**Definition**: The Kössel-Lewis approach explains chemical bonding in terms of valence electrons and the tendency of atoms to achieve stable noble gas configurations.
**Historical Context**:
**Lewis Model of Atomic Structure**:
**Kössel's Key Observations**:
**Example: Formation of NaCl**
**Example: Formation of CaF₂**
**Exam Important Points**:
---
**Definition**: Simple notations introduced by G.N. Lewis to represent valence electrons of atoms.
**Representation Method**:
**Lewis Symbols for Second Period Elements**:
**Significance**:
**Example**: Oxygen has 6 valence electrons, so group valence = 6 or (8-6) = 2. Oxygen typically shows valence of 2.
---
**Statement**: Atoms combine (either by transferring or sharing valence electrons) to achieve an octet (8 electrons) in their valence shell, thereby attaining the stable configuration of the nearest noble gas.
**Key Postulates**:
**Mechanisms**:
1. **Electron Transfer**: Formation of ionic bonds (discussed separately)
2. **Electron Sharing**: Formation of covalent bonds (discussed separately)
**Why Octet Rule Works**:
**Exam Important Points**:
---
**Definition**: A covalent bond is formed when two atoms share a pair of electrons. Both atoms contribute equally to the shared pair.
**Historical Development**:
**Formation Example: Cl₂ Molecule**
**Requirements for Covalent Bond Formation**:
**Example: Water (H₂O) Formation**
**Example: Carbon Tetrachloride (CCl₄)**
**Types of Covalent Bonds Based on Electron Pairs Shared**:
1. **Single Covalent Bond**: Sharing of one electron pair (one dash)
2. **Double Covalent Bond**: Sharing of two electron pairs (two dashes)
3. **Triple Covalent Bond**: Sharing of three electron pairs (three dashes)
**Exam Important Points**:
---
**Purpose**: Pictorial representation of bonding in molecules and ions using shared electron pairs and showing how atoms achieve octet rule.
**Step-by-Step Method for Writing Lewis Structures**:
**Step 1: Calculate Total Valence Electrons**
**Example for CH₄**:
**Example for CO₃²⁻**:
**Step 2: Determine Skeletal Structure**
**Step 3: Distribute Electrons for Single Bonds**
**Step 4: Complete Octets of Terminal Atoms**
**Step 5: Handle Remaining Electrons on Central Atom**
**Step 6: Form Multiple Bonds if Necessary**
**Worked Example: CO Molecule**
Step 1: Valence electrons = 4 (C) + 6 (O) = 10 electrons
Step 2: Skeletal structure: C-O
Step 3: Single bond accounts for 2 electrons; remaining = 8 electrons
Step 4: Complete O's octet: O needs 6 more electrons (3 lone pairs)
```
C : O:• (with 3 lone pairs on O)
• •
```
Step 5: Central C atom now has only 2 electrons; needs 6 more for octet
Step 6: Form triple bond between C and O (use O's lone pairs)
```
C ≡ O: (or C≡O with one lone pair on C)
```
**Worked Example: NO₂⁻ (Nitrite Ion)**
Step 1: Valence electrons = 5 (N) + 6 (O) × 2 + 1 (charge) = 18 electrons
Step 2: Skeletal structure: O-N-O
Step 3: Two single bonds account for 4 electrons; remaining = 14 electrons
Step 4: Complete O atoms' octets: each needs 3 lone pairs = 6 electrons per O = 12 electrons used; 2 remain
Step 5: N atom still needs electrons for octet; has only 2 electrons
Step 6: Convert one N-O single bond to double bond using one lone pair from that O
Result: Two resonance structures exist (shown with formal charges)
**Common Lewis Structures Table**:
| Molecule | Total e⁻ | Structure | Notes |
|----------|----------|-----------|-------|
| CH₄ | 8 | C surrounded by 4 H | All single bonds |
| C₂H₄ | 12 | C=C with 2 H on each | One double bond |
| C₂H₂ | 10 | C≡C with 1 H on each | One triple bond |
| H₂O | 8 | O with 2 H and 2 lone pairs | Two single bonds |
| NH₃ | 8 | N with 3 H and 1 lone pair | Three single bonds |
| CO₂ | 16 | O=C=O linear | Two double bonds |
| N₂ | 10 | N≡N | Triple bond, very stable |
**Exam Important Points**:
---
**Definition**: The formal charge on an atom in a Lewis structure is the difference between valence electrons in the free atom and electrons assigned to that atom in the structure.
**Formula**:
```
Formal Charge (F.C.) = V - N - (B/2)
```
Where:
**Calculation Logic**:
**Example 1: Ozone (O₃) Molecule**
Lewis structure shows: O₁-O₂=O₃ with lone pairs
For central O atom (O₁):
For terminal O atom (O₂):
For terminal O atom (O₃):
**Representation**: O₃ shown with +1 on one O and -1 on other terminal O
**Example 2: Nitrite Ion (NO₂⁻)**
For N atom:
For doubly-bonded O:
For singly-bonded O:
**Rules for Using Formal Charges**:
1. **Sum of formal charges**: Must equal overall charge of molecule/ion
2. **Selecting most reasonable Lewis structure**:
3. **Important clarification**:
**Exam Important Points**:
---
The octet rule, while useful, is NOT universal. Three major exceptions exist:
**Condition**: Some compounds have fewer than 8 electrons around central atom, especially with elements having less than 4 valence electrons.
**Examples**:
1. **Lithium Chloride (LiCl)**
2. **Beryllium Hydride (BeH₂)**
3. **Boron Trifluoride (BCl₃)**
4. **Aluminum Chloride (AlCl₃)**
**Why Octet Rule Fails**:
**Condition**: Molecules with odd total number of valence electrons cannot satisfy octet rule for all atoms (by definition, some atom must have unpaired electron).
**Examples**:
1. **Nitric Oxide (NO)**
2. **Nitrogen Dioxide (NO₂)**
3. **Chlorine Dioxide (ClO₂)**
**Characteristics**:
**Condition**: Elements from period 3 onwards (third period and beyond) can have more than 8 valence electrons due to availability of d orbitals.
**Why This Occurs**:
**Examples**:
1. **Phosphorus Pentafluoride (PF₅)**
2. **Sulfur Hexafluoride (SF₆)**
3. **Sulfuric Acid (H₂SO₄)**
4. **Coordination Compounds**
5. **Xenon Compounds**
**Mechanism**:
**Important Note about Sulfur**:
---
Beyond the three exceptions, the octet rule has fundamental limitations:
**Summary**: The octet rule is useful starting point for predicting bonding patterns in organic and main group chemistry but must be supplemented with other theories for complete understanding of molecular structure and properties.
---
**Definition**: An ionic (electrovalent) bond is formed through electrostatic attraction between positively charged ions (cations) and negatively charged ions (anions).
**Formation Mechanism**:
Based on Kössel and Lewis treatment, ionic bond formation requires:
1. **Ionization Energy (Removal of Electrons)**
```
M(g) → M⁺(g) + e⁻ ΔH = IE (endothermic)
```
Where M is metal atom, IE is ionization enthalpy
Example: Na(g) → Na⁺(g) + e⁻ ; IE = 495.8 kJ/mol
2. **Electron Gain Process (Addition of Electrons)**
```
X(g) + e⁻ → X⁻(g) ΔH = -Δₑₘ (exothermic)
```
Where X is non-metal atom, Δₑₘ is electron gain enthalpy
Example: Cl(g) + e⁻ → Cl⁻(g) ; Δ_egH = -348.7 kJ/mol
3. **Ionic Bond Formation (Electrostatic Attraction)**
```
M⁺(g) + X⁻(g) → MX(s) ΔH = -Lattice Energy
```
**Overall Process for NaCl**:
```
Na(g) + Cl(g) → NaCl(s)
```
Broken into steps:
**Conditions for Ionic Bond Formation**:
Ionic compounds form preferentially between:
**Important Note on Electron Gain Enthalpy (Δ_egH)**:
**Typical Ionic Compound Characteristics**:
1. **Composition**:
2. **Structure**:
3. **Rock Salt (NaCl) Structure**:
4. **Physical Properties**:
**Lattice Energy Definition**: Energy required to completely separate one mole of an ionic solid into gaseous ions.
**Mathematical Relationship**:
```
ΔH_formation = ΔH_IE + ΔH_EG + ΔH_lattice
ΔH_lattice is always negative (energy released)
```
**Critical Point**: Even when ionization enthalpy + electron gain enthalpy = positive (energetically unfavorable), ionic compound still forms spontaneously because lattice energy is large and negative.
**Example: NaCl Formation Analysis**
**Why Lattice Energy Dominates**:
**Electrovalence**:
**Not the same as oxidation state** (though usually identical for simple ionic compounds)
---
**Polar Ionic Bonds**:
**Fajan's Rules** (determine extent of covalent character):
1. **Small cation size**: High charge density → more covalent character
2. **Large cation charge**: Higher charge density → more covalent character
3. **Large anion size**: More easily polarized → more covalent character
4. **Electron configuration of cation**: d¹⁰ or d¹⁸ configurations show more covalent character
**Practical Impact**:
---
| Property | Ionic Bond | Covalent Bond |
|----------|-----------|--------------|
| **Formation** | Transfer of electrons | Sharing of electrons |
| **Electron distribution** | Complete transfer | Equal or unequal sharing |
| **Electronegativity difference** | Large (≥2.0) | Small (<2.0) |
| **Occurrence** | Metal + Non-metal | Non-metal + Non-metal |
| **State of matter** | Usually solid at RT | Liquid, gas, or solid |
| **Solubility** | Often soluble in polar solvents | Often soluble in non-polar solvents
Q1. The Lewis symbol of phosphorus (P) has five dots around it. What is the group valence of phosphorus?
Answer: C — Group valence equals either the number of dots (5) or 8 minus the number of dots (8−5=3), so both 5 and 3 are valid valences for phosphorus.
Q2. In the formation of CaF₂, calcium achieves a stable configuration by losing electrons and fluorine by gaining electrons. Which statement is correct?
Answer: B — Ca ([Ar]4s²) loses 2 electrons to form Ca²⁺ ([Ar]); each F ([He]2s²2p⁵) gains 1 electron to form F⁻ ([Ne]), so CaF₂ forms with one Ca²⁺ and two F⁻ ions.
Q3. According to the Kössel-Lewis approach, which of the following is NOT a reason why atoms form chemical bonds?
Answer: C — Increasing the number of orbitals is not a bonding motive; atoms bond to gain stability, lower energy, and achieve noble gas configurations.
Q4. Two atoms X and Y form a molecule X₂Y. X has 6 valence electrons and Y has 4 valence electrons. How many shared electron pairs are likely formed between X and each X-Y bond, assuming both atoms achieve octet or appropriate stable configuration?
Answer: C — For O₂S (analogy: X=O with 6 valence e⁻, Y=S with 4 valence e⁻), O atoms form a double bond (2 shared pairs) with S to both achieve octet, and the O-O bond would be single (1 pair) to satisfy the octet rule.
Q5. Which of the following compounds violates the octet rule?
Answer: C — In PCl₅, phosphorus (group 15) has 5 valence electrons and forms 5 P-Cl bonds, resulting in 10 electrons around P, violating the octet rule; NaCl, H₂O, and CO₂ all obey the octet rule.
Q6. Which statement best explains why the Kössel-Lewis approach successfully explains ionic compound formation but has limitations for covalent compounds?
Answer: B — The Kössel-Lewis model assumes electrons fully transfer (ionic) or achieve a fixed octet by sharing, but covalent bonding is dynamic and overlapping; some atoms achieve less than 8 electrons stably (e.g., BF₃ with 6 around B).
Q7. An element M in group 13 forms a compound with chlorine. Based on Lewis approach, which statement is correct?
Answer: B — Group 13 elements (like Al) have 3 valence electrons and lose all 3 to form covalent compounds (not purely ionic); MCl₃ forms with M having an oxidation state of +3 (electrovalence +3).
Q8. Assertion: The octet rule explains the formation of both ionic and covalent compounds. Reason: Both ionic and covalent bonding aim to achieve noble gas electronic configurations.
Answer: A — The octet rule does explain both bonding types: ionic compounds achieve it through complete electron transfer, and covalent compounds achieve it through electron sharing; the underlying reason in both cases is attaining stable noble gas configurations.
Q9. Calculate the electrovalence of S in the compound H₂S, assuming sulfur achieves an octet by gaining electrons.
Answer: B — In H₂S, sulfur ([Ne]3s²3p⁴) gains 2 electrons to form S²⁻ and achieve [Ar]; the electrovalence is −2 (negative because it gains electrons).
Q10. A compound XY₂ is formed where X is from group 16 and Y is from group 1. If both atoms follow the octet/duplet rule, how many total valence electrons are present in one molecule of XY₂?
Answer: B — X (group 16) has 6 valence electrons and needs 2 more for octet = 8; each Y (group 1) has 1 valence electron and achieves duplet with 2. Total = 6 (from X) + 1 (from first Y) + 1 (from second Y) = 8 in octet count, but counting all valence electrons: 6 + 1 + 1 = 8 for bonding; however, the compound has 6 (X) + 2 + 2 (Y₂) = 10 total valence electrons in the molecule.
What is a chemical bond?
The attractive force holding atoms, ions, or other constituents together in a chemical species.
Define the octet rule.
Atoms combine by transferring or sharing valence electrons to achieve eight electrons in their outer shell, like noble gases.
What are Lewis symbols and what do they show?
Simple notations representing an element's symbol with dots around it; the number of dots equals the number of valence electrons.
What is the relationship between group number and Lewis symbol dots for main-group elements?
The number of dots equals the group number (e.g., Group 17 halogens have 7 dots).
What is electrovalence?
The number of unit charges (positive or negative) on an ion formed by electron transfer.
How do Na and Cl achieve stable configurations in NaCl formation?
Na transfers one electron to Cl; Na becomes Na⁺ ([Ne]) and Cl becomes Cl⁻ ([Ar]).
What is the key difference between ionic and covalent bonding according to Lewis?
Ionic bonding involves transfer of electrons between atoms; covalent bonding involves sharing of electron pairs.
Name one limitation of the octet rule.
Compounds like PCl₅ and BF₃ have central atoms with fewer than eight valence electrons, violating the octet rule.
Who refined Lewis' theory in 1919 and introduced what key term?
Langmuir refined Lewis' theory and introduced the term 'covalent bond' to describe shared electron pairs.
How does Cl₂ formation demonstrate the covalent bond concept?
Each Cl atom ([Ne]3s²3p⁵) contributes one electron to form a shared pair, allowing both atoms to achieve the [Ar] octet.
Define chemical bonding. Why do atoms combine to form molecules? [2 marks]
Define as an attractive force between constituents. State that every system tends toward maximum stability by lowering energy, and bonding is nature's mechanism to achieve this stable state.
Explain the formation of NaCl using the Kössel-Lewis approach. Show how both sodium and chlorine achieve stable configurations and calculate their electrovalences. [5 marks]
Write electronic configurations: Na [Ne]3s¹ loses 1 e⁻ to form Na⁺[Ne]; Cl [Ne]3s²3p⁵ gains 1 e⁻ to form Cl⁻[Ar]. State electrovalence of Na is +1 (charge magnitude) and Cl is −1. Explain that electrostatic attraction stabilizes the ionic compound NaCl. Show both achieve noble gas octets.
State the octet rule and explain its limitations with specific examples. Why does the Kössel-Lewis model fail for some covalent compounds, and what does this reveal about chemical bonding? [6 marks]
Octet rule: atoms achieve 8 electrons in outer shell (2 for He/H). Limitations: PCl₅ (P has 10 e⁻), BF₃ (B has 6 e⁻), SF₆ (S has 12 e⁻), etc. Explain that the static octet model assumes bonding electrons arrange in a fixed cubical pattern (Lewis' model), but real covalent bonding involves dynamic electron sharing and molecular orbital overlap. This shows bonding is more complex than simple electron transfer/sharing for octet; it involves orbital interactions and variable electron densities.
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