefore, C₂H₆ has a higher boiling point.
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2.4 Metallic Bonding
2.4.1 Formation of Metallic Bonding
Metals have low ionization energies and tend to lose their valence electrons easily. In a metallic solid, the valence electrons are delocalized — they are not attached to any particular atom but move freely throughout the entire metal structure. This “sea” of delocalized electrons holds the positively charged metal ions together.
2.4.2 Electron-Sea Model
The electron-sea model (or sea of electrons model) pictures a metal as a lattice of positive metal ions surrounded by a “sea” of mobile, delocalized electrons. The attraction between the cations and the electron sea is the metallic bond.
e⁻ e⁻ e⁻ e⁻ e⁻ e⁻
M⁺ M⁺ M⁺ M⁺ M⁺ M⁺
e⁻ e⁻ e⁻ e⁻ e⁻ e⁻
M⁺ M⁺ M⁺ M⁺ M⁺ M⁺
e⁻ e⁻ e⁻ e⁻ e⁻ e⁻
(Positive ions in a “sea” of free electrons)
2.4.3 Properties of Metals and Bonding
The electron-sea model explains the characteristic properties of metals:
- Electrical conductivity: Delocalized electrons are free to move and carry electric current.
- Thermal conductivity: Delocalized electrons transfer kinetic energy rapidly through the metal.
- Malleability and ductility: When force is applied, metal ions can slide past each other. The delocalized electrons adjust and continue to hold the ions together. Unlike ionic solids, there are no like charges to cause repulsion.
- Luster (shininess): Delocalized electrons absorb and re-emit light at various wavelengths, giving metals their characteristic shine.
- Melting points vary: Depends on the strength of metallic bonding. More valence electrons and smaller ion size → stronger metallic bonding → higher melting point (e.g., W has very high m.p.; Na has low m.p.).
Key Exam Notes — Metallic Bonding:
• Metallic bond = attraction between metal cations and delocalized electron sea.
• Explains: conductivity, malleability, ductility, luster.
• Malleability is possible because shifting cations does not create like-charge repulsion (unlike ionic crystals).
• Stronger metallic bonding with more valence electrons and smaller ion size.
• Metallic bond = attraction between metal cations and delocalized electron sea.
• Explains: conductivity, malleability, ductility, luster.
• Malleability is possible because shifting cations does not create like-charge repulsion (unlike ionic crystals).
• Stronger metallic bonding with more valence electrons and smaller ion size.
Practice Questions:
1. Use the electron-sea model to explain why metals are malleable but ionic solids are brittle.
2. Why does tungsten (W) have a much higher melting point than sodium (Na)?
1. Use the electron-sea model to explain why metals are malleable but ionic solids are brittle.
2. Why does tungsten (W) have a much higher melting point than sodium (Na)?
Answer 1:
In metals, when layers of cations shift due to applied force, the delocalized electrons simply move with them and continue to attract all the cations. No like charges become adjacent. In ionic solids, shifting layers brings ions of the same charge next to each other, causing strong repulsion that shatters the crystal.
Answer 2:
Tungsten has more valence electrons available for delocalization and a smaller ionic radius compared to sodium. More delocalized electrons and shorter distances between cations and electrons create stronger metallic bonding. Sodium has only 1 valence electron and a large atomic radius, resulting in weak metallic bonding and low melting point.
In metals, when layers of cations shift due to applied force, the delocalized electrons simply move with them and continue to attract all the cations. No like charges become adjacent. In ionic solids, shifting layers brings ions of the same charge next to each other, causing strong repulsion that shatters the crystal.
Answer 2:
Tungsten has more valence electrons available for delocalization and a smaller ionic radius compared to sodium. More delocalized electrons and shorter distances between cations and electrons create stronger metallic bonding. Sodium has only 1 valence electron and a large atomic radius, resulting in weak metallic bonding and low melting point.
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2.5 Chemical Bonding Theories
2.5.1 Valence Bond (VB) Theory
The Valence Bond Theory explains covalent bond formation using the concept of orbital overlap:
- A covalent bond forms when the atomic orbitals of two atoms overlap.
- The overlapping orbitals must contain electrons with opposite spins (Pauli exclusion principle).
- The greater the overlap, the stronger the bond.
- Each overlapping orbital contributes one electron to the shared pair.
Types of Orbital Overlap
Sigma (σ) Bond: Formed by head-on (end-to-end) overlap of orbitals along the internuclear axis. A sigma bond allows free rotation. Examples: s-s overlap (H₂), s-p overlap (HCl), p-p overlap (Cl₂).
Pi (π) Bond: Formed by side-by-side (lateral) overlap of p-orbitals, with the electron density above and below the internuclear axis. A pi bond does NOT allow free rotation. Pi bonds are always formed in addition to a sigma bond (a double bond = 1 σ + 1 π; a triple bond = 1 σ + 2 π).
Key Rule: Every covalent bond has exactly ONE sigma bond. Additional bonds between the same two atoms are pi bonds.
• Single bond = 1 σ
• Double bond = 1 σ + 1 π
• Triple bond = 1 σ + 2 π
• Single bond = 1 σ
• Double bond = 1 σ + 1 π
• Triple bond = 1 σ + 2 π
Hybridization
How can carbon form four identical bonds (as in CH₄) when it only has two unpaired electrons (2s² 2p²)? The answer is hybridization — the mixing of atomic orbitals to form new, equivalent hybrid orbitals.
In hybridization, one s orbital mixes with some number of p (and sometimes d) orbitals to form hybrid orbitals that all have the same energy and shape. The number of hybrid orbitals equals the total number of atomic orbitals mixed.
| Hybridization | Atomic Orbitals Mixed | Hybrid Orbitals Formed | Geometry | Bond Angle |
|---|---|---|---|---|
| sp | 1 s + 1 p | 2 sp orbitals | Linear | 180° |
| sp² | 1 s + 2 p | 3 sp² orbitals | Trigonal planar | 120° |
| sp³ | 1 s + 3 p | 4 sp³ orbitals | Tetrahedral | 109.5° |
| sp³d | 1 s + 3 p + 1 d | 5 sp³d orbitals | Trigonal bipyramidal | 90°, 120° |
| sp³d² | 1 s + 3 p + 2 d | 6 sp³d² orbitals | Octahedral | 90° |
How to determine hybridization: Count the number of electron domains (steric number) around the central atom:
Electron domains: 2 3 4 5 6
Hybridization: sp sp² sp³ sp³d sp³d²
Worked Example 2.12: Determine the hybridization of the central atom in: (a) BeCl₂, (b) BF₃, (c) CH₄, (d) NH₃, (e) H₂O, (f) PCl₅, (g) SF₆.
Solution:
(a) BeCl₂: 2 bonding domains, 0 lone pairs = 2 domains → sp (linear)
(b) BF₃: 3 bonding domains, 0 lone pairs = 3 domains → sp² (trigonal planar)
(c) CH₄: 4 bonding domains, 0 lone pairs = 4 domains → sp³ (tetrahedral)
(d) NH₃: 3 bonding + 1 lone pair = 4 domains → sp³ (trigonal pyramidal)
(e) H₂O: 2 bonding + 2 lone pairs = 4 domains → sp³ (bent)
(f) PCl₅: 5 bonding + 0 lone pairs = 5 domains → sp³d (trigonal bipyramidal)
(g) SF₆: 6 bonding + 0 lone pairs = 6 domains → sp³d² (octahedral)
Solution:
(a) BeCl₂: 2 bonding domains, 0 lone pairs = 2 domains → sp (linear)
(b) BF₃: 3 bonding domains, 0 lone pairs = 3 domains → sp² (trigonal planar)
(c) CH₄: 4 bonding domains, 0 lone pairs = 4 domains → sp³ (tetrahedral)
(d) NH₃: 3 bonding + 1 lone pair = 4 domains → sp³ (trigonal pyramidal)
(e) H₂O: 2 bonding + 2 lone pairs = 4 domains → sp³ (bent)
(f) PCl₅: 5 bonding + 0 lone pairs = 5 domains → sp³d (trigonal bipyramidal)
(g) SF₆: 6 bonding + 0 lone pairs = 6 domains → sp³d² (octahedral)
Worked Example 2.13: Describe the bonding in ethene (C₂H₄, H₂C=CH₂) using hybridization and identify all σ and π bonds.
Solution:
Each C atom is bonded to 2 H atoms and 1 C atom (3 domains, no lone pairs) → sp² hybridization.
Each C has 3 sp² hybrid orbitals (for the 2 C—H bonds and 1 C—C σ bond) and 1 unhybridized p orbital.
The sp² orbitals form 4 C—H σ bonds and 1 C—C σ bond (total: 5 σ bonds).
The two unhybridized p orbitals overlap side-by-side to form 1 C—C π bond.
Total bonds: 5 σ + 1 π = 6 bonds (1 double bond + 4 single bonds).
The molecule is planar (trigonal planar around each C) with bond angles of about 120°.
Solution:
Each C atom is bonded to 2 H atoms and 1 C atom (3 domains, no lone pairs) → sp² hybridization.
Each C has 3 sp² hybrid orbitals (for the 2 C—H bonds and 1 C—C σ bond) and 1 unhybridized p orbital.
The sp² orbitals form 4 C—H σ bonds and 1 C—C σ bond (total: 5 σ bonds).
The two unhybridized p orbitals overlap side-by-side to form 1 C—C π bond.
Total bonds: 5 σ + 1 π = 6 bonds (1 double bond + 4 single bonds).
The molecule is planar (trigonal planar around each C) with bond angles of about 120°.
Worked Example 2.14: Describe the bonding in ethyne (C₂H₂, HC≡CH).
Solution:
Each C is bonded to 1 H and 1 C (2 domains) → sp hybridization.
Each C has 2 sp hybrid orbitals (for 1 C—H σ bond and 1 C—C σ bond) and 2 unhybridized p orbitals.
Total σ bonds: 2 C—H + 1 C—C = 3 σ bonds.
The two unhybridized p orbitals on each C overlap to form 2 π bonds.
Total bonds: 3 σ + 2 π = 5 bonds (1 triple bond + 2 single bonds).
Molecular geometry: linear (180°).
Solution:
Each C is bonded to 1 H and 1 C (2 domains) → sp hybridization.
Each C has 2 sp hybrid orbitals (for 1 C—H σ bond and 1 C—C σ bond) and 2 unhybridized p orbitals.
Total σ bonds: 2 C—H + 1 C—C = 3 σ bonds.
The two unhybridized p orbitals on each C overlap to form 2 π bonds.
Total bonds: 3 σ + 2 π = 5 bonds (1 triple bond + 2 single bonds).
Molecular geometry: linear (180°).
Key Exam Notes — VB Theory & Hybridization:
• σ bond = head-on overlap; π bond = side-by-side overlap.
• Every bond has 1 σ; extra bonds are π. Triple = 1σ + 2π.
• Hybridization is determined by electron domains (NOT by the number of bonds alone — count lone pairs too!).
• sp = linear (180°), sp² = trigonal planar (120°), sp³ = tetrahedral (109.5°).
• Unhybridized p orbitals form π bonds; hybrid orbitals form σ bonds.
• Double bond = 1 σ + 1 π; Triple bond = 1 σ + 2 π.
• NH₃ and H₂O both have sp³ hybridization (lone pairs count as domains!).
• σ bond = head-on overlap; π bond = side-by-side overlap.
• Every bond has 1 σ; extra bonds are π. Triple = 1σ + 2π.
• Hybridization is determined by electron domains (NOT by the number of bonds alone — count lone pairs too!).
• sp = linear (180°), sp² = trigonal planar (120°), sp³ = tetrahedral (109.5°).
• Unhybridized p orbitals form π bonds; hybrid orbitals form σ bonds.
• Double bond = 1 σ + 1 π; Triple bond = 1 σ + 2 π.
• NH₃ and H₂O both have sp³ hybridization (lone pairs count as domains!).
Practice Questions:
1. How many σ and π bonds are in a molecule of formaldehyde (H₂C=O)?
2. A molecule has the formula AB₄ with no lone pairs on A. What is the hybridization of A and the molecular geometry?
3. Why can’t free rotation occur around a C=C double bond but it can around a C—C single bond?
1. How many σ and π bonds are in a molecule of formaldehyde (H₂C=O)?
2. A molecule has the formula AB₄ with no lone pairs on A. What is the hybridization of A and the molecular geometry?
3. Why can’t free rotation occur around a C=C double bond but it can around a C—C single bond?
Answer 1:
H₂C=O: 2 C—H single bonds (2 σ), 1 C=O double bond (1 σ + 1 π). Total = 3 σ bonds + 1 π bond.
Answer 2:
AB₄ with 0 lone pairs: 4 electron domains → sp³ hybridization → tetrahedral geometry (109.5°). Example: CH₄.
Answer 3:
A C—C single bond has only a σ bond, which is formed by end-to-end overlap along the internuclear axis. This allows the atoms to rotate freely. A C=C double bond has 1 σ bond + 1 π bond. The π bond is formed by side-by-side overlap of p orbitals above and below the internuclear axis. If you try to rotate, the p orbitals would no longer overlap, and the π bond would break. This requires significant energy, so rotation is restricted.
H₂C=O: 2 C—H single bonds (2 σ), 1 C=O double bond (1 σ + 1 π). Total = 3 σ bonds + 1 π bond.
Answer 2:
AB₄ with 0 lone pairs: 4 electron domains → sp³ hybridization → tetrahedral geometry (109.5°). Example: CH₄.
Answer 3:
A C—C single bond has only a σ bond, which is formed by end-to-end overlap along the internuclear axis. This allows the atoms to rotate freely. A C=C double bond has 1 σ bond + 1 π bond. The π bond is formed by side-by-side overlap of p orbitals above and below the internuclear axis. If you try to rotate, the p orbitals would no longer overlap, and the π bond would break. This requires significant energy, so rotation is restricted.
2.5.2 Molecular Orbital Theory (MOT)
The Valence Bond Theory has limitations. The Molecular Orbital Theory takes a different approach: instead of combining atomic orbitals on individual atoms, it combines atomic orbitals from ALL atoms in the molecule to form molecular orbitals (MOs) that belong to the entire molecule.
Key principles of MOT:
- Number of MOs = Number of AOs combined. If two atoms each contribute one atomic orbital, two molecular orbitals are formed.
- Bonding MO (σ or π): Lower energy than the original AOs. Electron density is between the nuclei → stabilizes the molecule.
- Antibonding MO (σ* or π*): Higher energy than the original AOs. Electron density is away from the region between nuclei → destabilizes the molecule.
- Electrons fill MOs following the same rules as for atomic orbitals: Aufbau, Pauli, Hund’s.
MO Diagram for Homonuclear Diatomic Molecules (H₂ through N₂)
For molecules H₂ through N₂, the energy ordering is:
σ2s σ*2s π2p π2p σ2p π*2p π*2p σ*2p
Energy: low ————————→ high
(Note: π2p is lower than σ2p for H₂ through N₂)
For O₂ and F₂, the ordering changes: σ2p is lower than π2p.
Bond Order
The bond order predicts the stability of a molecule:
$$\text{Bond order} = \frac{1}{2}(N_b – N_a)$$
where $N_b$ = number of electrons in bonding orbitals, $N_a$ = number of electrons in antibonding orbitals.
- Bond order = 1 → single bond
- Bond order = 2 → double bond
- Bond order = 3 → triple bond
- Bond order = 0 → molecule does not exist (or is extremely unstable)
- Higher bond order → shorter bond length, higher bond energy
Worked Example 2.15: Construct the MO diagram for O₂ and determine its bond order. Is O₂ paramagnetic or diamagnetic?
Solution:
O₂ has 12 valence electrons (6 + 6).
For O₂ (and F₂), the ordering is: σ2s < σ*2s < σ2p < π2p = π2p < π*2p = π*2p < σ*2p
Filling (12 electrons):
σ2s: ↑↓ (2)
σ*2s: ↑↓ (2)
σ2p: ↑↓ (2)
π2p: ↑↓ ↑↓ (4)
π*2p: ↑ ↑ (2) — by Hund’s rule, electrons go into separate degenerate orbitals with parallel spins
$N_b$ = 2 + 2 + 2 + 4 = 10
$N_a$ = 2 + 2 = 4
Bond order = ½(10 − 4) = 3… Wait, that’s not right. Let me recount.
Bonding: σ2s(2) + σ2p(2) + π2p(4) = 8
Antibonding: σ*2s(2) + π*2p(2) = 4
Bond order = ½(8 − 4) = 2 (double bond character)
Since there are 2 unpaired electrons in π*2p orbitals, O₂ is paramagnetic. This is one of the great successes of MOT — it correctly predicts that liquid O₂ is attracted to a magnet, which Lewis theory cannot explain!
Solution:
O₂ has 12 valence electrons (6 + 6).
For O₂ (and F₂), the ordering is: σ2s < σ*2s < σ2p < π2p = π2p < π*2p = π*2p < σ*2p
Filling (12 electrons):
σ2s: ↑↓ (2)
σ*2s: ↑↓ (2)
σ2p: ↑↓ (2)
π2p: ↑↓ ↑↓ (4)
π*2p: ↑ ↑ (2) — by Hund’s rule, electrons go into separate degenerate orbitals with parallel spins
$N_b$ = 2 + 2 + 2 + 4 = 10
$N_a$ = 2 + 2 = 4
Bond order = ½(10 − 4) = 3… Wait, that’s not right. Let me recount.
Bonding: σ2s(2) + σ2p(2) + π2p(4) = 8
Antibonding: σ*2s(2) + π*2p(2) = 4
Bond order = ½(8 − 4) = 2 (double bond character)
Since there are 2 unpaired electrons in π*2p orbitals, O₂ is paramagnetic. This is one of the great successes of MOT — it correctly predicts that liquid O₂ is attracted to a magnet, which Lewis theory cannot explain!
Worked Example 2.16: Determine the bond order of N₂, O₂⁺, and O₂⁻ using MOT.
Solution:
N₂ (10 valence e⁻, ordering: σ2s < σ*2s < π2p < σ2p):
Bonding: σ2s(2) + π2p(4) + σ2p(2) = 8
Antibonding: σ*2s(2) = 2
Bond order = ½(8 − 2) = 3 (triple bond)
O₂⁺ (11 valence e⁻): Remove one e⁻ from π*2p compared to O₂.
Bonding: 8; Antibonding: 3
Bond order = ½(8 − 3) = 2.5
O₂⁻ (13 valence e⁻): Add one e⁻ to π*2p compared to O₂.
Bonding: 8; Antibonding: 5
Bond order = ½(8 − 5) = 1.5
Note: Bond length: O₂⁺ < O₂ < O₂⁻ (higher bond order = shorter bond).
Solution:
N₂ (10 valence e⁻, ordering: σ2s < σ*2s < π2p < σ2p):
Bonding: σ2s(2) + π2p(4) + σ2p(2) = 8
Antibonding: σ*2s(2) = 2
Bond order = ½(8 − 2) = 3 (triple bond)
O₂⁺ (11 valence e⁻): Remove one e⁻ from π*2p compared to O₂.
Bonding: 8; Antibonding: 3
Bond order = ½(8 − 3) = 2.5
O₂⁻ (13 valence e⁻): Add one e⁻ to π*2p compared to O₂.
Bonding: 8; Antibonding: 5
Bond order = ½(8 − 5) = 1.5
Note: Bond length: O₂⁺ < O₂ < O₂⁻ (higher bond order = shorter bond).
Worked Example 2.17: Will He₂ exist? Use MOT to explain.
Solution:
He₂ has 4 electrons (2 from each He, using 1s orbitals).
σ1s: ↑↓ (2 bonding)
σ*1s: ↑↓ (2 antibonding)
Bond order = ½(2 − 2) = 0
Bond order of 0 means He₂ does not exist as a stable molecule. The bonding and antibonding effects cancel out. This agrees with observation — helium exists as individual atoms, not as He₂ molecules.
Solution:
He₂ has 4 electrons (2 from each He, using 1s orbitals).
σ1s: ↑↓ (2 bonding)
σ*1s: ↑↓ (2 antibonding)
Bond order = ½(2 − 2) = 0
Bond order of 0 means He₂ does not exist as a stable molecule. The bonding and antibonding effects cancel out. This agrees with observation — helium exists as individual atoms, not as He₂ molecules.
Key Exam Notes — Molecular Orbital Theory:
• MOs belong to the entire molecule, not individual atoms.
• # bonding MOs + # antibonding MOs = # AOs combined.
• Bond order = ½(Nb − Na). BO > 0 for stability.
• For H₂ to N₂: π2p < σ2p. For O₂ to F₂: σ2p < π2p.
• O₂ is paramagnetic (2 unpaired e⁻ in π*2p) — MOT explains this, VB theory cannot!
• Removing electrons (cation) increases bond order and decreases bond length.
• Adding electrons (anion) decreases bond order and increases bond length.
• BO = 0 → molecule does not exist (e.g., He₂, Be₂).
• MOs belong to the entire molecule, not individual atoms.
• # bonding MOs + # antibonding MOs = # AOs combined.
• Bond order = ½(Nb − Na). BO > 0 for stability.
• For H₂ to N₂: π2p < σ2p. For O₂ to F₂: σ2p < π2p.
• O₂ is paramagnetic (2 unpaired e⁻ in π*2p) — MOT explains this, VB theory cannot!
• Removing electrons (cation) increases bond order and decreases bond length.
• Adding electrons (anion) decreases bond order and increases bond length.
• BO = 0 → molecule does not exist (e.g., He₂, Be₂).
Practice Questions:
1. Write the MO electron configuration for N₂ and determine its magnetic property.
2. Which has a shorter bond: N₂ or N₂⁺? Explain using bond order.
1. Write the MO electron configuration for N₂ and determine its magnetic property.
2. Which has a shorter bond: N₂ or N₂⁺? Explain using bond order.
Answer 1:
N₂ (10 valence e⁻):
σ2s² σ*2s² π2p⁴ σ2p²
All electrons are paired → N₂ is diamagnetic.
Answer 2:
N₂: BO = ½(8 − 2) = 3
N₂⁺ (9 e⁻, remove 1 from σ2p): Bonding = 7, Antibonding = 2, BO = ½(7 − 2) = 2.5
Since N₂ has a higher bond order (3 vs 2.5), N₂ has a shorter bond than N₂⁺.
N₂ (10 valence e⁻):
σ2s² σ*2s² π2p⁴ σ2p²
All electrons are paired → N₂ is diamagnetic.
Answer 2:
N₂: BO = ½(8 − 2) = 3
N₂⁺ (9 e⁻, remove 1 from σ2p): Bonding = 7, Antibonding = 2, BO = ½(7 − 2) = 2.5
Since N₂ has a higher bond order (3 vs 2.5), N₂ has a shorter bond than N₂⁺.
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2.6 Types of Crystal
Solids can be classified based on the type of particles at the lattice points and the forces holding them together:
| Crystal Type | Particles at Lattice Points | Bonding Force | Properties | Examples |
|---|---|---|---|---|
| Ionic | Cations and anions | Electrostatic (ionic) | High m.p., hard, brittle, conducts when molten/dissolved | NaCl, MgO |
| Covalent network | Atoms | Covalent bonds (continuous) | Very high m.p., extremely hard, insulator | Diamond, SiO₂ |
| Molecular | Molecules | Intermolecular forces (LDF, dipole, H-bonding) | Low m.p., soft, insulator | Ice (H₂O), I₂, CO₂ |
| Metallic | Metal cations | Metallic bonding (electron sea) | Variable m.p., malleable, ductile, conducts | Fe, Cu, Na |
Notice the pattern: the stronger the force between particles, the higher the melting point. Covalent network solids have the highest (continuous covalent bonds), while molecular solids have the lowest (weak IMFs).
Key Exam Notes — Crystal Types:
• Ionic → ionic bonds → high m.p., brittle, conducts when molten.
• Covalent network → continuous covalent bonds → very high m.p., extremely hard.
• Molecular → IMFs → low m.p., soft, does not conduct.
• Metallic → metallic bonds → variable m.p., malleable, conducts.
• Diamond and SiO₂ are covalent network, NOT molecular. Graphite is also covalent network (layers of sp² carbon with weak LDF between layers).
• Ionic → ionic bonds → high m.p., brittle, conducts when molten.
• Covalent network → continuous covalent bonds → very high m.p., extremely hard.
• Molecular → IMFs → low m.p., soft, does not conduct.
• Metallic → metallic bonds → variable m.p., malleable, conducts.
• Diamond and SiO₂ are covalent network, NOT molecular. Graphite is also covalent network (layers of sp² carbon with weak LDF between layers).
Practice Questions:
1. Why does diamond have a much higher melting point than sodium chloride?
2. Solid CO₂ (dry ice) sublimes at −78 °C. What type of crystal is it?
1. Why does diamond have a much higher melting point than sodium chloride?
2. Solid CO₂ (dry ice) sublimes at −78 °C. What type of crystal is it?
Answer 1:
Diamond is a covalent network solid — every carbon atom is bonded to four others by strong covalent bonds in a continuous 3D network. To melt diamond, you must break ALL these covalent bonds, which requires enormous energy. NaCl is ionic — to melt it, you only need to overcome the electrostatic attraction between Na⁺ and Cl⁻ ions (which is strong but much weaker than breaking covalent bonds).
Answer 2:
Solid CO₂ is a molecular crystal. The CO₂ molecules are held together by weak London dispersion forces (CO₂ is nonpolar). This is why it sublimes at such a low temperature — the IMFs are easily overcome.
Diamond is a covalent network solid — every carbon atom is bonded to four others by strong covalent bonds in a continuous 3D network. To melt diamond, you must break ALL these covalent bonds, which requires enormous energy. NaCl is ionic — to melt it, you only need to overcome the electrostatic attraction between Na⁺ and Cl⁻ ions (which is strong but much weaker than breaking covalent bonds).
Answer 2:
Solid CO₂ is a molecular crystal. The CO₂ molecules are held together by weak London dispersion forces (CO₂ is nonpolar). This is why it sublimes at such a low temperature — the IMFs are easily overcome.
Quick Revision Notes — Exam Focus
1. Important Definitions
• Octet rule: Atoms tend to have 8 electrons in their valence shell (2 for H, He).
• Ionic bond: Electrostatic attraction between cations and anions formed by electron transfer.
• Covalent bond: Sharing of electron pairs between non-metal atoms.
• Coordinate (dative) bond: Covalent bond where both electrons come from one atom.
• Lattice energy: Energy released when 1 mole of ionic compound forms from gaseous ions.
• Resonance: When multiple Lewis structures differ only in electron positions; real structure is a hybrid.
• Formal charge: FC = valence e⁻ − non-bonding e⁻ − ½(bonding e⁻).
• Hybridization: Mixing of atomic orbitals to form equivalent hybrid orbitals.
• Sigma (σ) bond: Head-on overlap; present in every bond.
• Pi (π) bond: Side-by-side overlap; additional bonds beyond the first.
• Bond order (MOT): ½(Nb − Na).
• Hydrogen bonding: Strong dipole-dipole interaction involving H bonded to F, O, or N.
• Ionic bond: Electrostatic attraction between cations and anions formed by electron transfer.
• Covalent bond: Sharing of electron pairs between non-metal atoms.
• Coordinate (dative) bond: Covalent bond where both electrons come from one atom.
• Lattice energy: Energy released when 1 mole of ionic compound forms from gaseous ions.
• Resonance: When multiple Lewis structures differ only in electron positions; real structure is a hybrid.
• Formal charge: FC = valence e⁻ − non-bonding e⁻ − ½(bonding e⁻).
• Hybridization: Mixing of atomic orbitals to form equivalent hybrid orbitals.
• Sigma (σ) bond: Head-on overlap; present in every bond.
• Pi (π) bond: Side-by-side overlap; additional bonds beyond the first.
• Bond order (MOT): ½(Nb − Na).
• Hydrogen bonding: Strong dipole-dipole interaction involving H bonded to F, O, or N.
2. Key Formulas and Relationships
$$\text{Formal charge} = V – N – \frac{B}{2}$$ $$\text{Bond order (MOT)} = \frac{1}{2}(N_b – N_a)$$ $$F = k\frac{q_1 q_2}{r^2} \quad \text{(Coulomb’s law for ionic bond strength)}$$
3. Hybridization Quick Reference
Domains: 2 3 4 5 6
Hybrid: sp sp² sp³ sp³d sp³d²
Shape: linear trig.pl tetra trig.bipyr octahedral
Angle: 180° 120° 109.5° 90°,120° 90°
4. VSEPR Shapes Summary
| Domains | Lone Pairs | Shape | Angle | Example |
|---|---|---|---|---|
| 2 | 0 | Linear | 180° | CO₂, BeCl₂ |
| 3 | 0 | Trigonal planar | 120° | BF₃ |
| 3 | 1 | Bent | <120° | SO₂ |
| 4 | 0 | Tetrahedral | 109.5° | CH₄ |
| 4 | 1 | Trigonal pyramidal | <109.5° | NH₃ |
| 4 | 2 | Bent | <109.5° | H₂O |
| 5 | 0 | Trigonal bipyramidal | 90°, 120° | PCl₅ |
| 5 | 1 | See-saw | <90°, <120° | SF₄ |
| 5 | 2 | T-shaped | <90° | ClF₃ |
| 5 | 3 | Linear | 180° | XeF₂ |
| 6 | 0 | Octahedral | 90° | SF₆ |
| 6 | 1 | Square pyramidal | <90° | BrF₅ |
| 6 | 2 | Square planar | 90° | XeF₄ |
5. Bond Type Summary
| Single Bond | Double Bond | Triple Bond | |
|---|---|---|---|
| σ bonds | 1 | 1 | 1 |
| π bonds | 0 | 1 | 2 |
| Strength | Weakest | Intermediate | Strongest |
| Length | Longest | Intermediate | Shortest |
| Rotation | Free | Restricted | Restricted |
6. MO Energy Ordering
H₂ through N₂ (fewer than 15 valence electrons):
σ2s < σ*2s < π2p = π2p < σ2p < π*2p = π*2p < σ*2p
O₂ through F₂ (15 or more valence electrons):
σ2s < σ*2s < σ2p < π2p = π2p < π*2p = π*2p < σ*2p
Mnemonic: “N₂ is special” — for N₂ and lighter, π2p is lower. For O₂ and heavier, σ2p is lower.
σ2s < σ*2s < π2p = π2p < σ2p < π*2p = π*2p < σ*2p
O₂ through F₂ (15 or more valence electrons):
σ2s < σ*2s < σ2p < π2p = π2p < π*2p = π*2p < σ*2p
Mnemonic: “N₂ is special” — for N₂ and lighter, π2p is lower. For O₂ and heavier, σ2p is lower.
7. Intermolecular Forces Ranking
Ionic bonding >>> H-bonding >> Dipole-dipole > London dispersion
(strongest) (strongest IMF) (weakest IMF)
8. Crystal Types Quick Guide
Crystal Type | Force | Melting Point
—————–|—————–|————-
Ionic | Ionic bonds | High
Covalent network | Covalent bonds | Very high
Molecular | IMFs | Low
Metallic | Metallic bonds | Variable
9. Common Mistakes to Avoid
❌ Saying H-bonding occurs with H-Cl, H-S, or H-C (ONLY H-F, H-O, H-N!).
❌ Forgetting to count lone pairs as electron domains for VSEPR.
❌ Confusing electron-pair geometry with molecular geometry (they differ when lone pairs exist).
❌ Using the wrong MO ordering for N₂ vs O₂.
❌ Forgetting that coordinate bonds, once formed, are identical to regular covalent bonds.
❌ Subtracting charge electrons incorrectly in Lewis structures for ions.
❌ Thinking double bonds count as 2 electron domains (they count as 1!).
❌ Saying “ionic compounds form molecules” — they form crystal lattices, not molecules.
❌ Confusing bond polarity with molecular polarity (shape matters!).
❌ Writing sp³ for NH₃ as “sp² because it has 3 bonds” — lone pairs count!
❌ Forgetting to count lone pairs as electron domains for VSEPR.
❌ Confusing electron-pair geometry with molecular geometry (they differ when lone pairs exist).
❌ Using the wrong MO ordering for N₂ vs O₂.
❌ Forgetting that coordinate bonds, once formed, are identical to regular covalent bonds.
❌ Subtracting charge electrons incorrectly in Lewis structures for ions.
❌ Thinking double bonds count as 2 electron domains (they count as 1!).
❌ Saying “ionic compounds form molecules” — they form crystal lattices, not molecules.
❌ Confusing bond polarity with molecular polarity (shape matters!).
❌ Writing sp³ for NH₃ as “sp² because it has 3 bonds” — lone pairs count!
Challenge Exam Questions
Test yourself thoroughly! Try each question before checking the answer.
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Section A: Multiple Choice Questions
Q1. Which molecule has a bent (V-shaped) molecular geometry?
A) CO₂ B) BF₃ C) H₂O D) CH₄
A) CO₂ B) BF₃ C) H₂O D) CH₄
Answer: C
H₂O has 2 bonding pairs and 2 lone pairs around O = 4 electron domains (tetrahedral electron-pair geometry). With 2 lone pairs, the molecular shape is bent. CO₂ is linear, BF₃ is trigonal planar, CH₄ is tetrahedral.
H₂O has 2 bonding pairs and 2 lone pairs around O = 4 electron domains (tetrahedral electron-pair geometry). With 2 lone pairs, the molecular shape is bent. CO₂ is linear, BF₃ is trigonal planar, CH₄ is tetrahedral.
Q2. How many sigma and pi bonds are present in a molecule of propyne (CH₃C≡CH)?
A) 5σ, 2π B) 6σ, 2π C) 5σ, 3π D) 6σ, 3π
A) 5σ, 2π B) 6σ, 2π C) 5σ, 3π D) 6σ, 3π
Answer: B
CH₃C≡CH has: 3 C—H bonds (3σ) + 1 C—C single bond (1σ) + 1 C≡C triple bond (1σ + 2π) + 1 C—H bond on the other C (1σ) = 6σ + 2π. Every single bond is 1σ, the triple bond contributes 1σ + 2π.
CH₃C≡CH has: 3 C—H bonds (3σ) + 1 C—C single bond (1σ) + 1 C≡C triple bond (1σ + 2π) + 1 C—H bond on the other C (1σ) = 6σ + 2π. Every single bond is 1σ, the triple bond contributes 1σ + 2π.
Q3. According to Molecular Orbital Theory, the bond order of O₂⁻ is:
A) 1.0 B) 1.5 C) 2.0 D) 2.5
A) 1.0 B) 1.5 C) 2.0 D) 2.5
Answer: B
O₂ has 12 valence e⁻. O₂⁻ has 13 valence e⁻ (one extra in π*2p).
For O₂ ordering: σ2s² σ*2s² σ2p² π2p⁴ π*2p³
Bonding electrons (Nb) = 2 + 2 + 4 = 8
Antibonding electrons (Na) = 2 + 3 = 5
Bond order = ½(8 − 5) = 1.5
O₂ has 12 valence e⁻. O₂⁻ has 13 valence e⁻ (one extra in π*2p).
For O₂ ordering: σ2s² σ*2s² σ2p² π2p⁴ π*2p³
Bonding electrons (Nb) = 2 + 2 + 4 = 8
Antibonding electrons (Na) = 2 + 3 = 5
Bond order = ½(8 − 5) = 1.5
Q4. The strongest type of intermolecular force in liquid NH₃ is:
A) London dispersion forces B) Dipole-dipole forces C) Hydrogen bonding D) Ionic bonding
A) London dispersion forces B) Dipole-dipole forces C) Hydrogen bonding D) Ionic bonding
Answer: C
NH₃ has N—H bonds. Since H is bonded directly to N (one of the three most electronegative elements), NH₃ exhibits hydrogen bonding. While NH₃ also has LDF and dipole-dipole forces, hydrogen bonding is the strongest.
NH₃ has N—H bonds. Since H is bonded directly to N (one of the three most electronegative elements), NH₃ exhibits hydrogen bonding. While NH₃ also has LDF and dipole-dipole forces, hydrogen bonding is the strongest.
Q5. Which of the following is an example of a coordinate covalent bond?
A) H₂ B) NH₄⁺ C) NaCl D) O₂
A) H₂ B) NH₄⁺ C) NaCl D) O₂
Answer: B
In NH₄⁺, the fourth bond between NH₃ and H⁺ is a coordinate bond — NH₃ donates its lone pair to H⁺ (which has no electrons). Once formed, this bond is identical to the other three N—H bonds. H₂ and O₂ have regular covalent bonds; NaCl has an ionic bond.
In NH₄⁺, the fourth bond between NH₃ and H⁺ is a coordinate bond — NH₃ donates its lone pair to H⁺ (which has no electrons). Once formed, this bond is identical to the other three N—H bonds. H₂ and O₂ have regular covalent bonds; NaCl has an ionic bond.
Q6. The hybridization of the central atom in SF₆ is:
A) sp³ B) sp³d C) sp³d² D) sp²
A) sp³ B) sp³d C) sp³d² D) sp²
Answer: C
SF₆: S is bonded to 6 F atoms with 0 lone pairs = 6 electron domains. 6 domains → sp³d² hybridization → octahedral geometry.
SF₆: S is bonded to 6 F atoms with 0 lone pairs = 6 electron domains. 6 domains → sp³d² hybridization → octahedral geometry.
Section B: Fill in the Blanks
Q7. The energy required to separate one mole of an ionic solid into its gaseous ions is called ________.
Answer: Lattice energy. It is the energy released when gaseous ions come together to form one mole of ionic solid (the reverse process). The larger the lattice energy, the stronger the ionic bond and the higher the melting point.
Q8. In the VSEPR model, a double bond is treated as ________ electron domain(s).
Answer: One. A double bond (or triple bond) occupies only one region of space around the central atom, so it counts as one electron domain for VSEPR purposes, just like a single bond.
Q9. The molecular geometry of a molecule with 5 bonding domains and 1 lone pair around the central atom is ________.
Answer: See-saw. With 6 total domains (5 bonding + 1 lone pair), the electron-pair geometry is octahedral. With 1 lone pair, the molecular shape is square pyramidal… Wait, I need to correct this. With 5 bonding + 1 lone pair = 6 domains → octahedral electron-pair geometry. With 1 lone pair in octahedral → square pyramidal. See-saw is 4 bonding + 1 lone pair = 5 domains (trigonal bipyramidal). Let me correct: the answer depends on whether we mean 5+1=6 or 4+1=5. Since the question says “5 bonding domains and 1 lone pair” = 6 total → square pyramidal.
Q10. A molecule with bond order of zero according to MOT is predicted to be ________.
Answer: Unstable / nonexistent. Bond order of zero means the number of bonding electrons equals the number of antibonding electrons — the bonding and antibonding effects cancel completely. Example: He₂ (BO = 0) does not exist as a stable molecule.
Q11. Diamond is an example of a ________ crystal, while solid CO₂ is an example of a ________ crystal.
Answer: Diamond: covalent network (continuous covalent bonds in 3D). Solid CO₂: molecular (CO₂ molecules held together by weak London dispersion forces).
Section C: Short Answer Questions
Q12. Explain why the boiling point of water (100 °C) is much higher than that of H₂S (−60 °C), even though H₂S has a larger molecular mass.
Answer:
Water (H₂O) can form hydrogen bonds because H is bonded directly to O (highly electronegative). Hydrogen bonding is the strongest type of IMF and requires significant energy to overcome, resulting in a high boiling point (100 °C).
H₂S cannot form hydrogen bonds because S is not electronegative enough (hydrogen bonding only occurs with F, O, N). H₂S has only dipole-dipole forces and London dispersion forces, which are much weaker. Even though H₂S has more electrons and stronger LDF than H₂O, the hydrogen bonding in water far outweighs this difference.
This is a classic example showing that hydrogen bonding dominates over molecular mass effects.
Water (H₂O) can form hydrogen bonds because H is bonded directly to O (highly electronegative). Hydrogen bonding is the strongest type of IMF and requires significant energy to overcome, resulting in a high boiling point (100 °C).
H₂S cannot form hydrogen bonds because S is not electronegative enough (hydrogen bonding only occurs with F, O, N). H₂S has only dipole-dipole forces and London dispersion forces, which are much weaker. Even though H₂S has more electrons and stronger LDF than H₂O, the hydrogen bonding in water far outweighs this difference.
This is a classic example showing that hydrogen bonding dominates over molecular mass effects.
Q13. Draw the Lewis structure of XeF₄. What is the molecular geometry? Why are the lone pairs positioned opposite each other?
Answer:
Xe has 8 valence e⁻ + 4 from F atoms = 36 total e⁻. Four Xe—F bonds use 8 e⁻. Each F needs 6 more e⁻ as lone pairs (24 e⁻). Remaining: 36 − 8 − 24 = 4 e⁻ = 2 lone pairs on Xe.
Total electron domains: 4 bonding + 2 lone = 6 → octahedral electron-pair geometry.
Molecular geometry: square planar.
The two lone pairs are positioned opposite each other (at 180°) because this arrangement minimizes repulsion. In an octahedron, placing lone pairs opposite each other gives each lone pair 4 bonding pairs at 90°. If the lone pairs were adjacent (at 90° to each other), there would be strong lone pair–lone pair repulsion at 90°, which is less favorable. The trans arrangement minimizes the number of 90° lone pair–lone pair interactions (zero instead of one).
Xe has 8 valence e⁻ + 4 from F atoms = 36 total e⁻. Four Xe—F bonds use 8 e⁻. Each F needs 6 more e⁻ as lone pairs (24 e⁻). Remaining: 36 − 8 − 24 = 4 e⁻ = 2 lone pairs on Xe.
Total electron domains: 4 bonding + 2 lone = 6 → octahedral electron-pair geometry.
Molecular geometry: square planar.
The two lone pairs are positioned opposite each other (at 180°) because this arrangement minimizes repulsion. In an octahedron, placing lone pairs opposite each other gives each lone pair 4 bonding pairs at 90°. If the lone pairs were adjacent (at 90° to each other), there would be strong lone pair–lone pair repulsion at 90°, which is less favorable. The trans arrangement minimizes the number of 90° lone pair–lone pair interactions (zero instead of one).
Q14. Compare and contrast Valence Bond Theory and Molecular Orbital Theory. Give one example of something that MOT can explain but VB Theory cannot.
Answer:
Valence Bond Theory: Focuses on individual atoms. Bonds form by overlap of atomic orbitals on two atoms. Electrons are localized between the two bonded atoms. Explains hybridization and molecular geometry well.
Molecular Orbital Theory: Focuses on the entire molecule. Atomic orbitals from all atoms combine to form molecular orbitals that belong to the whole molecule. Electrons are delocalized. Can calculate bond order and predict magnetic properties.
Key example MOT explains but VB cannot: The paramagnetism of O₂. MOT predicts that O₂ has two unpaired electrons in π*2p orbitals, making it paramagnetic (attracted to a magnet). VB theory predicts all electrons are paired in O₂, which would make it diamagnetic — contradicting experimental evidence.
Valence Bond Theory: Focuses on individual atoms. Bonds form by overlap of atomic orbitals on two atoms. Electrons are localized between the two bonded atoms. Explains hybridization and molecular geometry well.
Molecular Orbital Theory: Focuses on the entire molecule. Atomic orbitals from all atoms combine to form molecular orbitals that belong to the whole molecule. Electrons are delocalized. Can calculate bond order and predict magnetic properties.
Key example MOT explains but VB cannot: The paramagnetism of O₂. MOT predicts that O₂ has two unpaired electrons in π*2p orbitals, making it paramagnetic (attracted to a magnet). VB theory predicts all electrons are paired in O₂, which would make it diamagnetic — contradicting experimental evidence.
Q15. Explain why CO₂ is nonpolar but SO₂ is polar, even though both contain polar bonds.
Answer:
Both molecules have polar bonds (C=O and S=O) because of electronegativity differences. However, molecular polarity depends on BOTH bond polarity AND molecular shape.
CO₂: Linear geometry (180°). The two C=O bond dipoles are equal in magnitude but point in exactly opposite directions. They cancel each other out. Net dipole moment = 0 → nonpolar.
SO₂: Bent geometry (~119°). The two S=O bond dipoles do not cancel because they are not pointing in opposite directions. There is a net dipole moment → polar molecule.
This illustrates the critical role of molecular geometry in determining overall polarity.
Both molecules have polar bonds (C=O and S=O) because of electronegativity differences. However, molecular polarity depends on BOTH bond polarity AND molecular shape.
CO₂: Linear geometry (180°). The two C=O bond dipoles are equal in magnitude but point in exactly opposite directions. They cancel each other out. Net dipole moment = 0 → nonpolar.
SO₂: Bent geometry (~119°). The two S=O bond dipoles do not cancel because they are not pointing in opposite directions. There is a net dipole moment → polar molecule.
This illustrates the critical role of molecular geometry in determining overall polarity.
Section D: Calculation and Extended Questions
Q16. Draw the Lewis structure of the sulfate ion (SO₄²⁻). Calculate the formal charge on each atom. Is there a better structure with lower formal charges? If so, draw it.
Answer:
Total valence e⁻: S = 6, 4O = 24, 2− charge = 2 → Total = 32 e⁻.
Structure 1 (all single bonds): S connected to 4 O atoms with single bonds. Each O has 3 lone pairs. S has 0 lone pairs. Wait — that uses 8 + 24 = 32 e⁻, but S only has 8 e⁻ around it (4 bonds). Let me check: S gets 8 e⁻ from bonds but has no lone pair. FC on S = 6 − 0 − 4 = +2. Each O: FC = 6 − 6 − 1 = −1. Total FC = +2 + 4(−1) = −2 (matches ion charge).
Structure 2 (two double bonds): Two S=O double bonds and two S—O single bonds (with negative charges). S: FC = 6 − 0 − ½(10) = +1. Each double-bonded O: FC = 6 − 4 − 2 = 0. Each single-bonded O: FC = 6 − 6 − 1 = −1. Total FC = +1 + 2(0) + 2(−1) = −1. This doesn’t match. Let me use two S=O and two S—O⁻: FC on S = 6 − 0 − ½(12) = 0. Each S=O oxygen: FC = 0. Each S—O⁻ oxygen: FC = −1. Total = 0 + 0 + 0 + (−1) + (−1) = −2. ✓
Structure 2 has formal charges of 0, 0, −1, −1 which is much better than +2, −1, −1, −1, −1. The resonance hybrid has two equivalent S=O bonds and two equivalent S—O⁻ bonds (four resonance structures).
Total valence e⁻: S = 6, 4O = 24, 2− charge = 2 → Total = 32 e⁻.
Structure 1 (all single bonds): S connected to 4 O atoms with single bonds. Each O has 3 lone pairs. S has 0 lone pairs. Wait — that uses 8 + 24 = 32 e⁻, but S only has 8 e⁻ around it (4 bonds). Let me check: S gets 8 e⁻ from bonds but has no lone pair. FC on S = 6 − 0 − 4 = +2. Each O: FC = 6 − 6 − 1 = −1. Total FC = +2 + 4(−1) = −2 (matches ion charge).
Structure 2 (two double bonds): Two S=O double bonds and two S—O single bonds (with negative charges). S: FC = 6 − 0 − ½(10) = +1. Each double-bonded O: FC = 6 − 4 − 2 = 0. Each single-bonded O: FC = 6 − 6 − 1 = −1. Total FC = +1 + 2(0) + 2(−1) = −1. This doesn’t match. Let me use two S=O and two S—O⁻: FC on S = 6 − 0 − ½(12) = 0. Each S=O oxygen: FC = 0. Each S—O⁻ oxygen: FC = −1. Total = 0 + 0 + 0 + (−1) + (−1) = −2. ✓
Structure 2 has formal charges of 0, 0, −1, −1 which is much better than +2, −1, −1, −1, −1. The resonance hybrid has two equivalent S=O bonds and two equivalent S—O⁻ bonds (four resonance structures).
Q17. For the molecule PCl₅:
(a) Determine the hybridization of P.
(b) Predict the electron-pair geometry and molecular geometry.
(c) Identify the bond angles.
(d) Are all P—Cl bonds equivalent? Explain.
(a) Determine the hybridization of P.
(b) Predict the electron-pair geometry and molecular geometry.
(c) Identify the bond angles.
(d) Are all P—Cl bonds equivalent? Explain.
Answer:
(a) P has 5 bonding pairs and 0 lone pairs = 5 electron domains → sp³d hybridization.
(b) Electron-pair geometry: trigonal bipyramidal. Molecular geometry: also trigonal bipyramidal (no lone pairs).
(c) Two types of positions in trigonal bipyramid:
• Axial positions: 3 Cl atoms at 90° to equatorial, 180° to each other.
• Equatorial positions: 3 Cl atoms at 120° to each other, 90° to axial.
Angles: 90° and 120°.
(d) No, not all P—Cl bonds are equivalent. The axial bonds are longer and weaker than the equatorial bonds. This is because axial Cl atoms experience repulsion from 3 equatorial Cl atoms at 90° (total 3 × 90° repulsions), while equatorial Cl atoms experience repulsion from only 2 axial Cl atoms at 90° (total 2 × 90° repulsions). The greater repulsion on axial bonds makes them longer.
(a) P has 5 bonding pairs and 0 lone pairs = 5 electron domains → sp³d hybridization.
(b) Electron-pair geometry: trigonal bipyramidal. Molecular geometry: also trigonal bipyramidal (no lone pairs).
(c) Two types of positions in trigonal bipyramid:
• Axial positions: 3 Cl atoms at 90° to equatorial, 180° to each other.
• Equatorial positions: 3 Cl atoms at 120° to each other, 90° to axial.
Angles: 90° and 120°.
(d) No, not all P—Cl bonds are equivalent. The axial bonds are longer and weaker than the equatorial bonds. This is because axial Cl atoms experience repulsion from 3 equatorial Cl atoms at 90° (total 3 × 90° repulsions), while equatorial Cl atoms experience repulsion from only 2 axial Cl atoms at 90° (total 2 × 90° repulsions). The greater repulsion on axial bonds makes them longer.
Q18. Use Molecular Orbital Theory to determine the bond order and magnetic property of each of the following. Arrange them in order of increasing bond length:
(a) N₂ (b) N₂⁺ (c) N₂⁻
(a) N₂ (b) N₂⁺ (c) N₂⁻
Answer:
N₂ has 10 valence e⁻. Using the ordering for N₂ (π2p < σ2p):
Configuration: σ2s² σ*2s² π2p⁴ σ2p²
Nb = 2 + 4 + 2 = 8; Na = 2
(a) N₂: BO = ½(8 − 2) = 3. All electrons paired → diamagnetic.
(b) N₂⁺ (9 e⁻): Remove 1 from σ2p. Nb = 7; Na = 2. BO = ½(7 − 2) = 2.5. 1 unpaired e⁻ → paramagnetic.
(c) N₂⁻ (11 e⁻): Add 1 to π*2p. Nb = 8; Na = 3. BO = ½(8 − 3) = 2.5. 1 unpaired e⁻ → paramagnetic.
Increasing bond length (decreasing bond order):
N₂ (BO=3) < N₂⁺ (BO=2.5) = N₂⁻ (BO=2.5)
N₂⁺ and N₂⁻ have the same bond order (2.5), so they have approximately equal bond lengths (both longer than N₂).
N₂ has 10 valence e⁻. Using the ordering for N₂ (π2p < σ2p):
Configuration: σ2s² σ*2s² π2p⁴ σ2p²
Nb = 2 + 4 + 2 = 8; Na = 2
(a) N₂: BO = ½(8 − 2) = 3. All electrons paired → diamagnetic.
(b) N₂⁺ (9 e⁻): Remove 1 from σ2p. Nb = 7; Na = 2. BO = ½(7 − 2) = 2.5. 1 unpaired e⁻ → paramagnetic.
(c) N₂⁻ (11 e⁻): Add 1 to π*2p. Nb = 8; Na = 3. BO = ½(8 − 3) = 2.5. 1 unpaired e⁻ → paramagnetic.
Increasing bond length (decreasing bond order):
N₂ (BO=3) < N₂⁺ (BO=2.5) = N₂⁻ (BO=2.5)
N₂⁺ and N₂⁻ have the same bond order (2.5), so they have approximately equal bond lengths (both longer than N₂).
Q19. For each of the following molecules, determine: (i) Lewis structure, (ii) electron domains on central atom, (iii) hybridization, (iv) molecular geometry, (v) bond polarity of each bond, (vi) overall molecular polarity.
(a) CH₃Cl (b) CCl₄ (c) NF₃
(a) CH₃Cl (b) CCl₄ (c) NF₃
Answer:
(a) CH₃Cl:
(i) C central, bonded to 3H and 1Cl. C has 4 bonds, 0 lone pairs. Total valence e⁻ = 4 + 3 + 7 = 14.
(ii) 4 electron domains.
(iii) sp³ hybridization.
(iv) Tetrahedral molecular geometry.
(v) C—H: nearly nonpolar (ΔEN ≈ 0.4). C—Cl: polar (ΔEN = 0.5).
(vi) Polar — the C—Cl dipole does not cancel with the C—H dipoles (different atoms, different magnitudes).
(b) CCl₄:
(i) C central, bonded to 4Cl. 0 lone pairs on C.
(ii) 4 electron domains.
(iii) sp³ hybridization.
(iv) Tetrahedral molecular geometry.
(v) Each C—Cl bond is polar (ΔEN = 0.5).
(vi) Nonpolar — the four identical C—Cl bond dipoles point symmetrically and cancel out perfectly in the tetrahedral geometry.
(c) NF₃:
(i) N central, bonded to 3F, 1 lone pair. Total valence e⁻ = 5 + 21 = 26.
(ii) 4 electron domains (3 bonding + 1 lone pair).
(iii) sp³ hybridization.
(iv) Trigonal pyramidal molecular geometry.
(v) Each N—F bond is polar (ΔEN = 1.0), pointing from N(δ+) to F(δ−).
(vi) Polar — the three N—F bond dipoles do not fully cancel in the pyramidal shape (the lone pair also contributes to asymmetry). Net dipole points away from the lone pair.
(a) CH₃Cl:
(i) C central, bonded to 3H and 1Cl. C has 4 bonds, 0 lone pairs. Total valence e⁻ = 4 + 3 + 7 = 14.
(ii) 4 electron domains.
(iii) sp³ hybridization.
(iv) Tetrahedral molecular geometry.
(v) C—H: nearly nonpolar (ΔEN ≈ 0.4). C—Cl: polar (ΔEN = 0.5).
(vi) Polar — the C—Cl dipole does not cancel with the C—H dipoles (different atoms, different magnitudes).
(b) CCl₄:
(i) C central, bonded to 4Cl. 0 lone pairs on C.
(ii) 4 electron domains.
(iii) sp³ hybridization.
(iv) Tetrahedral molecular geometry.
(v) Each C—Cl bond is polar (ΔEN = 0.5).
(vi) Nonpolar — the four identical C—Cl bond dipoles point symmetrically and cancel out perfectly in the tetrahedral geometry.
(c) NF₃:
(i) N central, bonded to 3F, 1 lone pair. Total valence e⁻ = 5 + 21 = 26.
(ii) 4 electron domains (3 bonding + 1 lone pair).
(iii) sp³ hybridization.
(iv) Trigonal pyramidal molecular geometry.
(v) Each N—F bond is polar (ΔEN = 1.0), pointing from N(δ+) to F(δ−).
(vi) Polar — the three N—F bond dipoles do not fully cancel in the pyramidal shape (the lone pair also contributes to asymmetry). Net dipole points away from the lone pair.
Q20. Explain the following observations using concepts from this unit:
(a) MgO has a much higher melting point than NaCl.
(b) Graphite is a good conductor of electricity but diamond is not.
(c) The boiling point increases in the order: CH₄ < SiH₄ < GeH₄ < SnH₄.
(a) MgO has a much higher melting point than NaCl.
(b) Graphite is a good conductor of electricity but diamond is not.
(c) The boiling point increases in the order: CH₄ < SiH₄ < GeH₄ < SnH₄.
Answer:
(a) Both are ionic crystals, but Mg²⁺ and O²⁻ have charges of +2 and −2, while Na⁺ and Cl⁻ have charges of +1 and −1. The lattice energy is proportional to the product of the charges: MgO (2 × 2 = 4) has roughly four times the lattice energy of NaCl (1 × 1 = 1). Also, Mg²⁺ and O²⁻ are smaller ions, giving a shorter distance and further increasing lattice energy. Higher lattice energy → higher melting point.
(b) Both diamond and graphite are covalent network solids of carbon, but they differ in structure. In diamond, each C is sp³ hybridized with 4 single bonds in a 3D network — all electrons are localized in σ bonds, leaving no mobile electrons for conduction. In graphite, each C is sp² hybridized with 3 σ bonds in flat hexagonal layers, and one unhybridized p electron per carbon forms a delocalized π system across the layers. These delocalized π electrons are free to move and carry electric current, making graphite a conductor along the layers.
(c) All four are nonpolar molecules, so the only IMF is London dispersion force. LDF increases with the number of electrons (molecular mass): CH₄ (16 g/mol) < SiH₄ (32) < GeH₄ (77) < SnH₄ (123). Larger electron clouds are more polarizable, leading to stronger LDF and higher boiling points.
(a) Both are ionic crystals, but Mg²⁺ and O²⁻ have charges of +2 and −2, while Na⁺ and Cl⁻ have charges of +1 and −1. The lattice energy is proportional to the product of the charges: MgO (2 × 2 = 4) has roughly four times the lattice energy of NaCl (1 × 1 = 1). Also, Mg²⁺ and O²⁻ are smaller ions, giving a shorter distance and further increasing lattice energy. Higher lattice energy → higher melting point.
(b) Both diamond and graphite are covalent network solids of carbon, but they differ in structure. In diamond, each C is sp³ hybridized with 4 single bonds in a 3D network — all electrons are localized in σ bonds, leaving no mobile electrons for conduction. In graphite, each C is sp² hybridized with 3 σ bonds in flat hexagonal layers, and one unhybridized p electron per carbon forms a delocalized π system across the layers. These delocalized π electrons are free to move and carry electric current, making graphite a conductor along the layers.
(c) All four are nonpolar molecules, so the only IMF is London dispersion force. LDF increases with the number of electrons (molecular mass): CH₄ (16 g/mol) < SiH₄ (32) < GeH₄ (77) < SnH₄ (123). Larger electron clouds are more polarizable, leading to stronger LDF and higher boiling points.
Q21. Construct the MO diagram for the C₂ molecule. Determine the bond order and predict whether C₂ is paramagnetic or diamagnetic.
Answer:
C₂ has 8 valence electrons (4 + 4). Since C₂ has fewer than 15 valence electrons, we use the ordering where π2p < σ2p.
Filling: σ2s² (2) → σ*2s² (2) → π2p² (2) → π2p² (2)
Total: 8 electrons filled.
Nb = 2 (σ2s) + 4 (π2p) = 6
Na = 2 (σ*2s) = 2
Bond order = ½(6 − 2) = 2 (double bond character)
All electrons are paired → diamagnetic.
Note: With the N₂ ordering, the π2p orbitals fill before σ2p, giving a bond order of 2. If we had used the O₂ ordering (σ2p before π2p), we would still get BO = 2, but the configuration would differ. The N₂ ordering is correct for C₂ since it has fewer than 15 valence electrons.
C₂ has 8 valence electrons (4 + 4). Since C₂ has fewer than 15 valence electrons, we use the ordering where π2p < σ2p.
Filling: σ2s² (2) → σ*2s² (2) → π2p² (2) → π2p² (2)
Total: 8 electrons filled.
Nb = 2 (σ2s) + 4 (π2p) = 6
Na = 2 (σ*2s) = 2
Bond order = ½(6 − 2) = 2 (double bond character)
All electrons are paired → diamagnetic.
Note: With the N₂ ordering, the π2p orbitals fill before σ2p, giving a bond order of 2. If we had used the O₂ ordering (σ2p before π2p), we would still get BO = 2, but the configuration would differ. The N₂ ordering is correct for C₂ since it has fewer than 15 valence electrons.
Q22. Describe the complete bonding picture of the carbonate ion (CO₃²⁻) including: Lewis structure, resonance, hybridization of C, and identification of all σ and π bonds.
Answer:
Lewis structure: Total valence e⁻ = 4 + 18 + 2 = 24. Central C bonded to 3 O atoms. One C=O double bond, two C—O single bonds (each with 3 lone pairs, and the single-bonded O atoms carry a negative charge).
Resonance: Three equivalent resonance structures — the double bond can be with any of the three O atoms. The real structure is a resonance hybrid where all three C—O bonds are identical (intermediate between single and double, each with a bond order of 1⅓).
Hybridization of C: C has 3 bonding domains (each C—O counts as 1 domain regardless of bond order) + 0 lone pairs = 3 electron domains → sp² hybridization.
Bonds: The three sp² hybrid orbitals on C form three σ bonds (one with each O atom). The unhybridized p orbital on C overlaps with a p orbital on one O to form a π bond. Since the π bond is delocalized across all three C—O bonds (resonance), each C—O bond has: 1 σ bond + ⅓ π bond. Total: 3 σ bonds and 1 delocalized π bond.
Lewis structure: Total valence e⁻ = 4 + 18 + 2 = 24. Central C bonded to 3 O atoms. One C=O double bond, two C—O single bonds (each with 3 lone pairs, and the single-bonded O atoms carry a negative charge).
Resonance: Three equivalent resonance structures — the double bond can be with any of the three O atoms. The real structure is a resonance hybrid where all three C—O bonds are identical (intermediate between single and double, each with a bond order of 1⅓).
Hybridization of C: C has 3 bonding domains (each C—O counts as 1 domain regardless of bond order) + 0 lone pairs = 3 electron domains → sp² hybridization.
Bonds: The three sp² hybrid orbitals on C form three σ bonds (one with each O atom). The unhybridized p orbital on C overlaps with a p orbital on one O to form a π bond. Since the π bond is delocalized across all three C—O bonds (resonance), each C—O bond has: 1 σ bond + ⅓ π bond. Total: 3 σ bonds and 1 delocalized π bond.