Q: Q16. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?\nBF3 + NH3 → F3B.NH3

Yes, there is a change in the hybridisation of both Boron (B) and Nitrogen (N) atoms as a result of the reaction.

Q: Q20. Use molecular orbital theory to explain why the Be₂ molecule does not exist.

According to Molecular Orbital Theory (MOT), the existence and stability of a molecule are determined by its bond order. A molecule is stable if its bond order is positive, and it does not exist if its bond order is zero or negative.\n\n### Explanation for Non-existence of Be₂ Molecule:\n\n1. **Electronic Configuration of Beryllium (Be):**\n * The atomic number of Beryllium (Be) is 4. Its electronic configuration is 1s² 2s².\n\n2. **Total Electrons in Be₂ Molecule:**\n * A Be₂ molecule consists of two Beryllium atoms. Therefore, the total number of electrons in the Be₂ molecule is 4 (from first Be) + 4 (from second Be) = 8 electrons.\n\n3. **Molecular Orbital (MO) Configuration of Be₂:**\n * The 8 electrons in Be₂ are filled into the molecular orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. The order of filling MOs for elements with Z ≤ 7 (like Be) is typically: σ1s, σ*1s, σ2s, σ*2s, (π2px = π2py), σ2pz, (π*2px = π*2py), σ*2pz.\n * For Be₂, the MO configuration will be:\n (σ1s)² (σ*1s)² (σ2s)² (σ*2s)²\n\n4. **Calculation of Bond Order:**\n * Bond Order (B.O.) is calculated using the formula: B.O. = ½ (N_b - N_a)\n Where, N_b = number of electrons in bonding molecular orbitals\n N_a = number of electrons in antibonding molecular orbitals\n * From the MO configuration of Be₂:\n * N_b (electrons in σ1s and σ2s) = 2 + 2 = 4\n * N_a (electrons in σ*1s and σ*2s) = 2 + 2 = 4\n * Therefore, B.O. = ½ (4 - 4) = ½ (0) = 0\n\n5. **Conclusion:**\n * Since the bond order for the Be₂ molecule is zero, it indicates that there is no net attractive force between the two Beryllium atoms. The number of electrons in bonding orbitals is exactly cancelled out by the number of electrons in antibonding orbitals. Consequently, the Be₂ molecule is unstable and does not exist under normal conditions.

Question 1

Q1. What is the significance of a chemical bond?

Accepted Answer

A chemical bond is a lasting attraction between atoms, ions, or molecules that enables the formation of chemical compounds. The significance of a chemical bond lies in several fundamental aspects of chemistry and the stability of matter:

### 1. Attainment of Stability
Atoms form chemical bonds primarily to achieve a more stable electronic configuration, typically resembling that of the nearest noble gas. This usually involves completing their valence shell, often by achieving an octet (eight electrons) or a duplet (two electrons for elements like H, Li, Be). By forming bonds, atoms lower their potential energy and become more stable.

### 2. Formation of Molecules and Compounds
Chemical bonds are the fundamental forces that hold atoms together to form molecules (e.g., H₂O, CO₂) and ionic compounds (e.g., NaCl, MgO). Without chemical bonds, matter would exist only as individual atoms, and the vast diversity of substances we observe would not be possible.

### 3. Energy Changes
Bond formation is generally an exothermic process, meaning energy is released when a bond is formed, leading to a more stable state. Conversely, energy must be supplied (endothermic process) to break a chemical bond. This energy change is crucial for understanding chemical reactions, thermodynamics, and the energy content of substances.

### 4. Determination of Properties
The type of chemical bond (ionic, covalent, metallic, hydrogen bond, van der Waals forces) dictates many physical and chemical properties of a substance, including:
    *   **Melting and Boiling Points:** Stronger bonds (e.g., ionic, metallic) lead to higher melting/boiling points.
    *   **Solubility:** Polar molecules tend to dissolve in polar solvents, and non-polar in non-polar solvents.
    *   **Electrical Conductivity:** Ionic compounds conduct electricity in molten or aqueous states, metals conduct in solid/molten states, while most covalent compounds are insulators.
    *   **Hardness and Brittleness:** Ionic compounds are typically hard and brittle, while metals are malleable and ductile.
    *   **Reactivity:** The strength and nature of bonds influence how readily a substance undergoes chemical reactions.

### 5. Structure and Geometry
Chemical bonds determine the three-dimensional arrangement of atoms within a molecule, known as molecular geometry. This geometry, in turn, influences the molecule's polarity, reactivity, and biological function (e.g., enzyme-substrate interactions, drug binding).

In essence, chemical bonds are the glue that holds the universe together at the atomic level, enabling the formation of all matter and dictating its behavior and properties.

Question 2

Q2. Write the Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Accepted Answer

The Lewis dot symbol for an atom represents the element's symbol surrounded by dots, where each dot represents a valence electron. The number of valence electrons is determined by the group number of the element (for main group elements).

1.  **Magnesium (Mg)**:
    *   Atomic number = 12
    *   Electronic configuration = [Ne] 3s²
    *   Valence electrons = 2
    *   Lewis dot symbol: Mg··

2.  **Sodium (Na)**:
    *   Atomic number = 11
    *   Electronic configuration = [Ne] 3s¹
    *   Valence electrons = 1
    *   Lewis dot symbol: Na·

3.  **Boron (B)**:
    *   Atomic number = 5
    *   Electronic configuration = [He] 2s² 2p¹
    *   Valence electrons = 3
    *   Lewis dot symbol: ·B·
                                  ·

4.  **Oxygen (O)**:
    *   Atomic number = 8
    *   Electronic configuration = [He] 2s² 2p⁴
    *   Valence electrons = 6
    *   Lewis dot symbol: ·O··
                                  ··
                                  ··

5.  **Nitrogen (N)**:
    *   Atomic number = 7
    *   Electronic configuration = [He] 2s² 2p³
    *   Valence electrons = 5
    *   Lewis dot symbol: ·N··
                                  ··
                                  ·

6.  **Bromine (Br)**:
    *   Atomic number = 35
    *   Electronic configuration = [Ar] 3d¹⁰ 4s² 4p⁵
    *   Valence electrons = 7
    *   Lewis dot symbol: ·Br··
                                  ··
                                  ··
                                  ·

Question 3

Q3. Write the Lewis dot symbols for the following ions: Li+, O2–, Mg2+, N3–, F–.

Accepted Answer

The Lewis dot symbol for an ion represents the element's symbol, with dots representing the remaining valence electrons (after gaining or losing electrons), enclosed in square brackets, and the charge written as a superscript outside the brackets.

1.  **Lithium ion (Li⁺)**:
    *   Atomic number = 3
    *   Electronic configuration of Li = [He] 2s¹
    *   Valence electrons of Li = 1
    *   Li loses 1 electron to form Li⁺.
    *   Electronic configuration of Li⁺ = [He]
    *   Valence electrons of Li⁺ = 0
    *   Lewis dot symbol: [Li]⁺

2.  **Oxide ion (O²⁻)**:
    *   Atomic number = 8
    *   Electronic configuration of O = [He] 2s² 2p⁴
    *   Valence electrons of O = 6
    *   O gains 2 electrons to form O²⁻.
    *   Electronic configuration of O²⁻ = [He] 2s² 2p⁶ (like Neon)
    *   Valence electrons of O²⁻ = 8 (octet)
    *   Lewis dot symbol: [:Ö:]²⁻
                                  ¨

3.  **Magnesium ion (Mg²⁺)**:
    *   Atomic number = 12
    *   Electronic configuration of Mg = [Ne] 3s²
    *   Valence electrons of Mg = 2
    *   Mg loses 2 electrons to form Mg²⁺.
    *   Electronic configuration of Mg²⁺ = [Ne]
    *   Valence electrons of Mg²⁺ = 0
    *   Lewis dot symbol: [Mg]²⁺

4.  **Nitride ion (N³⁻)**:
    *   Atomic number = 7
    *   Electronic configuration of N = [He] 2s² 2p³
    *   Valence electrons of N = 5
    *   N gains 3 electrons to form N³⁻.
    *   Electronic configuration of N³⁻ = [He] 2s² 2p⁶ (like Neon)
    *   Valence electrons of N³⁻ = 8 (octet)
    *   Lewis dot symbol: [:N:]³⁻
                                  ¨

5.  **Fluoride ion (F⁻)**:
    *   Atomic number = 9
    *   Electronic configuration of F = [He] 2s² 2p⁵
    *   Valence electrons of F = 7
    *   F gains 1 electron to form F⁻.
    *   Electronic configuration of F⁻ = [He] 2s² 2p⁶ (like Neon)
    *   Valence electrons of F⁻ = 8 (octet)
    *   Lewis dot symbol: [:F:]⁻
                                  ¨

Question 4

Q4. Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO3^2–, HCOOH.

Accepted Answer

Drawing Lewis structures involves determining the total number of valence electrons, identifying the central atom, forming single bonds, distributing lone pairs to satisfy octets, and forming multiple bonds if necessary. For ions, the charge is added/subtracted from the total valence electrons.

### 1. H₂S (Hydrogen Sulfide)
*   **Total Valence Electrons:** 2(H) + 6(S) = 2(1) + 6 = 8 valence electrons.
*   **Central Atom:** Sulfur (S) is less electronegative than hydrogen and can form more bonds.
*   **Structure:**
    1.  Place S in the center, bond to two H atoms: H—S—H (4 electrons used).
    2.  Remaining electrons = 8 - 4 = 4 electrons.
    3.  Distribute remaining electrons as lone pairs on S to complete its octet. S already has 4 electrons from bonds, needs 4 more. Add two lone pairs to S.
    4.  H atoms achieve duplet (2 electrons) with one bond each.
*   **Lewis Structure Description:** A central sulfur atom (S) is singly bonded to two hydrogen atoms (H). The sulfur atom also has two lone pairs of electrons. Each hydrogen atom has no lone pairs. The structure is H-S-H with two lone pairs on S.

### 2. SiCl₄ (Silicon Tetrachloride)
*   **Total Valence Electrons:** 4(Si) + 4(Cl) = 4 + 4(7) = 4 + 28 = 32 valence electrons.
*   **Central Atom:** Silicon (Si) is less electronegative than chlorine.
*   **Structure:**
    1.  Place Si in the center, bond to four Cl atoms: Cl—Si—Cl (8 electrons used).
                                                                    |   |
                                                                    Cl  Cl
    2.  Remaining electrons = 32 - 8 = 24 electrons.
    3.  Distribute remaining electrons as lone pairs on terminal Cl atoms to complete their octets. Each Cl needs 6 more electrons (3 lone pairs). 4 Cl atoms × 6 electrons/Cl = 24 electrons. All 24 electrons are used.
    4.  The central Si atom has 8 electrons (4 bonds) and satisfies its octet.
*   **Lewis Structure Description:** A central silicon atom (Si) is singly bonded to four chlorine atoms (Cl). Each chlorine atom has three lone pairs of electrons. The silicon atom has no lone pairs.

### 3. BeF₂ (Beryllium Fluoride)
*   **Total Valence Electrons:** 2(Be) + 2(F) = 2 + 2(7) = 2 + 14 = 16 valence electrons.
*   **Central Atom:** Beryllium (Be).
*   **Structure:**
    1.  Place Be in the center, bond to two F atoms: F—Be—F (4 electrons used).
    2.  Remaining electrons = 16 - 4 = 12 electrons.
    3.  Distribute remaining electrons as lone pairs on terminal F atoms to complete their octets. Each F needs 6 more electrons (3 lone pairs). 2 F atoms × 6 electrons/F = 12 electrons. All 12 electrons are used.
    4.  The central Be atom has only 4 electrons (2 bonds). Beryllium is an exception to the octet rule and can have an incomplete octet.
*   **Lewis Structure Description:** A central beryllium atom (Be) is singly bonded to two fluorine atoms (F). Each fluorine atom has three lone pairs of electrons. The beryllium atom has no lone pairs and an incomplete octet (4 electrons).

### 4. CO₃²⁻ (Carbonate ion)
*   **Total Valence Electrons:** 4(C) + 3(O) + 2(charge) = 4 + 3(6) + 2 = 4 + 18 + 2 = 24 valence electrons.
*   **Central Atom:** Carbon (C).
*   **Structure:**
    1.  Place C in the center, bond to three O atoms: O—C—O (6 electrons used).
                                                                    |
                                                                    O
    2.  Remaining electrons = 24 - 6 = 18 electrons.
    3.  Distribute remaining electrons as lone pairs on terminal O atoms. Each O needs 6 electrons (3 lone pairs). 3 O atoms × 6 electrons/O = 18 electrons. All 18 electrons are used.
    4.  The central C atom has only 6 electrons (3 bonds), so it does not satisfy its octet. To complete the octet of C, one lone pair from an oxygen atom must be converted into a double bond.
    5.  This leads to **resonance structures** because the double bond can be formed with any of the three oxygen atoms.
*   **Lewis Structure Description (Resonance):** The carbonate ion has a central carbon atom bonded to three oxygen atoms. In one resonance structure, one oxygen atom forms a double bond with carbon and has two lone pairs, while the other two oxygen atoms form single bonds with carbon and each have three lone pairs. The entire structure is enclosed in square brackets with a 2- charge. There are three equivalent resonance structures, where the double bond shifts between the three C-O positions. The actual structure is a resonance hybrid.

### 5. HCOOH (Formic Acid)
*   **Total Valence Electrons:** 1(H) + 4(C) + 2(O) + 1(H) = 1 + 4 + 2(6) + 1 = 1 + 4 + 12 + 1 = 18 valence electrons.
*   **Central Atom:** Carbon (C) is typically the central atom in organic acids. The structure is H-C(=O)-OH.
*   **Structure:**
    1.  Connect atoms: H—C—O—H (6 electrons used).
                                  ||
                                  O
    2.  Remaining electrons = 18 - 6 = 12 electrons.
    3.  Distribute lone pairs. The oxygen in the C=O group needs 4 electrons (2 lone pairs). The oxygen in the O-H group needs 4 electrons (2 lone pairs). Total 4 + 4 = 8 electrons used.
    4.  Remaining electrons = 12 - 8 = 4 electrons.
    5.  The central carbon has 6 electrons (3 bonds). To complete its octet, one lone pair from the double-bonded oxygen is converted into a double bond. This means the C=O bond is already formed in the initial step, so we need to re-evaluate.
    *   **Corrected approach for HCOOH:**
        1.  Skeleton: H—C—O—H. One O is double-bonded to C, the other is single-bonded to C and H. So, H—C—O—H. One O is double-bonded to C. Let's assume the C is bonded to one H, one O (double bond), and one O (single bond, which is then bonded to another H).
        2.  Skeleton: H—C—O—H (6 electrons for single bonds: C-H, C-O, O-H). Let's place the double bond now. H—C(=O)—O—H. (2 for C-H, 4 for C=O, 2 for C-O, 2 for O-H = 10 electrons used).
        3.  Remaining electrons = 18 - 10 = 8 electrons.
        4.  Distribute lone pairs: The double-bonded O needs 4 electrons (2 lone pairs). The single-bonded O needs 4 electrons (2 lone pairs). Total 4 + 4 = 8 electrons used. All electrons are used.
        5.  Check octets: H (2), C (8), O (8), O (8). All octets are satisfied.
*   **Lewis Structure Description:** A central carbon atom (C) is singly bonded to one hydrogen atom (H) and one oxygen atom (O). This carbon atom is also double-bonded to another oxygen atom. The oxygen atom that is single-bonded to carbon is also singly bonded to a hydrogen atom. The double-bonded oxygen has two lone pairs. The single-bonded oxygen has two lone pairs. The hydrogen atoms have no lone pairs.

Question 5

Q5. Define octet rule. Write its significance and limitations.

Accepted Answer

The octet rule is a fundamental concept in chemical bonding that helps explain the formation of many compounds. However, it also has several limitations.

### Definition of Octet Rule
The **octet rule** states that atoms tend to combine in such a way that each atom has eight electrons in its valence shell, acquiring a stable electron configuration similar to that of a noble gas (typically ns²np⁶). This tendency is observed primarily for elements in the second period of the periodic table.

### Significance of Octet Rule
1.  **Explains Chemical Bonding:** The octet rule provides a simple explanation for the formation of ionic and covalent bonds. Atoms gain, lose, or share electrons to achieve a stable octet, leading to the formation of compounds.
2.  **Predicts Valency:** It helps in understanding and predicting the combining capacity (valency) of elements, as atoms tend to form bonds that allow them to complete their octet.
3.  **Explains Stability:** Achieving an octet is associated with enhanced stability, similar to the inert noble gases, which explains why chemical reactions occur.

### Limitations of Octet Rule
Despite its usefulness, the octet rule is not universally applicable and has several significant limitations:

1.  **Incomplete Octet of the Central Atom:** In some compounds, the central atom has fewer than eight electrons in its valence shell. These are often referred to as electron-deficient molecules. Examples include:
    *   **LiCl:** Lithium has only 2 electrons in its valence shell.
    *   **BeH₂:** Beryllium has 4 electrons in its valence shell.
    *   **BCl₃:** Boron has 6 electrons in its valence shell.
    *   **AlCl₃:** Aluminium has 6 electrons in its valence shell.

2.  **Expanded Octet:** Elements in the third period and beyond have vacant d-orbitals available for bonding. This allows the central atom to accommodate more than eight electrons in its valence shell. Such molecules are said to have an expanded octet. Examples include:
    *   **PF₅:** Phosphorus has 10 electrons in its valence shell.
    *   **SF₆:** Sulfur has 12 electrons in its valence shell.
    *   **H₂SO₄:** Sulfur can have 12 electrons in its valence shell.
    *   **IF₇:** Iodine has 14 electrons in its valence shell.

3.  **Odd-electron Molecules:** Molecules with an odd number of valence electrons cannot satisfy the octet rule for all atoms. The central atom often ends up with an odd number of electrons. Examples include:
    *   **Nitric oxide (NO):** Total valence electrons = 5 (N) + 6 (O) = 11.
    *   **Nitrogen dioxide (NO₂):** Total valence electrons = 5 (N) + 2*6 (O) = 17.

4.  **The Octet Rule does not explain the shape of molecules:** It provides no information about the geometry or three-dimensional arrangement of atoms in a molecule. For example, it doesn't explain why H₂O is bent and CO₂ is linear.

5.  **The Octet Rule does not explain the relative stability of molecules:** It cannot explain why certain compounds are more stable than others, nor does it account for the energy changes involved in bond formation.

6.  **Noble Gas Compounds:** The formation of compounds of noble gases like Xenon (e.g., XeF₂, XeF₄, XeOF₄) contradicts the assumption that noble gases are chemically inert and have complete octets, as these compounds involve the expansion of the octet for the noble gas atom.

Question 6

Q6. Write the favourable factors for the formation of an ionic bond.

Accepted Answer

The formation of an ionic bond, which involves the complete transfer of electrons from one atom to another, is favoured by the following factors:

1.  **Low Ionization Enthalpy of the Metal Atom:** The atom that loses electrons (typically a metal) to form a cation should have a low ionization enthalpy. This means that a relatively small amount of energy is required to remove one or more electrons from its outermost shell, making cation formation energetically favourable.
    *   *Example:* Alkali metals (Group 1) and alkaline earth metals (Group 2) have low ionization enthalpies.

2.  **High Electron Gain Enthalpy (more negative) of the Non-metal Atom:** The atom that gains electrons (typically a non-metal) to form an anion should have a high (more negative) electron gain enthalpy. This means that a significant amount of energy is released when an electron is added to its outermost shell, making anion formation energetically favourable.
    *   *Example:* Halogens (Group 17) have very high negative electron gain enthalpies.

3.  **High Lattice Enthalpy of the Ionic Compound:** The lattice enthalpy is the energy released when gaseous ions combine to form one mole of a solid ionic compound. A high lattice enthalpy indicates strong electrostatic forces of attraction between the oppositely charged ions in the crystal lattice. This strong attraction provides the driving force for the formation of the ionic solid and contributes significantly to its stability, compensating for the energy required for ionization and electron gain.

Question 7

Q7. Write the Lewis dot structures of the following atoms and ions: S, S2–, Al, Al3+, H, H–

Accepted Answer

The Lewis dot structure represents the valence electrons of an atom or ion as dots around its chemical symbol. The number of valence electrons is determined by the group number for main group elements.

1.  **S (Sulfur):** Sulfur is in Group 16, so it has 6 valence electrons.
    `..`
    `:S:`
    `..`

2.  **S2– (Sulfide ion):** Sulfur gains 2 electrons to form S2–. So, it has 6 + 2 = 8 valence electrons (a complete octet).
    `..`
    `:S:` `2-`
    `..`

3.  **Al (Aluminum):** Aluminum is in Group 13, so it has 3 valence electrons.
    `.Al.`

4.  **Al3+ (Aluminum ion):** Aluminum loses 3 electrons to form Al3+. It has 3 - 3 = 0 valence electrons (the electrons from the outermost shell are lost, leaving a stable inner shell configuration, but no valence electrons to show).
    `[Al]3+`

5.  **H (Hydrogen):** Hydrogen is in Group 1, so it has 1 valence electron.
    `H.`

6.  **H– (Hydride ion):** Hydrogen gains 1 electron to form H–. So, it has 1 + 1 = 2 valence electrons (a complete duplet, like Helium).
    `[H:]-`

Question 8

Q8. Draw the Lewis structures for the following molecules and ions:
(i) H2S (ii) SiCl4 (iii) BeF2 (iv) CO3^2- (v) HCOOH

Accepted Answer

The Lewis structure shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. The goal is to satisfy the octet rule (or duplet for hydrogen) for most atoms.

**General Steps to Draw Lewis Structures:**
1.  Calculate the total number of valence electrons.
2.  Identify the central atom (usually the least electronegative atom, or the unique atom).
3.  Draw single bonds between the central atom and terminal atoms.
4.  Distribute the remaining electrons as lone pairs to satisfy the octet rule for terminal atoms first, then the central atom.
5.  If the central atom does not have an octet, form multiple bonds (double or triple) using lone pairs from terminal atoms.

**(i) H2S (Hydrogen Sulfide)**
*   Total valence electrons: H (1) × 2 + S (6) = 8 electrons.
*   Central atom: S (Sulfur).
*   Single bonds: H—S—H (4 electrons used).
*   Remaining electrons: 8 - 4 = 4 electrons.
*   Distribute remaining electrons: Place 2 lone pairs on S to complete its octet.
    `H—S—H`
    `..`
    `..`

**(ii) SiCl4 (Silicon Tetrachloride)**
*   Total valence electrons: Si (4) + Cl (7) × 4 = 4 + 28 = 32 electrons.
*   Central atom: Si (Silicon).
*   Single bonds: Cl—Si—Cl (8 electrons used, 4 bonds).
    `Cl`
    `|`
    `Cl—Si—Cl`
    `|`
    `Cl`
*   Remaining electrons: 32 - 8 = 24 electrons.
*   Distribute remaining electrons: Place 3 lone pairs on each of the 4 Cl atoms (3 × 4 = 12 lone pairs, 24 electrons used) to complete their octets. Silicon already has an octet.
    `..Cl..`
    `|`
    `..Cl—Si—Cl..`
    `|`
    `..Cl..`

**(iii) BeF2 (Beryllium Fluoride)**
*   Total valence electrons: Be (2) + F (7) × 2 = 2 + 14 = 16 electrons.
*   Central atom: Be (Beryllium).
*   Single bonds: F—Be—F (4 electrons used).
*   Remaining electrons: 16 - 4 = 12 electrons.
*   Distribute remaining electrons: Place 3 lone pairs on each of the 2 F atoms (3 × 2 = 6 lone pairs, 12 electrons used) to complete their octets. Beryllium has only 4 electrons, which is an incomplete octet, but it is stable in this configuration.
    `..F—Be—F..`

**(iv) CO3^2- (Carbonate ion)**
*   Total valence electrons: C (4) + O (6) × 3 + 2 (for 2- charge) = 4 + 18 + 2 = 24 electrons.
*   Central atom: C (Carbon).
*   Single bonds: O—C—O (6 electrons used, 3 bonds).
    `O`
    `|`
    `O—C—O`
*   Remaining electrons: 24 - 6 = 18 electrons.
*   Distribute remaining electrons: Place 3 lone pairs on each of the 3 O atoms (3 × 3 = 9 lone pairs, 18 electrons used). Now each O has an octet (2 from bond + 6 from lone pairs). Carbon has only 6 electrons.
*   Form multiple bonds: To give Carbon an octet, convert one lone pair from one of the oxygen atoms into a double bond. This results in resonance structures.
    `[ O ]`
    `||`
    `[ O—C—O ]^2-`
    `..` `..` `..`
    (with 2 lone pairs on the double-bonded O, and 3 lone pairs on each single-bonded O. The 2- charge is delocalized over the two single-bonded oxygen atoms).

**(v) HCOOH (Formic acid)**
*   Total valence electrons: H (1) + C (4) + O (6) × 2 + H (1) = 1 + 4 + 12 + 1 = 18 electrons.
*   Central atom: C (Carbon).
*   Connectivity: The carbon atom is bonded to one H, one O (which is double-bonded), and another O (which is single-bonded to another H).
    `O`
    `||`
    `H—C—O—H`
*   Single bonds: C-H, C-O, O-H (6 electrons used).
*   Remaining electrons: 18 - 6 = 12 electrons.
*   Distribute remaining electrons: Place 2 lone pairs on the double-bonded O (4 electrons used), and 2 lone pairs on the single-bonded O (4 electrons used). This completes octets for both oxygen atoms. Carbon has 6 electrons (from C-H, C-O single bonds).
*   Form multiple bonds: Convert one lone pair from the double-bonded oxygen into a double bond with carbon. This gives carbon an octet.
    `O`
    `||`
    `H—C—O—H`
    (The double-bonded O has 2 lone pairs, the single-bonded O has 2 lone pairs).

Question 9

Q9. Define octet rule. Write its significance and limitations.

Accepted Answer

### Definition of Octet Rule

The **octet rule** states that atoms tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas (specifically, the s²p⁶ configuration). For hydrogen, the equivalent is a 'duplet' (two electrons in its valence shell, like helium).

This rule is based on the observation that noble gases (except helium) have eight electrons in their outermost shell and are exceptionally stable and unreactive. Atoms achieve this stable octet by gaining, losing, or sharing electrons through the formation of ionic or covalent bonds.

### Significance of the Octet Rule

The octet rule has been a cornerstone in understanding chemical bonding and molecular structure for many years. Its significance lies in the following points:

1.  **Explains Chemical Bonding:** It provides a simple and intuitive explanation for why atoms form chemical bonds. Atoms bond to achieve a stable electron configuration, typically an octet.
2.  **Predicts Valency:** It helps in predicting the combining capacity (valency) of elements, especially for main group elements, by indicating how many electrons an atom needs to gain, lose, or share to achieve an octet.
3.  **Explains Stability of Compounds:** Compounds formed according to the octet rule are generally more stable. The formation of an octet leads to a lower energy state for the atoms involved.
4.  **Foundation for Lewis Structures:** It is the basis for drawing Lewis dot structures, which are essential for visualizing electron distribution in molecules and predicting molecular geometry.

### Limitations of the Octet Rule

Despite its usefulness, the octet rule is not universally applicable and has several significant limitations:

1.  **Incomplete Octet of the Central Atom:** In some compounds, the central atom has fewer than eight valence electrons, yet the molecule is stable. This is particularly observed for elements in Group 1, 2, and 13.
    *   **Examples:**
        *   **LiCl:** Lithium has only 2 electrons in its valence shell.
        *   **BeH2:** Beryllium has only 4 electrons in its valence shell.
        *   **BF3:** Boron has only 6 electrons in its valence shell.

2.  **Expanded Octet (Hypervalent Molecules):** Elements in and beyond the third period (e.g., P, S, Cl, Xe) have vacant d-orbitals available in their valence shell. They can utilize these d-orbitals to accommodate more than eight electrons in their valence shell, forming stable compounds with an expanded octet.
    *   **Examples:**
        *   **PCl5:** Phosphorus has 10 electrons in its valence shell.
        *   **SF6:** Sulfur has 12 electrons in its valence shell.
        *   **H2SO4:** Sulfur has 12 electrons in its valence shell.
        *   **IF7:** Iodine has 14 electrons in its valence shell.

3.  **Odd-Electron Molecules:** Molecules with an odd number of valence electrons cannot satisfy the octet rule for all atoms, as it's impossible to distribute an odd number of electrons to achieve octets for all atoms.
    *   **Examples:**
        *   **NO (Nitric Oxide):** Total valence electrons = 5 (N) + 6 (O) = 11 electrons.
        *   **NO2 (Nitrogen Dioxide):** Total valence electrons = 5 (N) + 6 (O) × 2 = 17 electrons.

4.  **Noble Gas Compounds:** The octet rule implies that noble gases are chemically inert due to their stable octet configuration. However, some noble gases, particularly Xenon (Xe) and Krypton (Kr), are known to form compounds with highly electronegative elements like fluorine and oxygen, contradicting the rule.
    *   **Examples:**
        *   **XeF2, XeF4, XeF6**
        *   **KrF2**

5.  **Transition Metals:** The octet rule is generally not applicable to compounds involving transition metals, as their bonding often involves d-orbitals and is better explained by more advanced theories like Crystal Field Theory or Ligand Field Theory.

In conclusion, while the octet rule is a useful generalization for understanding bonding in many simple molecules, it is not a fundamental law and has significant exceptions, especially for elements beyond the second period.

Question 10

Q10. Describe the change in hybridisation (if any) of the Al atom in the following reaction. 
AlCl₃ + Cl⁻ → AlCl₄⁻

Accepted Answer

To determine the change in hybridisation of the Al atom, we need to find the hybridisation of Al in both AlCl₃ and AlCl₄⁻.

**1. Hybridisation of Al in AlCl₃:**
*   The central atom is Aluminium (Al), which belongs to Group 13 and has 3 valence electrons.
*   Al forms 3 single bonds with 3 chlorine atoms.
*   Number of bond pairs = 3
*   Number of lone pairs on Al = 0 (Al uses all its 3 valence electrons for bonding)
*   Steric number = Number of bond pairs + Number of lone pairs = 3 + 0 = 3
*   A steric number of 3 corresponds to **sp² hybridisation**.
*   The geometry of AlCl₃ is trigonal planar.

**2. Hybridisation of Al in AlCl₄⁻:**
*   In AlCl₄⁻, the Al atom is bonded to 4 chlorine atoms. The negative charge indicates that Al has accepted an electron pair from Cl⁻ to form a fourth bond.
*   Number of bond pairs = 4
*   Number of lone pairs on Al = 0
*   Steric number = Number of bond pairs + Number of lone pairs = 4 + 0 = 4
*   A steric number of 4 corresponds to **sp³ hybridisation**.
*   The geometry of AlCl₄⁻ is tetrahedral.

**Conclusion:**
During the reaction AlCl₃ + Cl⁻ → AlCl₄⁻, the hybridisation of the Al atom changes from **sp² to sp³**.

Question 11

Q11. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF₃ + NH₃ → F₃B.NH₃

Accepted Answer

To determine the change in hybridisation, we will analyze the hybridisation of B and N atoms in the reactants (BF₃ and NH₃) and the product (F₃B.NH₃).

**1. Hybridisation in Reactants:**
*   **BF₃:**
    *   Central atom: Boron (B), Group 13, 3 valence electrons.
    *   Bonds: 3 B-F single bonds.
    *   Lone pairs on B: 0 (all 3 valence electrons are used in bonding).
    *   Steric number = 3 (3 bond pairs + 0 lone pairs) = 3.
    *   Hybridisation of B in BF₃ is **sp²**.
    *   Geometry: Trigonal planar.
*   **NH₃:**
    *   Central atom: Nitrogen (N), Group 15, 5 valence electrons.
    *   Bonds: 3 N-H single bonds.
    *   Lone pairs on N: 1 (5 valence electrons - 3 used in bonding = 2 electrons, forming 1 lone pair).
    *   Steric number = 3 (bond pairs) + 1 (lone pair) = 4.
    *   Hybridisation of N in NH₃ is **sp³**.
    *   Geometry: Pyramidal.

**2. Hybridisation in Product (F₃B.NH₃ - Boron trifluoride ammonia adduct):**
*   This is an adduct formed by a Lewis acid-base reaction, where NH₃ (Lewis base) donates its lone pair to BF₃ (Lewis acid).
*   **Boron (B) in F₃B.NH₃:**
    *   B now forms 4 bonds (3 B-F bonds and 1 dative B←N bond).
    *   Lone pairs on B: 0.
    *   Steric number = 4 (4 bond pairs + 0 lone pairs) = 4.
    *   Hybridisation of B in F₃B.NH₃ is **sp³**.
    *   Geometry: Tetrahedral.
*   **Nitrogen (N) in F₃B.NH₃:**
    *   N still forms 3 N-H bonds and now forms 1 dative N→B bond (using its lone pair).
    *   Lone pairs on N: 0 (the lone pair is used in forming the dative bond).
    *   Steric number = 4 (3 N-H bond pairs + 1 N→B bond pair + 0 lone pairs) = 4.
    *   Hybridisation of N in F₃B.NH₃ is **sp³**.
    *   Geometry: Tetrahedral.

**Conclusion:**
*   The hybridisation of the **Boron (B) atom changes from sp² to sp³**.
*   The hybridisation of the **Nitrogen (N) atom remains sp³** (though its geometry changes from pyramidal to tetrahedral due to the lone pair being converted into a bond pair).

Question 12

Q12. Explain the formation of H₂⁺ molecular ion.

Accepted Answer

The formation of the H₂⁺ molecular ion can be explained using the Linear Combination of Atomic Orbitals (LCAO) approach of Molecular Orbital (MO) Theory.

**1. Composition of H₂⁺:**
H₂⁺ is a diatomic molecular ion consisting of two hydrogen nuclei (protons) and only one electron. It is the simplest molecular ion.

**2. Atomic Orbitals Involved:**
Each hydrogen atom contributes one 1s atomic orbital. When these two hydrogen atoms approach each other to form a molecule, their 1s atomic orbitals combine.

**3. Linear Combination of Atomic Orbitals (LCAO):**
According to LCAO-MO theory, when two atomic orbitals combine, they form two molecular orbitals:
*   **Bonding Molecular Orbital (BMO):** Formed by the constructive interference (addition) of the atomic orbitals. It has lower energy than the original atomic orbitals and concentrates electron density between the nuclei, leading to attraction and bond formation. For H₂⁺, this is denoted as **σ1s**.
*   **Antibonding Molecular Orbital (ABMO):** Formed by the destructive interference (subtraction) of the atomic orbitals. It has higher energy than the original atomic orbitals and has a nodal plane between the nuclei, leading to repulsion. For H₂⁺, this is denoted as **σ*1s**.

**4. Electron Configuration:**
H₂⁺ has only one electron. According to the Aufbau principle, this single electron will occupy the lowest energy molecular orbital, which is the bonding σ1s orbital.
*   The electron configuration of H₂⁺ is **(σ1s)¹**.

**5. Bond Order Calculation:**
Bond order (B.O.) is calculated as: B.O. = ½ (Number of electrons in BMO - Number of electrons in ABMO)
*   For H₂⁺: B.O. = ½ (1 - 0) = ½

**6. Stability and Nature of Bond:**
*   A positive bond order (½) indicates that the H₂⁺ molecular ion is stable and can exist. The energy of the electron in the σ1s orbital is lower than its energy in an isolated 1s atomic orbital, providing stability.
*   The bond in H₂⁺ is a **one-electron bond**. Although it's a weaker bond compared to a typical two-electron covalent bond (like in H₂ with bond order 1), it is sufficient to hold the two hydrogen nuclei together.

In summary, the H₂⁺ ion forms because the single electron occupies the lower energy bonding molecular orbital (σ1s), resulting in a net attractive force between the two hydrogen nuclei, despite having only half a bond.

Question 13

Q13. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Accepted Answer

For the linear combination of atomic orbitals (LCAO) to effectively form molecular orbitals, certain conditions must be met. These conditions ensure that the atomic orbitals can interact constructively and destructively to produce stable and distinct molecular orbitals. The three important conditions are:

**1. The combining atomic orbitals must have comparable energies:**
*   Atomic orbitals that are combining to form molecular orbitals must have energies that are close to each other. For example, a 1s orbital of one atom can combine effectively with a 1s orbital of another atom, and a 2s orbital can combine with a 2s orbital. Similarly, 2p orbitals combine with 2p orbitals.
*   However, a 1s orbital of one atom generally cannot combine effectively with a 2s or 2p orbital of another atom because their energy difference is too large. If the energy difference is significant, the overlap will be minimal, and the resulting molecular orbitals will not be stable.

**2. The combining atomic orbitals must have proper symmetry about the molecular axis:**
*   The atomic orbitals must have the same symmetry with respect to the molecular axis (the internuclear axis). This means that the regions of positive and negative overlap must align correctly.
*   For example, an s orbital (spherically symmetric) can combine with another s orbital or with a pz orbital (if the molecular axis is defined as the z-axis) to form sigma (σ) molecular orbitals. However, an s orbital cannot combine with a px or py orbital to form a sigma bond because their symmetries do not match along the internuclear axis, leading to zero net overlap.
*   Similarly, pz orbitals combine head-on along the internuclear axis to form σ molecular orbitals, while px and py orbitals combine sideways to form pi (π) molecular orbitals.

**3. The combining atomic orbitals must overlap to the maximum possible extent:**
*   For effective combination and the formation of strong molecular orbitals, the atomic orbitals must overlap significantly. Greater the extent of overlap, the stronger the bond formed and the more stable the resulting molecular orbital.
*   The extent of overlap depends on the distance between the nuclei and the orientation of the atomic orbitals. Optimal internuclear distance allows for maximum overlap without excessive repulsion between nuclei.

These conditions are crucial for the successful application of the LCAO-MO theory in predicting the bonding and properties of molecules.

Question 14

Q14. Use molecular orbital theory to explain why the B₂ molecule is paramagnetic.

Accepted Answer

To explain why the B₂ molecule is paramagnetic, we need to apply the Molecular Orbital (MO) Theory and determine its electron configuration.

**1. Determine Total Number of Electrons:**
*   Boron (B) has an atomic number of 5, meaning it has 5 electrons.
*   Therefore, the B₂ molecule has a total of 5 + 5 = 10 electrons.

**2. Write the Molecular Orbital Energy Level Diagram:**
For diatomic molecules with 14 or fewer electrons (like B₂), the order of filling molecular orbitals is as follows:
σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz

*Note: For molecules with more than 14 electrons, the order of σ2pz and π2p orbitals is swapped: σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz.*

**3. Fill the Molecular Orbitals with Electrons:**
We need to fill 10 electrons into these molecular orbitals according to Hund's rule and the Pauli exclusion principle:
*   (σ1s)² (2 electrons)
*   (σ*1s)² (4 electrons total)
*   (σ2s)² (6 electrons total)
*   (σ*2s)² (8 electrons total)
*   The next two electrons (9th and 10th) will enter the degenerate π2px and π2py molecular orbitals. According to Hund's rule, electrons will occupy degenerate orbitals singly with parallel spins before pairing up.
    *   (π2px)¹ (9 electrons total)
    *   (π2py)¹ (10 electrons total)

**4. Molecular Orbital Electron Configuration of B₂:**
(σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2px)¹ (π2py)¹

**5. Determine Magnetic Nature:**
*   A substance is **paramagnetic** if it contains one or more unpaired electrons.
*   A substance is **diamagnetic** if all its electrons are paired.

From the electron configuration of B₂, we observe that there are **two unpaired electrons**, one in the π2px molecular orbital and one in the π2py molecular orbital.

**Conclusion:**
Since the B₂ molecule contains two unpaired electrons, it is **paramagnetic**.

Question 15

Q15. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl– → AlCl4–

Accepted Answer

The hybridisation of the Al atom changes from sp² to sp³ in the given reaction.

Question 16

Q16. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B.NH3

Accepted Answer

Yes, there is a change in the hybridisation of both Boron (B) and Nitrogen (N) atoms as a result of the reaction.

Question 17

Q17. Draw diagrams showing (i) sp hybridisation, (ii) sp2 hybridisation and (iii) sp3 hybridisation of carbon atom and describe the orbital pictures of ethylene and acetylene.

Accepted Answer

### (i) sp Hybridisation of Carbon Atom

**Description:**
In sp hybridisation, one s-orbital and one p-orbital of the carbon atom mix to form two new equivalent hybrid orbitals called sp hybrid orbitals. These two sp hybrid orbitals are oriented at an angle of 180° to each other, resulting in a linear geometry. The remaining two unhybridised p-orbitals are perpendicular to the sp hybrid orbitals and also perpendicular to each other.

**Orbital Picture:**
*   The carbon atom in its excited state has the electronic configuration 1s² 2s¹ 2pₓ¹ 2py¹ 2pz¹.
*   One 2s orbital and one 2p orbital (e.g., 2pₓ) combine to form two sp hybrid orbitals.
*   These two sp orbitals are oriented linearly.
*   The remaining two 2p orbitals (2py and 2pz) remain unhybridised and are perpendicular to the sp orbitals and to each other.

### (ii) sp² Hybridisation of Carbon Atom

**Description:**
In sp² hybridisation, one s-orbital and two p-orbitals of the carbon atom mix to form three new equivalent hybrid orbitals called sp² hybrid orbitals. These three sp² hybrid orbitals are oriented at an angle of 120° to each other in a trigonal planar arrangement. The remaining one unhybridised p-orbital is perpendicular to the plane of the sp² hybrid orbitals.

**Orbital Picture:**
*   The carbon atom in its excited state has the electronic configuration 1s² 2s¹ 2pₓ¹ 2py¹ 2pz¹.
*   One 2s orbital and two 2p orbitals (e.g., 2pₓ and 2py) combine to form three sp² hybrid orbitals.
*   These three sp² orbitals are oriented in a trigonal planar fashion.
*   The remaining one 2p orbital (2pz) remains unhybridised and is perpendicular to the plane of the sp² orbitals.

### (iii) sp³ Hybridisation of Carbon Atom

**Description:**
In sp³ hybridisation, one s-orbital and all three p-orbitals of the carbon atom mix to form four new equivalent hybrid orbitals called sp³ hybrid orbitals. These four sp³ hybrid orbitals are oriented at an angle of 109.5° to each other, pointing towards the corners of a regular tetrahedron. This results in a tetrahedral geometry.

**Orbital Picture:**
*   The carbon atom in its excited state has the electronic configuration 1s² 2s¹ 2pₓ¹ 2py¹ 2pz¹.
*   One 2s orbital and all three 2p orbitals (2pₓ, 2py, 2pz) combine to form four sp³ hybrid orbitals.
*   These four sp³ orbitals are oriented tetrahedrally.
*   There are no unhybridised p-orbitals left.

### Orbital Picture of Ethylene (C₂H₄)

**Structure:** Ethylene has a double bond between two carbon atoms and two hydrogen atoms attached to each carbon atom. Each carbon atom is sp² hybridised.

**Bonding:**
1.  **Carbon-Carbon Sigma (σ) Bond:** Each carbon atom undergoes sp² hybridisation. One sp² hybrid orbital from each carbon atom overlaps axially (head-on) to form a C-C sigma bond.
2.  **Carbon-Hydrogen Sigma (σ) Bonds:** The remaining two sp² hybrid orbitals on each carbon atom overlap axially with the 1s orbitals of two hydrogen atoms to form two C-H sigma bonds per carbon.
3.  **Carbon-Carbon Pi (π) Bond:** Each carbon atom has one unhybridised 2p orbital (perpendicular to the sp² plane). These two unhybridised 2p orbitals (one from each carbon) overlap laterally (sideways) to form one C-C pi bond. This pi bond lies above and below the plane of the sigma bonds.

**Geometry:** The molecule is planar, with bond angles approximately 120° around each carbon atom.

### Orbital Picture of Acetylene (C₂H₂)

**Structure:** Acetylene has a triple bond between two carbon atoms and one hydrogen atom attached to each carbon atom. Each carbon atom is sp hybridised.

**Bonding:**
1.  **Carbon-Carbon Sigma (σ) Bond:** Each carbon atom undergoes sp hybridisation. One sp hybrid orbital from each carbon atom overlaps axially (head-on) to form a C-C sigma bond.
2.  **Carbon-Hydrogen Sigma (σ) Bonds:** The remaining one sp hybrid orbital on each carbon atom overlaps axially with the 1s orbital of one hydrogen atom to form one C-H sigma bond per carbon.
3.  **Carbon-Carbon Pi (π) Bonds:** Each carbon atom has two unhybridised 2p orbitals (perpendicular to the sp hybrid orbitals and to each other). These two unhybridised 2p orbitals from one carbon overlap laterally with the corresponding two unhybridised 2p orbitals from the other carbon atom to form two C-C pi bonds. These two pi bonds are mutually perpendicular and surround the C-C sigma bond.

**Geometry:** The molecule is linear, with bond angles of 180°.

Question 18

Q18. Explain the formation of sigma (σ) and pi (π) bonds using suitable diagrams.

Accepted Answer

Covalent bonds are formed by the overlap of atomic orbitals. Depending on the pattern of overlap, two main types of covalent bonds are formed: sigma (σ) bonds and pi (π) bonds.

### 1. Sigma (σ) Bond

**Definition:** A sigma (σ) bond is formed by the axial (head-on or end-to-end) overlap of atomic orbitals along the internuclear axis. This type of overlap results in a strong bond where the electron density is concentrated symmetrically around the internuclear axis.

**Formation:**
Sigma bonds can be formed by the following types of orbital overlap:

*   **s-s overlap:** When two half-filled s-orbitals overlap axially. (e.g., H₂ molecule: H(1s) - H(1s) overlap)
*   **s-p overlap:** When a half-filled s-orbital overlaps axially with a half-filled p-orbital. (e.g., HF molecule: H(1s) - F(2p) overlap)
*   **p-p overlap:** When two half-filled p-orbitals overlap axially along the internuclear axis. (e.g., Cl₂ molecule: Cl(3p) - Cl(3p) overlap)
*   **Hybrid orbital overlap:** When hybrid orbitals (sp, sp², sp³) overlap axially with other hybrid orbitals or atomic orbitals. (e.g., C-C and C-H bonds in alkanes)

**Characteristics of Sigma Bonds:**
*   They are strong bonds due to extensive head-on overlap.
*   Electron density is concentrated along the internuclear axis.
*   They allow free rotation of atoms around the bond axis because the overlap remains effective during rotation.
*   A single bond is always a sigma bond.

### 2. Pi (π) Bond

**Definition:** A pi (π) bond is formed by the lateral (sideways) overlap of two half-filled unhybridised p-orbitals (or sometimes d-orbitals) perpendicular to the internuclear axis. The electron density is concentrated above and below the internuclear axis, not directly on it.

**Formation:**
Pi bonds are typically formed by the sideways overlap of two unhybridised p-orbitals. For a pi bond to form, there must already be a sigma bond between the two atoms, as the p-orbitals must be parallel to each other for effective sideways overlap.

*   **p-p sideways overlap:** When two half-filled unhybridised p-orbitals overlap laterally. (e.g., C=C in ethene, C≡C in ethyne)

**Characteristics of Pi Bonds:**
*   They are generally weaker than sigma bonds because the sideways overlap is less effective than head-on overlap.
*   Electron density is concentrated above and below the internuclear axis.
*   They restrict free rotation around the bond axis because rotation would break the sideways overlap.
*   Multiple bonds (double or triple bonds) consist of one sigma bond and one or two pi bonds. A double bond has one σ and one π bond. A triple bond has one σ and two π bonds.

Question 19

Q19. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Accepted Answer

The formation of molecular orbitals (MOs) from atomic orbitals (AOs) through the Linear Combination of Atomic Orbitals (LCAO) method is governed by certain essential conditions. These conditions ensure effective overlap and proper interaction between the atomic orbitals to form stable molecular orbitals.

### Conditions for Linear Combination of Atomic Orbitals (LCAO):

1.  **The combining atomic orbitals must have comparable energies:**
    *   For effective overlap and significant interaction, the atomic orbitals combining to form molecular orbitals must have energies that are close to each other. For example, a 1s orbital of one atom can combine effectively with a 1s orbital of another atom, but not with a 2s orbital, due to the large energy difference between 1s and 2s orbitals. Similarly, 2s orbitals combine with 2s orbitals, and 2p orbitals combine with 2p orbitals.

2.  **The combining atomic orbitals must have proper symmetry about the molecular axis:**
    *   The atomic orbitals must have the same symmetry with respect to the internuclear axis (molecular axis). For instance, if the z-axis is considered the internuclear axis, a 2pz orbital of one atom can combine with a 2pz orbital of another atom to form sigma (σ) molecular orbitals. However, a 2pz orbital cannot combine with a 2px or 2py orbital because their symmetries are different with respect to the internuclear axis, leading to zero net overlap.

3.  **The combining atomic orbitals must overlap to the maximum possible extent:**
    *   The extent of overlap between the atomic orbitals is directly related to the strength of the molecular orbital formed. Greater the overlap between the atomic orbitals, stronger will be the bond formed and more stable will be the resulting molecular orbital. This condition implies that the atoms must be sufficiently close to each other for their orbitals to interact effectively.

Question 20

Q20. Use molecular orbital theory to explain why the Be₂ molecule does not exist.

Accepted Answer

According to Molecular Orbital Theory (MOT), the existence and stability of a molecule are determined by its bond order. A molecule is stable if its bond order is positive, and it does not exist if its bond order is zero or negative.

### Explanation for Non-existence of Be₂ Molecule:

1.  **Electronic Configuration of Beryllium (Be):**
    *   The atomic number of Beryllium (Be) is 4. Its electronic configuration is 1s² 2s².

2.  **Total Electrons in Be₂ Molecule:**
    *   A Be₂ molecule consists of two Beryllium atoms. Therefore, the total number of electrons in the Be₂ molecule is 4 (from first Be) + 4 (from second Be) = 8 electrons.

3.  **Molecular Orbital (MO) Configuration of Be₂:**
    *   The 8 electrons in Be₂ are filled into the molecular orbitals according to the Aufbau principle, Pauli exclusion principle, and Hund's rule. The order of filling MOs for elements with Z ≤ 7 (like Be) is typically: σ1s, σ*1s, σ2s, σ*2s, (π2px = π2py), σ2pz, (π*2px = π*2py), σ*2pz.
    *   For Be₂, the MO configuration will be:
        (σ1s)² (σ*1s)² (σ2s)² (σ*2s)²

4.  **Calculation of Bond Order:**
    *   Bond Order (B.O.) is calculated using the formula: B.O. = ½ (N_b - N_a)
        Where, N_b = number of electrons in bonding molecular orbitals
        N_a = number of electrons in antibonding molecular orbitals
    *   From the MO configuration of Be₂:
        *   N_b (electrons in σ1s and σ2s) = 2 + 2 = 4
        *   N_a (electrons in σ*1s and σ*2s) = 2 + 2 = 4
    *   Therefore, B.O. = ½ (4 - 4) = ½ (0) = 0

5.  **Conclusion:**
    *   Since the bond order for the Be₂ molecule is zero, it indicates that there is no net attractive force between the two Beryllium atoms. The number of electrons in bonding orbitals is exactly cancelled out by the number of electrons in antibonding orbitals. Consequently, the Be₂ molecule is unstable and does not exist under normal conditions.

Question 21

Q21. Compare the relative stability of the following species and indicate their magnetic properties; O₂⁺, O₂, O₂⁻ (superoxide), O₂²⁻ (peroxide)

Accepted Answer

To compare the relative stability and determine the magnetic properties of the given oxygen species, we will use Molecular Orbital Theory (MOT). Stability is directly proportional to bond order, and magnetic properties (paramagnetic or diamagnetic) are determined by the presence or absence of unpaired electrons in the molecular orbitals. For oxygen and heavier diatomic molecules (total electrons > 14), the energy order of molecular orbitals is: σ1s < σ*1s < σ2s < σ*2s < σ2pz < (π2px = π2py) < (π*2px = π*2py) < σ*2pz Let's analyze each species: ### 1. O₂⁺ (Oxygen Cation) * **Total electrons:** Oxygen (O) has 8 electrons. So, O₂ has 16 electrons. O₂⁺ means one electron is removed, so 16 - 1 = 15 electrons. * **Molecular Orbital (MO) Configuration:** (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)¹ * **Bond Order (B.O.):** N_b (bonding electrons) = 2 (σ1s) + 2 (σ2s) + 2 (σ2pz) + 2 (π2px) + 2 (π2py) = 10 N_a (antibonding electrons) = 2 (σ*1s) + 2 (σ*2s) + 1 (π*2px) = 5 B.O. = ½ (N_b - N_a) = ½ (10 - 5) = ½ (5) = 2.5 * **Magnetic Property:** Contains one unpaired electron in the π*2px molecular orbital. Therefore, O₂⁺ is **paramagnetic**. ### 2. O₂ (Oxygen Molecule) * **Total electrons:** 8 (from one O) + 8 (from another O) = 16 electrons. * **Molecular Orbital (MO) Configuration:** (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)¹ (π*2py)¹ * **Bond Order (B.O.):** N_b = 2 + 2 + 2 + 2 + 2 = 10 N_a = 2 + 2 + 1 + 1 = 6 B.O. = ½ (10 - 6) = ½ (4) = 2.0 * **Magnetic Property:** Contains two unpaired electrons (one in π*2px and one in π*2py molecular orbitals). Therefore, O₂ is **paramagnetic**. ### 3. O₂⁻ (Superoxide Ion) * **Total electrons:** O₂ has 16 electrons. O₂⁻ means one electron is added, so 16 + 1 = 17 electrons. * **Molecular Orbital (MO) Configuration:** (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)² (π*2py)¹ * **Bond Order (B.O.):** N_b = 2 + 2 + 2 + 2 + 2 = 10 N_a = 2 + 2 + 2 + 1 = 7 B.O. = ½ (10 - 7) = ½ (3) = 1.5 * **Magnetic Property:** Contains one unpaired electron in the π*2py molecular orbital. Therefore, O₂⁻ is **paramagnetic**. ### 4. O₂²⁻ (Peroxide Ion) * **Total electrons:** O₂ has 16 electrons. O₂²⁻ means two electrons are added, so 16 + 2 = 18 electrons. * **Molecular Orbital (MO) Configuration:** (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2pz)² (π2px)² (π2py)² (π*2px)² (π*2py)² * **Bond Order (B.O.):** N_b = 2 + 2 + 2 + 2 + 2 = 10 N_a = 2 + 2 + 2 + 2 = 8 B.O. = ½ (10 - 8) = ½ (2) = 1.0 * **Magnetic Property:** All electrons are paired in the molecular orbitals. Therefore, O₂²⁻ is **diamagnetic**. ### Comparison of Relative Stability: Stability is directly proportional to the bond order. Higher the bond order, greater the stability. * O₂⁺: Bond Order = 2.5 * O₂: Bond Order = 2.0 * O₂⁻: Bond Order = 1.5 * O₂²⁻: Bond Order = 1.0 Therefore, the order of relative stability is: **O₂⁺ > O₂ > O₂⁻ > O₂²⁻** ### Summary of Magnetic Properties: * O₂⁺: Paramagnetic * O₂: Paramagnetic * O₂⁻: Paramagnetic * O₂²⁻: Diamagnetic

Question 22

Q22. What is the significance of a plus and a minus sign in representing the orbitals?

Accepted Answer

In the context of atomic and molecular orbitals, the plus (+) and minus (-) signs do **not** represent an electrical charge. Instead, they represent the **phase (or sign) of the wave function (ψ)** that describes the orbital.

An atomic orbital is a mathematical function (wave function) whose square gives the probability of finding an electron in a particular region of space. Like any wave, a wave function can have positive or negative amplitudes (phases) in different regions of space. These regions are separated by nodes where the wave function is zero.

### Significance:

1.  **Phase of the Wave Function:** The signs indicate the algebraic sign of the wave function in a particular region of space. For example, a p-orbital has two lobes, one with a positive phase and the other with a negative phase.

2.  **Orbital Overlap and Bonding:** When atomic orbitals combine to form molecular orbitals (according to the LCAO method):
    *   **Constructive Interference (Bonding):** If two atomic orbitals with the **same phase** (e.g., + with + or - with -) overlap, their wave functions add up. This leads to constructive interference, resulting in an increased electron density between the nuclei and the formation of a **bonding molecular orbital** (e.g., σ or π).
    *   **Destructive Interference (Antibonding):** If two atomic orbitals with **opposite phases** (e.g., + with -) overlap, their wave functions cancel each other out. This leads to destructive interference, resulting in a region of zero electron density (a nodal plane) between the nuclei and the formation of an **antibonding molecular orbital** (e.g., σ* or π*).

Therefore, the plus and minus signs are crucial for understanding how atomic orbitals interact to form molecular orbitals and whether a bonding (stabilizing) or antibonding (destabilizing) interaction will occur.

Question 23

Q23. Describe the change in hybridisation (if any) of the Al atom and N atom in the reaction. 
AlCl3 + NH3 → Cl3Al.NH3

Accepted Answer

The reaction involves the formation of an adduct between Aluminium chloride (AlCl3) and Ammonia (NH3). AlCl3 acts as a Lewis acid and NH3 acts as a Lewis base. Let's analyze the hybridisation of Al and N atoms before and after the reaction.

**1. Hybridisation of Al in AlCl3 (Reactant):**
*   Aluminium (Al) is the central atom. Its electronic configuration is [Ne] 3s² 3p¹.
*   In AlCl3, Al forms three single bonds with three chlorine atoms. There are no lone pairs on Al.
*   To form three bonds, Al undergoes sp² hybridisation. One 3s orbital and two 3p orbitals mix to form three sp² hybrid orbitals.
*   The geometry of AlCl3 is trigonal planar.

**2. Hybridisation of N in NH3 (Reactant):**
*   Nitrogen (N) is the central atom. Its electronic configuration is [He] 2s² 2p³.
*   In NH3, N forms three single bonds with three hydrogen atoms and has one lone pair of electrons.
*   To accommodate three bond pairs and one lone pair, N undergoes sp³ hybridisation. One 2s orbital and three 2p orbitals mix to form four sp³ hybrid orbitals.
*   The geometry of NH3 is trigonal pyramidal.

**3. Hybridisation of Al in Cl3Al.NH3 (Product):**
*   In the adduct Cl3Al.NH3, the nitrogen atom of NH3 donates its lone pair to the empty p-orbital of Al in AlCl3, forming a coordinate covalent bond (dative bond).
*   Initially, Al in AlCl3 was sp² hybridised with one empty unhybridised p-orbital. Upon accepting the lone pair from NH3, this empty p-orbital becomes involved in bonding.
*   Now, Al is surrounded by four electron domains (three Al-Cl bonds and one Al←N coordinate bond). To accommodate these four electron domains, Al undergoes sp³ hybridisation.
*   The geometry around Al changes from trigonal planar to tetrahedral.

**4. Hybridisation of N in Cl3Al.NH3 (Product):**
*   In the adduct, the nitrogen atom still forms three single bonds with hydrogen atoms and one coordinate bond with the aluminium atom. The lone pair on nitrogen is now shared with aluminium.
*   The nitrogen atom still has four electron domains (three N-H bonds and one N→Al coordinate bond, where the lone pair is now a bond pair). Therefore, N remains sp³ hybridised.
*   The geometry around N remains trigonal pyramidal (though slightly distorted due to the new bond to Al).

**Conclusion:**
*   **Aluminium (Al) atom:** Changes from **sp²** hybridisation in AlCl3 to **sp³** hybridisation in Cl3Al.NH3.
*   **Nitrogen (N) atom:** Remains **sp³** hybridised in both NH3 and Cl3Al.NH3. There is no change in the hybridisation of the N atom.

Question 24

Q24. Is there any change in the hybridisation of B and N atoms as a result of the following reaction? 
BF3 + NH3 → F3B.NH3

Accepted Answer

This reaction is similar to the previous one, involving the formation of an adduct between Boron trifluoride (BF3) and Ammonia (NH3). BF3 acts as a Lewis acid and NH3 acts as a Lewis base. Let's analyze the hybridisation of B and N atoms before and after the reaction.

**1. Hybridisation of B in BF3 (Reactant):**
*   Boron (B) is the central atom. Its electronic configuration is [He] 2s² 2p¹.
*   In BF3, B forms three single bonds with three fluorine atoms. There are no lone pairs on B.
*   To form three bonds, B undergoes sp² hybridisation. One 2s orbital and two 2p orbitals mix to form three sp² hybrid orbitals.
*   The geometry of BF3 is trigonal planar.

**2. Hybridisation of N in NH3 (Reactant):**
*   Nitrogen (N) is the central atom. Its electronic configuration is [He] 2s² 2p³.
*   In NH3, N forms three single bonds with three hydrogen atoms and has one lone pair of electrons.
*   To accommodate three bond pairs and one lone pair, N undergoes sp³ hybridisation. One 2s orbital and three 2p orbitals mix to form four sp³ hybrid orbitals.
*   The geometry of NH3 is trigonal pyramidal.

**3. Hybridisation of B in F3B.NH3 (Product):**
*   In the adduct F3B.NH3, the nitrogen atom of NH3 donates its lone pair to the empty p-orbital of B in BF3, forming a coordinate covalent bond (dative bond).
*   Initially, B in BF3 was sp² hybridised with one empty unhybridised p-orbital. Upon accepting the lone pair from NH3, this empty p-orbital becomes involved in bonding.
*   Now, B is surrounded by four electron domains (three B-F bonds and one B←N coordinate bond). To accommodate these four electron domains, B undergoes sp³ hybridisation.
*   The geometry around B changes from trigonal planar to tetrahedral.

**4. Hybridisation of N in F3B.NH3 (Product):**
*   In the adduct, the nitrogen atom still forms three single bonds with hydrogen atoms and one coordinate bond with the boron atom. The lone pair on nitrogen is now shared with boron.
*   The nitrogen atom still has four electron domains (three N-H bonds and one N→B coordinate bond, where the lone pair is now a bond pair). Therefore, N remains sp³ hybridised.
*   The geometry around N remains trigonal pyramidal (though slightly distorted due to the new bond to B).

**Conclusion:**
*   **Boron (B) atom:** Changes from **sp²** hybridisation in BF3 to **sp³** hybridisation in F3B.NH3.
*   **Nitrogen (N) atom:** Remains **sp³** hybridised in both NH3 and F3B.NH3. There is no change in the hybridisation of the N atom.

Question 25

Q25. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.

Accepted Answer

**1. Formation of Double Bond in C2H4 (Ethene):**
*   **Hybridisation:** Each carbon atom in C2H4 undergoes sp² hybridisation. One 2s orbital and two 2p orbitals mix to form three sp² hybrid orbitals. One 2p orbital remains unhybridised.
*   **Sigma (σ) Bonds:**
    *   One sp² hybrid orbital from each carbon atom overlaps axially to form a C-C sigma bond.
    *   The remaining two sp² hybrid orbitals on each carbon atom overlap axially with the 1s orbitals of two hydrogen atoms to form two C-H sigma bonds per carbon.
*   **Pi (π) Bond:**
    *   The unhybridised 2p orbital on each carbon atom lies perpendicular to the plane of the sp² hybrid orbitals.
    *   These two unhybridised 2p orbitals (one from each carbon) overlap laterally (sideways) to form one C-C pi bond.
*   **Overall Structure:** The C=C double bond consists of one sigma bond and one pi bond. The molecule is planar with H-C-H and H-C-C bond angles approximately 120°.

**2. Formation of Triple Bond in C2H2 (Ethyne):**
*   **Hybridisation:** Each carbon atom in C2H2 undergoes sp hybridisation. One 2s orbital and one 2p orbital mix to form two sp hybrid orbitals. Two 2p orbitals remain unhybridised.
*   **Sigma (σ) Bonds:**
    *   One sp hybrid orbital from each carbon atom overlaps axially to form a C-C sigma bond.
    *   The other sp hybrid orbital on each carbon atom overlaps axially with the 1s orbital of one hydrogen atom to form one C-H sigma bond per carbon.
*   **Pi (π) Bonds:**
    *   The two unhybridised 2p orbitals on each carbon atom are perpendicular to each other and to the sp hybrid orbitals.
    *   One set of unhybridised 2p orbitals (e.g., 2py from each carbon) overlaps laterally to form the first C-C pi bond.
    *   The other set of unhybridised 2p orbitals (e.g., 2pz from each carbon) overlaps laterally to form the second C-C pi bond.
*   **Overall Structure:** The C≡C triple bond consists of one sigma bond and two pi bonds. The molecule is linear with H-C-C bond angles of 180°.

Question 26

Q26. What is the total number of sigma and pi bonds in the following molecules? 
(a) C2H2 
(b) C2H4

Accepted Answer

To determine the number of sigma (σ) and pi (π) bonds, we first need to draw the Lewis structure or condensed structural formula for each molecule, keeping in mind that:
*   A single bond consists of 1 σ bond.
*   A double bond consists of 1 σ bond and 1 π bond.
*   A triple bond consists of 1 σ bond and 2 π bonds.

**(a) C2H2 (Ethyne)**
*   **Structural Formula:** H—C≡C—H
*   **Counting Bonds:**
    *   C-H single bonds: 2 (each is 1 σ bond)
    *   C≡C triple bond: 1 (which is 1 σ bond + 2 π bonds)
*   **Total Sigma (σ) bonds:** 2 (from C-H) + 1 (from C≡C) = **3 σ bonds**
*   **Total Pi (π) bonds:** 0 (from C-H) + 2 (from C≡C) = **2 π bonds**

**(b) C2H4 (Ethene)**
*   **Structural Formula:** H₂C=CH₂ (or H-C=C-H with H's attached to each C)
*   **Counting Bonds:**
    *   C-H single bonds: 4 (each is 1 σ bond)
    *   C=C double bond: 1 (which is 1 σ bond + 1 π bond)
*   **Total Sigma (σ) bonds:** 4 (from C-H) + 1 (from C=C) = **5 σ bonds**
*   **Total Pi (π) bonds:** 0 (from C-H) + 1 (from C=C) = **1 π bond**

Question 27

Q27. Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + Cl– → AlCl4–

Accepted Answer

The hybridisation of the Al atom changes from sp² to sp³ in the given reaction.

**1. In AlCl₃ (Aluminium trichloride):**
*   The central atom is Aluminium (Al).
*   Aluminium is bonded to three chlorine atoms. There are no lone pairs on the Al atom.
*   Number of electron domains (bond pairs + lone pairs) = 3 + 0 = 3.
*   According to VSEPR theory, a steric number of 3 corresponds to sp² hybridisation.
*   The geometry of AlCl₃ is trigonal planar.

**2. In AlCl₄⁻ (Tetrachloroaluminate ion):**
*   The central atom is Aluminium (Al).
*   Aluminium is bonded to four chlorine atoms. There are no lone pairs on the Al atom.
*   Number of electron domains (bond pairs + lone pairs) = 4 + 0 = 4.
*   According to VSEPR theory, a steric number of 4 corresponds to sp³ hybridisation.
*   The geometry of AlCl₄⁻ is tetrahedral.

**Conclusion:** The Al atom undergoes a change in hybridisation from sp² in AlCl₃ to sp³ in AlCl₄⁻.

Question 28

Q28. Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF3 + NH3 → F3B.NH3

Accepted Answer

Yes, there is a change in the hybridisation of the Boron (B) atom, but no change in the hybridisation of the Nitrogen (N) atom as a result of the reaction.

**1. In BF₃ (Boron trifluoride):**
*   The central atom is Boron (B).
*   Boron is bonded to three fluorine atoms. There are no lone pairs on the B atom.
*   Number of electron domains (bond pairs + lone pairs) = 3 + 0 = 3.
*   This corresponds to sp² hybridisation. The geometry of BF₃ is trigonal planar.

**2. In NH₃ (Ammonia):**
*   The central atom is Nitrogen (N).
*   Nitrogen is bonded to three hydrogen atoms and has one lone pair of electrons.
*   Number of electron domains (bond pairs + lone pairs) = 3 + 1 = 4.
*   This corresponds to sp³ hybridisation. The geometry of NH₃ is pyramidal.

**3. In F₃B.NH₃ (Boron trifluoride-ammonia adduct):**
*   This is an adduct formed by a coordinate bond between N (donor) and B (acceptor).
*   **For Boron (B):** In the adduct, Boron forms three bonds with fluorine atoms and one coordinate bond with the nitrogen atom. Thus, it is surrounded by four electron domains (four bond pairs) and has no lone pairs. The steric number is 4.
    *   This corresponds to sp³ hybridisation. The geometry around B is tetrahedral.
    *   **Change for B: sp² → sp³**
*   **For Nitrogen (N):** In the adduct, Nitrogen forms three bonds with hydrogen atoms and one coordinate bond with the boron atom. The lone pair on nitrogen is used to form this coordinate bond, but it still contributes to the electron domains around nitrogen. Thus, it is surrounded by four electron domains (four bond pairs) and has no lone pairs remaining as a 'free' lone pair. The steric number is 4.
    *   This corresponds to sp³ hybridisation. The geometry around N is tetrahedral.
    *   **Change for N: No change in hybridisation (sp³ → sp³),** although its geometry changes from pyramidal to tetrahedral due to the formation of the fourth bond.

Question 29

Q29. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.

Accepted Answer

The formation of double and triple bonds between carbon atoms involves the hybridisation of carbon orbitals and subsequent sigma (σ) and pi (π) bond formation through orbital overlap.

**1. Formation of a Double Bond in C₂H₄ (Ethene):**
*   **Hybridisation:** Each carbon atom in ethene undergoes sp² hybridisation. This means one 2s orbital mixes with two 2p orbitals to form three sp² hybrid orbitals. One 2p orbital remains unhybridised.
*   **Sigma (σ) Bonds:**
    *   Each carbon atom uses two of its sp² hybrid orbitals to form sigma bonds with two hydrogen atoms (sp²-s overlap).
    *   The remaining sp² hybrid orbital on each carbon atom overlaps axially with an sp² hybrid orbital from the other carbon atom to form a C-C sigma bond (sp²-sp² overlap).
*   **Pi (π) Bond:**
    *   The unhybridised 2p orbital on each carbon atom lies perpendicular to the plane of the sp² hybrid orbitals.
    *   These two unhybridised 2p orbitals (one from each carbon) overlap laterally (sideways) to form one C-C pi bond.
*   **Result:** The C=C double bond consists of one sigma bond and one pi bond. The molecule is planar with bond angles approximately 120°.

**2. Formation of a Triple Bond in C₂H₂ (Ethyne):**
*   **Hybridisation:** Each carbon atom in ethyne undergoes sp hybridisation. This means one 2s orbital mixes with one 2p orbital to form two sp hybrid orbitals. Two 2p orbitals (e.g., 2py and 2pz) remain unhybridised.
*   **Sigma (σ) Bonds:**
    *   Each carbon atom uses one of its sp hybrid orbitals to form a sigma bond with one hydrogen atom (sp-s overlap).
    *   The remaining sp hybrid orbital on each carbon atom overlaps axially with an sp hybrid orbital from the other carbon atom to form a C-C sigma bond (sp-sp overlap).
*   **Pi (π) Bonds:**
    *   The two unhybridised 2p orbitals on each carbon atom (e.g., 2py and 2pz) are perpendicular to each other and to the sp hybrid orbitals.
    *   One set of unhybridised 2p orbitals (e.g., 2py from each carbon) overlaps laterally to form the first C-C pi bond.
    *   The other set of unhybridised 2p orbitals (e.g., 2pz from each carbon) also overlaps laterally to form the second C-C pi bond.
*   **Result:** The C≡C triple bond consists of one sigma bond and two pi bonds. The molecule is linear with bond angles of 180°.

**Diagram Description:**

**For C₂H₄ (Ethene):**
*   Show two carbon atoms, each with three sp² hybrid orbitals (arranged trigonal planar, 120° apart) and one unhybridised p-orbital (perpendicular to the sp² plane).
*   Illustrate the axial overlap of one sp² orbital from each carbon to form the C-C sigma bond.
*   Show the axial overlap of the remaining two sp² orbitals on each carbon with 1s orbitals of hydrogen atoms to form C-H sigma bonds.
*   Depict the lateral (sideways) overlap of the two unhybridised p-orbitals (one from each carbon) above and below the internuclear axis to form the C-C pi bond.
*   The overall structure should show a planar molecule.

**For C₂H₂ (Ethyne):**
*   Show two carbon atoms, each with two sp hybrid orbitals (arranged linearly, 180° apart) and two unhybridised p-orbitals (perpendicular to each other and to the sp orbitals).
*   Illustrate the axial overlap of one sp orbital from each carbon to form the C-C sigma bond.
*   Show the axial overlap of the remaining sp orbital on each carbon with a 1s orbital of a hydrogen atom to form C-H sigma bonds.
*   Depict the lateral (sideways) overlap of one set of unhybridised p-orbitals (e.g., px) above and below the internuclear axis to form the first C-C pi bond.
*   Depict the lateral (sideways) overlap of the second set of unhybridised p-orbitals (e.g., py) in front and behind the internuclear axis to form the second C-C pi bond.
*   The overall structure should show a linear molecule.

Question 30

Q30. Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + Cl– → AlCl4–

Accepted Answer

To determine the change in hybridisation of the Al atom, we need to analyze its bonding environment in both AlCl3 and AlCl4–.

**1. Hybridisation of Al in AlCl3:**
*   The central atom is Aluminium (Al).
*   Aluminium is bonded to three chlorine (Cl) atoms.
*   There are no lone pairs on the central Al atom (Al is in Group 13, has 3 valence electrons, all used in bonding).
*   The number of electron domains (bond pairs + lone pairs) around Al is 3 + 0 = 3.
*   According to VSEPR theory and hybridisation rules, a steric number of 3 corresponds to **sp2 hybridisation**.
*   The geometry of AlCl3 is trigonal planar.

**2. Hybridisation of Al in AlCl4–:**
*   The central atom is Aluminium (Al).
*   Aluminium is bonded to four chlorine (Cl) atoms. The negative charge indicates that Al has accepted a lone pair from Cl– to form a coordinate bond.
*   There are no lone pairs on the central Al atom.
*   The number of electron domains (bond pairs + lone pairs) around Al is 4 + 0 = 4.
*   According to VSEPR theory and hybridisation rules, a steric number of 4 corresponds to **sp3 hybridisation**.
*   The geometry of AlCl4– is tetrahedral.

**Conclusion:**
In the reaction AlCl3 + Cl– → AlCl4–, the hybridisation of the Al atom changes from **sp2 to sp3**.

Question 31

Q31. Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF3 + NH3 → F3B.NH3

Accepted Answer

To determine the change in hybridisation, we will analyze the hybridisation of B and N atoms before and after the reaction.

**1. Hybridisation of B and N in reactants (BF3 and NH3):**

*   **In BF3:**
    *   Central atom: Boron (B).
    *   Boron is bonded to three fluorine (F) atoms.
    *   Number of valence electrons on B = 3. All are used in bonding, so no lone pairs.
    *   Number of electron domains (bond pairs + lone pairs) = 3 + 0 = 3.
    *   Hybridisation of B in BF3 is **sp2**.
    *   Geometry: Trigonal planar.

*   **In NH3:**
    *   Central atom: Nitrogen (N).
    *   Nitrogen is bonded to three hydrogen (H) atoms.
    *   Number of valence electrons on N = 5. Three are used in bonding, leaving one lone pair.
    *   Number of electron domains (bond pairs + lone pairs) = 3 + 1 = 4.
    *   Hybridisation of N in NH3 is **sp3**.
    *   Geometry: Pyramidal.

**2. Hybridisation of B and N in product (F3B.NH3, an adduct):**

*   In the adduct F3B.NH3, a coordinate (dative) bond is formed between the lone pair on Nitrogen of NH3 and the vacant p-orbital of Boron in BF3.

*   **In F3B.NH3 (for B atom):**
    *   Central atom: Boron (B).
    *   Boron is now bonded to three fluorine atoms and one nitrogen atom (via a coordinate bond).
    *   Number of electron domains (bond pairs + lone pairs) = 4 + 0 = 4.
    *   Hybridisation of B in F3B.NH3 is **sp3**.
    *   Geometry: Tetrahedral.

*   **In F3B.NH3 (for N atom):**
    *   Central atom: Nitrogen (N).
    *   Nitrogen is bonded to three hydrogen atoms and one boron atom (the lone pair on N is now shared as a bond pair).
    *   Number of electron domains (bond pairs + lone pairs) = 4 + 0 = 4.
    *   Hybridisation of N in F3B.NH3 is **sp3**.
    *   Geometry: Tetrahedral.

**Conclusion:**
*   The hybridisation of the **Boron (B) atom changes from sp2 to sp3**.
*   The hybridisation of the **Nitrogen (N) atom remains sp3** (although its geometry changes from pyramidal to tetrahedral as the lone pair becomes a bond pair).

Question 32

Q32. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.

Accepted Answer

The formation of double and triple bonds between carbon atoms involves the concept of hybridisation and the overlap of atomic orbitals (both hybridized and unhybridized) to form sigma (σ) and pi (π) bonds.

### 1. Formation of a Double Bond in C2H4 (Ethene)

In C2H4, each carbon atom is sp2 hybridised.

**Steps for bond formation:**
1.  **Hybridisation:** Each carbon atom undergoes sp2 hybridisation, forming three sp2 hybrid orbitals and leaving one unhybridised p-orbital perpendicular to the plane of the sp2 orbitals.
2.  **C-C Sigma (σ) bond:** One sp2 hybrid orbital from each carbon atom overlaps axially to form a strong C-C sigma bond.
3.  **C-H Sigma (σ) bonds:** The remaining two sp2 hybrid orbitals on each carbon atom overlap axially with the 1s orbital of two hydrogen atoms to form two C-H sigma bonds.
4.  **C-C Pi (π) bond:** The unhybridised p-orbital on one carbon atom overlaps laterally (sideways) with the unhybridised p-orbital on the other carbon atom to form one C-C pi bond. This pi bond consists of two electron clouds, one above and one below the plane of the C-C sigma bond.

**Overall structure:** Ethene has one C-C double bond (one σ and one π bond) and four C-H single bonds (all σ bonds). The molecule is planar.

### 2. Formation of a Triple Bond in C2H2 (Ethyne)

In C2H2, each carbon atom is sp hybridised.

**Steps for bond formation:**
1.  **Hybridisation:** Each carbon atom undergoes sp hybridisation, forming two sp hybrid orbitals and leaving two unhybridised p-orbitals (e.g., px and py) perpendicular to each other and to the axis of the sp hybrid orbitals.
2.  **C-C Sigma (σ) bond:** One sp hybrid orbital from each carbon atom overlaps axially to form a strong C-C sigma bond.
3.  **C-H Sigma (σ) bonds:** The remaining sp hybrid orbital on each carbon atom overlaps axially with the 1s orbital of one hydrogen atom to form one C-H sigma bond.
4.  **C-C Pi (π) bonds:** The two unhybridised p-orbitals on one carbon atom (e.g., px and py) overlap laterally with the corresponding two unhybridised p-orbitals on the other carbon atom (px and py) to form two C-C pi bonds. These two pi bonds are perpendicular to each other and surround the C-C sigma bond.

**Overall structure:** Ethyne has one C-C triple bond (one σ and two π bonds) and two C-H single bonds (all σ bonds). The molecule is linear.

Question 33

Q33. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl₃ + Cl⁻ → AlCl₄⁻

Accepted Answer

To determine the change in hybridisation of the Al atom, we need to analyze the hybridisation of Al in both AlCl₃ and AlCl₄⁻.

**1. Hybridisation of Al in AlCl₃:**
*   Aluminum (Al) is the central atom. Its electronic configuration is [Ne] 3s²3p¹. It has 3 valence electrons.
*   In AlCl₃, Al forms three single covalent bonds with three chlorine atoms.
*   There are no lone pairs on the Al atom.
*   The number of electron domains (bond pairs + lone pairs) around Al is 3 (3 bond pairs + 0 lone pairs).
*   According to VSEPR theory, a steric number of 3 corresponds to sp² hybridisation and a trigonal planar geometry.

**2. Hybridisation of Al in AlCl₄⁻:**
*   In AlCl₄⁻, the Al atom is bonded to four chlorine atoms. The negative charge is on the complex.
*   The Al atom forms four single covalent bonds with four chlorine atoms.
*   There are no lone pairs on the Al atom.
*   The number of electron domains (bond pairs + lone pairs) around Al is 4 (4 bond pairs + 0 lone pairs).
*   According to VSEPR theory, a steric number of 4 corresponds to sp³ hybridisation and a tetrahedral geometry.

**Conclusion:**
During the reaction AlCl₃ + Cl⁻ → AlCl₄⁻, the hybridisation of the Al atom changes from **sp² to sp³**.

Question 34

Q34. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF₃ + NH₃ → F₃B.NH₃

Accepted Answer

Let's analyze the hybridisation of B and N atoms before and after the reaction.

**1. Hybridisation in reactants (BF₃ and NH₃):**
*   **In BF₃:**
    *   Boron (B) is the central atom. Its electronic configuration is [He] 2s²2p¹. It has 3 valence electrons.
    *   In BF₃, B forms three single covalent bonds with three fluorine atoms.
    *   There are no lone pairs on the B atom.
    *   The number of electron domains (bond pairs + lone pairs) around B is 3 (3 bond pairs + 0 lone pairs).
    *   This corresponds to **sp² hybridisation** and a trigonal planar geometry.
*   **In NH₃:**
    *   Nitrogen (N) is the central atom. Its electronic configuration is [He] 2s²2p³. It has 5 valence electrons.
    *   In NH₃, N forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons.
    *   The number of electron domains around N is 4 (3 bond pairs + 1 lone pair).
    *   This corresponds to **sp³ hybridisation** and a pyramidal geometry.

**2. Hybridisation in product (F₃B.NH₃, an adduct):**
*   In the adduct F₃B.NH₃, a dative (coordinate) bond is formed between the nitrogen atom of NH₃ and the boron atom of BF₃, where nitrogen donates its lone pair to boron.
*   **For Boron (B) in F₃B.NH₃:**
    *   Boron is now bonded to three fluorine atoms and one nitrogen atom (via a dative bond). It forms a total of four bonds.
    *   There are no lone pairs on the B atom.
    *   The number of electron domains around B is 4 (4 bond pairs + 0 lone pairs).
    *   This corresponds to **sp³ hybridisation** and a tetrahedral geometry.
*   **For Nitrogen (N) in F₃B.NH₃:**
    *   Nitrogen is still bonded to three hydrogen atoms and now also to one boron atom (via a dative bond). The lone pair that was present on N in NH₃ is now shared with B.
    *   The number of electron domains around N is 4 (3 N-H bond pairs + 1 N-B bond pair).
    *   This corresponds to **sp³ hybridisation** and a tetrahedral geometry around the nitrogen atom.

**Conclusion:**
*   The hybridisation of the **Boron (B) atom changes from sp² to sp³**.
*   The hybridisation of the **Nitrogen (N) atom remains sp³** (no change).

Question 35

Q35. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.

Accepted Answer

The formation of double and triple bonds involves both sigma (σ) and pi (π) bonds. Sigma bonds are formed by head-on (axial) overlap of atomic orbitals, while pi bonds are formed by lateral (sideways) overlap of unhybridised p-orbitals.

**1. Formation of a Double Bond in C₂H₄ (Ethene):**
*   Each carbon atom in ethene is sp² hybridised. This means each carbon has three sp² hybrid orbitals and one unhybridised p-orbital.
*   **Sigma (σ) bonds:**
    *   One C-C sigma bond is formed by the axial overlap of an sp² hybrid orbital from each carbon atom.
    *   Four C-H sigma bonds are formed by the axial overlap of an sp² hybrid orbital from each carbon atom with the 1s orbital of a hydrogen atom (two C-H bonds per carbon).
*   **Pi (π) bond:**
    *   The unhybridised p-orbital on each carbon atom lies perpendicular to the plane of the sp² hybrid orbitals.
    *   These two unhybridised p-orbitals undergo lateral (sideways) overlap above and below the plane of the C-C sigma bond, forming one pi (π) bond.
*   The double bond consists of one sigma bond and one pi bond.

**2. Formation of a Triple Bond in C₂H₂ (Ethyne):**
*   Each carbon atom in ethyne is sp hybridised. This means each carbon has two sp hybrid orbitals and two unhybridised p-orbitals.
*   **Sigma (σ) bonds:**
    *   One C-C sigma bond is formed by the axial overlap of an sp hybrid orbital from each carbon atom.
    *   Two C-H sigma bonds are formed by the axial overlap of an sp hybrid orbital from each carbon atom with the 1s orbital of a hydrogen atom (one C-H bond per carbon).
*   **Pi (π) bonds:**
    *   Each carbon atom has two unhybridised p-orbitals (e.g., 2p_y and 2p_z) that are perpendicular to each other and to the C-C bond axis.
    *   These two sets of p-orbitals undergo lateral (sideways) overlap. One pair of p-orbitals (e.g., 2p_y from each carbon) forms one pi bond, and the other pair of p-orbitals (e.g., 2p_z from each carbon) forms a second pi bond.
    *   These two pi bonds are perpendicular to each other and surround the C-C sigma bond.
*   The triple bond consists of one sigma bond and two pi bonds.

Question 36

Q36. What is the total number of sigma and pi bonds in the following molecules?
(a) C2H2 (b) C2H4

Accepted Answer

To determine the number of sigma (σ) and pi (π) bonds, we first draw the Lewis structure for each molecule.

**(a) C2H2 (Ethyne/Acetylene)**
Structure: H—C≡C—H
- There are two C—H single bonds. Each single bond is a sigma (σ) bond.
- There is one C≡C triple bond. A triple bond consists of one sigma (σ) bond and two pi (π) bonds.
Total sigma bonds = 2 (C—H) + 1 (C—C) = 3 σ bonds
Total pi bonds = 2 (C≡C) = 2 π bonds

**(b) C2H4 (Ethene/Ethylene)**
Structure: H2C=CH2
- There are four C—H single bonds. Each single bond is a sigma (σ) bond.
- There is one C=C double bond. A double bond consists of one sigma (σ) bond and one pi (π) bond.
Total sigma bonds = 4 (C—H) + 1 (C—C) = 5 σ bonds
Total pi bonds = 1 (C=C) = 1 π bond

Question 37

Q37. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Accepted Answer

The formation of molecular orbitals (MOs) from atomic orbitals (AOs) through the Linear Combination of Atomic Orbitals (LCAO) method is governed by several important conditions. These conditions ensure effective overlap and stable molecular orbital formation.

### Important Conditions for LCAO:

1.  **The combining atomic orbitals must have comparable energies:**
    *   For effective overlap, the atomic orbitals participating in the combination must have energies that are close to each other. For example, a 1s orbital of one atom can combine effectively with a 1s orbital of another atom, or a 2s orbital with a 2s orbital, or a 2p orbital with a 2p orbital. However, a 1s orbital will not combine effectively with a 2s or 2p orbital because their energy difference is too large. This condition is particularly crucial for heteronuclear diatomic molecules where atomic orbitals of different atoms combine.

2.  **The combining atomic orbitals must have proper symmetry about the molecular axis:**
    *   The atomic orbitals must have the same symmetry properties with respect to the molecular axis. For instance, a 2s orbital can combine with a 2pz orbital (if z is the internuclear axis) because both are cylindrically symmetrical around the z-axis. However, a 2pz orbital cannot combine with a 2px or 2py orbital to form sigma molecular orbitals because their symmetries do not match along the internuclear axis. This ensures maximum overlap and constructive interference.

3.  **The combining atomic orbitals must overlap to the maximum possible extent:**
    *   The greater the extent of overlap between the atomic orbitals, the stronger will be the bond formed and the greater will be the electron density in the bonding molecular orbital. This leads to a more stable molecular orbital. The atoms must approach each other to a distance where their orbitals can effectively overlap. If the overlap is zero or very small, no molecular orbital will be formed, or it will be very weak.

These three conditions are fundamental to the success of the LCAO approach in explaining chemical bonding and predicting molecular properties.

Question 38

Q38. Predict the hybridization of each carbon atom in the following molecules:
(a) CH2=C=O (b) CH3-CH=CH2 (c) CH3-C≡CH (d) C6H6

Accepted Answer

To determine the hybridization of a carbon atom, we count the number of sigma bonds and lone pairs around it. The steric number (sum of sigma bonds and lone pairs) determines the hybridization:
*   Steric number 4: sp3 hybridization
*   Steric number 3: sp2 hybridization
*   Steric number 2: sp hybridization

**(a) CH2=C=O (Ketene)**
Structure: H₂C=C=O
*   **Carbon 1 (CH2=):** This carbon forms two C-H single bonds (2σ) and one C=C double bond (1σ, 1π). Total sigma bonds = 2 + 1 = 3. No lone pairs. Steric number = 3. **Hybridization: sp2**
*   **Carbon 2 (=C=O):** This carbon forms one C=C double bond (1σ, 1π) and one C=O double bond (1σ, 1π). Total sigma bonds = 1 + 1 = 2. No lone pairs. Steric number = 2. **Hybridization: sp**

**(b) CH3-CH=CH2 (Propene)**
Structure: CH₃—CH=CH₂
*   **Carbon 1 (CH3-):** This carbon forms three C-H single bonds (3σ) and one C-C single bond (1σ). Total sigma bonds = 3 + 1 = 4. No lone pairs. Steric number = 4. **Hybridization: sp3**
*   **Carbon 2 (-CH=):** This carbon forms one C-C single bond (1σ), one C-H single bond (1σ), and one C=C double bond (1σ, 1π). Total sigma bonds = 1 + 1 + 1 = 3. No lone pairs. Steric number = 3. **Hybridization: sp2**
*   **Carbon 3 (=CH2):** This carbon forms two C-H single bonds (2σ) and one C=C double bond (1σ, 1π). Total sigma bonds = 2 + 1 = 3. No lone pairs. Steric number = 3. **Hybridization: sp2**

**(c) CH3-C≡CH (Propyne)**
Structure: CH₃—C≡CH
*   **Carbon 1 (CH3-):** This carbon forms three C-H single bonds (3σ) and one C-C single bond (1σ). Total sigma bonds = 3 + 1 = 4. No lone pairs. Steric number = 4. **Hybridization: sp3**
*   **Carbon 2 (-C≡):** This carbon forms one C-C single bond (1σ) and one C≡C triple bond (1σ, 2π). Total sigma bonds = 1 + 1 = 2. No lone pairs. Steric number = 2. **Hybridization: sp**
*   **Carbon 3 (≡CH):** This carbon forms one C≡C triple bond (1σ, 2π) and one C-H single bond (1σ). Total sigma bonds = 1 + 1 = 2. No lone pairs. Steric number = 2. **Hybridization: sp**

**(d) C6H6 (Benzene)**
Structure: A cyclic ring of six carbon atoms, each bonded to one hydrogen atom, with alternating single and double bonds (or delocalized pi system).
*   **Each Carbon Atom:** In benzene, each carbon atom is identical. Each carbon atom forms two C-C bonds (one single, one double in a resonance structure, or 1.5 bonds in reality, but for hybridization, we count sigma bonds) and one C-H single bond. For hybridization, we consider each carbon to be bonded to three other atoms (two carbons and one hydrogen) via sigma bonds. Total sigma bonds = 3. No lone pairs. Steric number = 3. **Hybridization: sp2** (for all six carbon atoms)

Question 39

Q39. Describe the change in hybridisation (if any) of the Al atom in the following reaction.
AlCl3 + Cl– → AlCl4–

Accepted Answer

The hybridisation of the Al atom changes from sp2 to sp3 in the given reaction.

Question 40

Q40. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF3 + NH3 → F3B.NH3

Accepted Answer

Yes, there is a change in the hybridisation of the Boron (B) atom, but no change in the hybridisation of the Nitrogen (N) atom as a result of the reaction.

Chapter 4: Chemical Bonding and Molecular Structure