Question 1

Q1. Discuss the general characteristics of p-block elements with reference to their electronic configuration, oxidation states, atomic radii, ionisation enthalpy and electronegativity.

Accepted Answer

The p-block elements are located on the right side of the periodic table, encompassing groups 13 to 18. Their general characteristics are primarily influenced by the filling of the p-orbitals.

### 1. Electronic Configuration
*   **General Outer Electronic Configuration**: The general outer electronic configuration of p-block elements is ns²np¹⁻⁶. This means that the differentiating electron enters the p-subshell of the outermost shell.
*   **Group-wise Configuration**: 
    *   Group 13 (Boron family): ns²np¹
    *   Group 14 (Carbon family): ns²np²
    *   Group 15 (Nitrogen family): ns²np³
    *   Group 16 (Oxygen family): ns²np⁴
    *   Group 17 (Halogen family): ns²np⁵
    *   Group 18 (Noble gases): ns²np⁶ (except Helium, which is 1s²)

### 2. Oxidation States
*   **Variety of Oxidation States**: p-block elements exhibit a variety of oxidation states, both positive and negative, due to the availability of multiple valence electrons (s and p electrons) and sometimes empty d-orbitals in heavier elements.
*   **Group Oxidation State**: The maximum positive oxidation state is generally equal to the group number minus 10 (e.g., Group 13: +3, Group 14: +4, Group 15: +5, Group 16: +6, Group 17: +7). This corresponds to the total number of valence electrons (ns + np electrons).
*   **Inert Pair Effect**: For heavier elements in each group (especially from Group 13 onwards), the tendency to exhibit the group oxidation state decreases down the group. Instead, they prefer to show an oxidation state two units less than the group oxidation state (e.g., +1 for Tl, +2 for Pb, +3 for Bi). This is due to the 'inert pair effect', where the ns² electrons become increasingly reluctant to participate in bonding because of poor shielding by intervening d and f electrons, leading to increased effective nuclear charge and stronger attraction of ns electrons to the nucleus.
*   **Negative Oxidation States**: Elements in groups 15, 16, and 17 can also exhibit negative oxidation states by gaining electrons to complete their octet (e.g., -3 for N, P; -2 for O, S; -1 for F, Cl).

### 3. Atomic Radii
*   **Trend Across a Period**: Atomic radii generally decrease across a period from left to right in the p-block. This is because the effective nuclear charge increases due to the addition of electrons into the same principal shell, pulling the valence electrons closer to the nucleus.
*   **Trend Down a Group**: Atomic radii generally increase down a group. This is due to the addition of new principal energy shells, which increases the distance of the valence electrons from the nucleus and also increases shielding by inner electrons.
*   **Exceptions/Anomalies**: 
    *   In Group 13, there is a slight anomaly: Ga has a slightly smaller atomic radius than Al. This is due to the poor shielding effect of the 3d electrons in Ga, which leads to a higher effective nuclear charge.
    *   Similar effects are observed in other groups due to the presence of d and f electrons in heavier elements.

### 4. Ionisation Enthalpy
*   **Trend Across a Period**: Ionisation enthalpy generally increases across a period from left to right. This is because the atomic size decreases and the effective nuclear charge increases, making it harder to remove an electron.
*   **Trend Down a Group**: Ionisation enthalpy generally decreases down a group. This is due to the increase in atomic size and shielding effect, which reduces the attraction between the nucleus and the outermost electrons, making them easier to remove.
*   **Exceptions/Anomalies**: 
    *   Small irregularities are observed due to factors like stable half-filled or fully-filled orbitals (e.g., Group 15 elements have higher IE than Group 16 elements due to stable half-filled p-orbitals).
    *   The inert pair effect in heavier elements can also influence ionisation enthalpies, as the ns² electrons are held more tightly.

### 5. Electronegativity
*   **Trend Across a Period**: Electronegativity generally increases across a period from left to right. This is because the effective nuclear charge increases and atomic size decreases, leading to a stronger attraction for shared electrons.
*   **Trend Down a Group**: Electronegativity generally decreases down a group. This is due to the increase in atomic size and shielding effect, which reduces the attraction of the nucleus for shared electrons.
*   **Highest Electronegativity**: Fluorine (F) is the most electronegative element in the entire periodic table, located in the p-block.
*   **Metallic Character**: The metallic character increases down a group and decreases across a period in the p-block. The elements on the left side of the p-block (e.g., Al, Ga) are metals, while those on the right (e.g., F, Cl, O) are non-metals. Metalloids (e.g., B, Si, Ge, As, Sb, Te) are found along the diagonal separating metals and non-metals.

Question 2

Q2. How would you explain the anomalous behaviour of the first member of p-block elements with respect to other members of the corresponding group?

Accepted Answer

The first member of each group in the p-block (e.g., Boron in Group 13, Carbon in Group 14, Nitrogen in Group 15, Oxygen in Group 16, Fluorine in Group 17) exhibits anomalous or different behaviour compared to the rest of the members of its respective group. This anomalous behaviour can be attributed to three main reasons:

### 1. Small Size
*   The first member of each group has an exceptionally small atomic and ionic size compared to other elements in the group. This small size leads to a very high charge density (charge/radius ratio).
*   **Consequence**: High polarizing power (tendency to distort electron clouds of anions) and a greater tendency to form covalent bonds rather than ionic bonds (e.g., Li, Be, B form predominantly covalent compounds).

### 2. High Electronegativity and High Ionisation Enthalpy
*   Due to their small size, the first members have a very high electronegativity and high ionisation enthalpy.
*   **Consequence**: They have a strong tendency to attract electrons (high electronegativity) and require a large amount of energy to remove an electron (high ionisation enthalpy). This influences their bonding characteristics and reactivity.

### 3. Absence of d-orbitals in their Valence Shell
*   This is the most significant reason for anomalous behaviour. The first member of each p-block group only has s and p orbitals in its valence shell (e.g., 2s and 2p for second period elements). They do not have vacant d-orbitals.
*   **Consequences**:
    *   **Maximum Covalency**: The maximum covalency of the first member is limited to four (one s and three p orbitals). They cannot expand their octet. For example, nitrogen can form NF₃ but not NF₅, whereas phosphorus can form PCl₅. Boron forms BF₄⁻ but not BF₆³⁻, while aluminium forms AlF₆³⁻.
    *   **Formation of pπ-pπ Multiple Bonds**: Due to their small size and the absence of d-orbitals, the first members can form effective pπ-pπ multiple bonds with themselves and with other small, highly electronegative elements (e.g., C=C, C≡C, N≡N, C=O, C≡N, O=O). Heavier elements, due to their larger size, cannot form effective pπ-pπ bonds; instead, they prefer to form single bonds or dπ-pπ/dπ-dπ bonds (if d-orbitals are available).
    *   **Differences in Allotropy**: The ability to form pπ-pπ bonds leads to different allotropic forms and molecular structures (e.g., N₂ is a gas with a triple bond, while P₄ is a solid with single bonds; O₂ is a gas, S₈ is a solid).
    *   **Differences in Hydrides/Oxides**: The stability and nature of hydrides and oxides can also differ significantly (e.g., water is a liquid due to hydrogen bonding, H₂S is a gas).

**Examples of Anomalous Behaviour:**
*   **Boron (Group 13)**: Forms only covalent compounds, maximum covalency is 4 (e.g., BF₄⁻). Does not form B³⁺ ions. Forms electron-deficient compounds. Aluminium forms Al³⁺ ions and AlF₆³⁻.
*   **Carbon (Group 14)**: Exhibits catenation to a much greater extent than other group members. Forms stable C=C and C≡C bonds. Silicon forms Si=Si bonds only under specific conditions and catenates less.
*   **Nitrogen (Group 15)**: Exists as a diatomic gas (N≡N) due to strong pπ-pπ bonding. Does not form pentahalides (e.g., NF₅). Phosphorus exists as P₄ and forms PCl₅.
*   **Oxygen (Group 16)**: Exists as a diatomic gas (O=O). Exhibits hydrogen bonding in compounds like H₂O. Sulphur exists as S₈ and does not form hydrogen bonds to the same extent.

Question 3

Q3. What is the oxidation state of Al in AlCl₃?

Accepted Answer

To determine the oxidation state of Al in AlCl₃, we use the following rules:
1.  The sum of the oxidation states of all atoms in a neutral compound is zero.
2.  Chlorine (Cl) typically has an oxidation state of -1 in its compounds, especially with less electronegative elements like metals.

Let the oxidation state of Al be 'x'.
There is one Al atom and three Cl atoms.

So, x + 3 × (-1) = 0
x - 3 = 0
x = +3

Therefore, the oxidation state of Al in AlCl₃ is +3.

Question 4

Q4. Why does boron trifluoride behave as a Lewis acid?

Accepted Answer

Boron trifluoride (BF₃) behaves as a Lewis acid because it is an electron-deficient molecule, meaning it has an incomplete octet in its valence shell.

**Explanation:**
1.  **Electronic Configuration of Boron**: Boron (B) is in Group 13 and has an atomic number of 5. Its electronic configuration is 1s²2s²2p¹.
2.  **Valence Electrons**: In BF₃, boron forms three covalent bonds with three fluorine atoms. Each bond involves sharing one electron from boron and one from fluorine.
3.  **Electron Count Around Boron**: After forming three covalent bonds, the boron atom has only 3 × 2 = 6 electrons in its outermost shell. This is less than the stable octet (8 electrons).
4.  **Empty Orbital**: Since boron has only used its 2s and two 2p orbitals for bonding (sp² hybridization), there is still one vacant 2p orbital available in the valence shell of the boron atom.
5.  **Lewis Acid Definition**: A Lewis acid is defined as an electron-pair acceptor. Due to the incomplete octet and the presence of a vacant 2p orbital, the boron atom in BF₃ can readily accept a lone pair of electrons from an electron-rich species (a Lewis base) to complete its octet.

For example, BF₃ reacts with ammonia (NH₃), where nitrogen donates its lone pair to boron, forming an adduct:
BF₃ + :NH₃ → F₃B ← NH₃ (adduct formation)

Question 5

Q5. Consider the compounds, BCl₃ and CCl₄. Predict the hybridisation and state whether the molecule is planar or non-planar.

Accepted Answer

Let's analyze each compound separately:

### 1. Boron trichloride (BCl₃)

**a. Hybridisation of Boron:**
*   **Central Atom**: Boron (B)
*   **Valence Electrons of Boron**: 3 (Group 13)
*   **Number of Chlorine Atoms Bonded**: 3
*   **Lone Pairs on Boron**: Boron forms three single bonds with three chlorine atoms. Since boron has 3 valence electrons and uses all of them in bonding, there are no lone pairs on the central boron atom.
*   **Steric Number**: Number of bond pairs + Number of lone pairs = 3 + 0 = 3
*   **Hybridisation**: A steric number of 3 corresponds to **sp² hybridisation**.

**b. Geometry and Planarity:**
*   **Electron Geometry**: Trigonal planar (due to sp² hybridisation).
*   **Molecular Geometry**: With no lone pairs on the central atom, the molecular geometry is also **trigonal planar**.
*   **Planarity**: Since all four atoms (one B and three Cl) lie in the same plane, BCl₃ is a **planar molecule**.
*   **Bond Angle**: The Cl-B-Cl bond angles are 120°.

### 2. Carbon tetrachloride (CCl₄)

**a. Hybridisation of Carbon:**
*   **Central Atom**: Carbon (C)
*   **Valence Electrons of Carbon**: 4 (Group 14)
*   **Number of Chlorine Atoms Bonded**: 4
*   **Lone Pairs on Carbon**: Carbon forms four single bonds with four chlorine atoms. Since carbon has 4 valence electrons and uses all of them in bonding, there are no lone pairs on the central carbon atom.
*   **Steric Number**: Number of bond pairs + Number of lone pairs = 4 + 0 = 4
*   **Hybridisation**: A steric number of 4 corresponds to **sp³ hybridisation**.

**b. Geometry and Planarity:**
*   **Electron Geometry**: Tetrahedral (due to sp³ hybridisation).
*   **Molecular Geometry**: With no lone pairs on the central atom, the molecular geometry is also **tetrahedral**.
*   **Planarity**: In a tetrahedral geometry, the central atom and the four surrounding atoms do not lie in the same plane. The molecule has a 3D structure. Therefore, CCl₄ is a **non-planar molecule**.
*   **Bond Angle**: The Cl-C-Cl bond angles are 109.5°.

Question 6

Q6. Explain the difference in properties of diamond and graphite on the basis of their structures.

Accepted Answer

Diamond and graphite are two important allotropes of carbon. Their strikingly different physical properties arise directly from their distinct crystal structures and bonding patterns.

### 1. Diamond

**Structure:**
*   In diamond, each carbon atom is sp³ hybridized and is tetrahedrally bonded to four other carbon atoms. This forms a giant covalent network structure, extending in three dimensions.
*   The C-C bond length is 154 pm.
*   This arrangement results in a rigid, compact, and highly stable lattice.

**Properties:**
*   **Hardness:** Extremely hard (hardest known natural substance) due to strong covalent bonds throughout the entire crystal and the rigid 3D network structure. It is used in cutting tools and abrasives.
*   **Electrical Conductivity:** A poor conductor of electricity (insulator) because all valence electrons are involved in forming strong covalent bonds, leaving no free electrons to carry current.
*   **Density:** High density (3.51 g/cm³) due to the compact packing of atoms.
*   **Transparency:** Transparent and has a high refractive index, making it sparkle.
*   **Melting Point:** Very high melting point (sublimes at ~3700 °C) due to the strong covalent bonds that require a large amount of energy to break.
*   **Chemical Reactivity:** Chemically inert under normal conditions.

### 2. Graphite

**Structure:**
*   In graphite, each carbon atom is sp² hybridized and is covalently bonded to three other carbon atoms in the same plane, forming hexagonal rings. These rings are arranged in flat, two-dimensional layers.
*   Within each layer, the C-C bond length is 141.5 pm, which is intermediate between a single and a double bond, indicating partial double bond character.
*   The layers are held together by weak van der Waals forces, with an interlayer distance of 340 pm.
*   Each carbon atom has one unhybridized p-orbital perpendicular to the plane of the layers. These p-orbitals overlap to form a delocalized electron cloud above and below each layer.

**Properties:**
*   **Hardness:** Soft and slippery. The weak van der Waals forces between the layers allow them to slide over each other easily. This property makes it a good lubricant and the main component of pencil leads.
*   **Electrical Conductivity:** A good conductor of electricity (unlike diamond) because of the presence of delocalized π-electrons within each layer, which are free to move and carry charge.
*   **Density:** Lower density (2.22 g/cm³) compared to diamond due to the larger interlayer spacing.
*   **Opacity:** Opaque and black.
*   **Melting Point:** Very high melting point (sublimes at ~3700 °C), similar to diamond, due to the strong covalent bonds within the layers.
*   **Chemical Reactivity:** More reactive than diamond due to the presence of delocalized electrons.

In summary, the sp³ hybridization and 3D network in diamond lead to its extreme hardness and insulating nature, while the sp² hybridization and layered structure with delocalized electrons in graphite result in its softness, slipperiness, and electrical conductivity.

Question 7

Q7. What are the uses of borax?

Accepted Answer

Borax (Na₂B₄O₇·10H₂O) is an important compound of boron with a wide range of applications:

1.  **Glass and Ceramics Industry:** It is extensively used in the manufacture of heat-resistant borosilicate glass (e.g., Pyrex), optical glasses, and glazes for pottery and ceramics. It acts as a flux and lowers the melting point of silica.
2.  **Metallurgy:** Used as a flux in soldering and welding to remove oxide layers from metal surfaces, creating a clean surface for joining.
3.  **Borax Bead Test:** It is used in qualitative analysis for identifying colored metal ions (e.g., Co²⁺, Ni²⁺, Cr³⁺, Cu²⁺) by forming characteristic colored metaborates when heated with a metal salt in a flame.
4.  **Cleaning Agent:** It is a component of many detergents, soaps, and cleaning products due to its ability to soften water and act as a mild antiseptic.
5.  **Antiseptic and Medicinal Uses:** Dilute solutions of borax are used as mild antiseptics for eyes and skin, and in some pharmaceutical preparations.
6.  **Fire Retardant:** Used as a fire retardant for wood and textile materials.
7.  **Preservative:** Used as a food preservative in some countries (though its use is restricted in many due to health concerns).
8.  **Agriculture:** Used as a micronutrient fertilizer for boron-deficient soils.

Question 8

Q8. Explain the structure of diborane.

Accepted Answer

Diborane (B₂H₆) is the simplest boron hydride and is an electron-deficient compound. Its structure cannot be explained by conventional two-centre two-electron (2c-2e) bonds. Instead, it involves unique three-centre two-electron (3c-2e) bonds, often called 'banana bonds' or 'bridge bonds'.

### 1. Molecular Formula and Electron Deficiency
*   Diborane has the molecular formula B₂H₆. If we consider conventional 2c-2e bonds, each boron atom needs 3 valence electrons to form 3 bonds, and each hydrogen atom needs 1 valence electron to form 1 bond. For B₂H₆, this would require 2×3 (for B) + 6×1 (for H) = 12 valence electrons. However, diborane only has 2×3 (for B) + 6×1 (for H) = 12 valence electrons in total. If we try to draw a structure with 2c-2e bonds, we would need 7 bonds (one B-B and six B-H), requiring 14 electrons. Since it only has 12 electrons, it is electron-deficient.

### 2. Hybridization and Geometry
*   Each boron atom in diborane is sp³ hybridized. This means each boron atom has four sp³ hybrid orbitals.

### 3. Bonding in Diborane
*   **Terminal B-H bonds:** Each boron atom forms two normal 2c-2e covalent bonds with two hydrogen atoms. These are called terminal hydrogen atoms. There are four such terminal B-H bonds in diborane. The bond length for terminal B-H is approximately 119 pm.
*   **Bridge B-H-B bonds (Banana Bonds):** The remaining two hydrogen atoms act as bridge atoms, connecting the two boron atoms. Each bridge hydrogen atom forms a three-centre two-electron (3c-2e) bond with the two boron atoms. In these bonds, two electrons are shared among three atoms (B-H-B). These are also known as 'banana bonds' because of their curved appearance in molecular orbital diagrams.
    *   Each boron atom uses one sp³ hybrid orbital to overlap with the 1s orbital of a bridge hydrogen atom, which in turn overlaps with another sp³ hybrid orbital from the second boron atom. This forms a B-H-B bridge bond.
    *   There are two such B-H-B bridge bonds in diborane.
    *   The bond length for bridge B-H is approximately 133 pm, which is longer than the terminal B-H bond, indicating a weaker bond.

### 4. Overall Structure
*   The two boron atoms and the four terminal hydrogen atoms lie in one plane. The two bridge hydrogen atoms lie above and below this plane.
*   The molecule has a D₂h symmetry.
*   The H-B-H bond angle for terminal hydrogens is approximately 120°.
*   The B-H-B bond angle for bridge hydrogens is approximately 97°.
*   The B-B distance is approximately 177 pm, which is longer than a typical B-B single bond, indicating that there is no direct B-B bond.

### Summary of Bonding:
*   **4 terminal B-H bonds:** Each is a 2-centre, 2-electron (2c-2e) bond.
*   **2 bridge B-H-B bonds:** Each is a 3-centre, 2-electron (3c-2e) bond.

This unique bonding explains the electron-deficient nature and the stability of diborane.

Question 9

Q9. What are the important characteristics of boranes?

Accepted Answer

Boranes are compounds of boron and hydrogen. They exhibit several unique characteristics due to boron's electron deficiency and tendency to form multi-centre bonds:

1.  **Electron-Deficient Compounds:** Boranes are electron-deficient, meaning they do not have enough valence electrons to form conventional two-centre two-electron (2c-2e) bonds between all adjacent atoms. This leads to the formation of unique bonding types.
2.  **Multi-centre Bonding:** They typically feature three-centre two-electron (3c-2e) bonds, often called 'banana bonds' or 'bridge bonds', in addition to regular 2c-2e bonds. For example, in diborane (B₂H₆), there are two B-H-B bridge bonds.
3.  **Variety of Structures:** Boranes exist in a wide range of structural types, from simple molecules like diborane (B₂H₆) to complex cage-like structures (e.g., B₁₀H₁₄, B₁₂H₁₂²⁻). They are often classified as nido-, arachno-, and closo-boranes based on their skeletal structures.
4.  **High Reactivity:** Many boranes are highly reactive, particularly with air (oxygen) and water. Lower boranes like diborane are spontaneously flammable in air and hydrolyze readily.
    *   B₂H₆ + 3O₂ → B₂O₃ + 3H₂O (highly exothermic)
    *   B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂
5.  **Lewis Acidity:** Due to their electron-deficient nature, boranes act as strong Lewis acids, readily accepting electron pairs from Lewis bases (e.g., NH₃, ethers) to form adducts.
    *   B₂H₆ + 2NMe₃ → 2H₃B·NMe₃
6.  **Reducing Agents:** Many boranes are powerful reducing agents, capable of reducing various organic and inorganic compounds.
7.  **Thermal Stability:** Their thermal stability generally increases with increasing molecular complexity. Lower boranes are less stable than higher boranes.
8.  **Polymerization:** Smaller boranes can polymerize at higher temperatures to form larger, more complex boranes.

Question 10

Q10. Write balanced equations for the following reactions:
(a) BF₃ + LiH →
(b) B₂H₆ + H₂O →
(c) NaH + B₂H₆ →
(d) H₃BO₃ →
(e) Al + NaOH →
(f) B₂H₆ + NH₃ →

Accepted Answer

Here are the balanced chemical equations for the given reactions:

**(a) BF₃ + LiH →**
*   **Reaction:** Boron trifluoride reacts with lithium hydride to form diborane and lithium fluoride. This is a method for the preparation of diborane.
*   **Balanced Equation:** 2BF₃ + 6LiH → B₂H₆ + 6LiF

**(b) B₂H₆ + H₂O →**
*   **Reaction:** Diborane undergoes hydrolysis (reaction with water) to produce boric acid and hydrogen gas. This reaction is vigorous.
*   **Balanced Equation:** B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂

**(c) NaH + B₂H₆ →**
*   **Reaction:** Sodium hydride reacts with diborane to form sodium borohydride (NaBH₄), a common reducing agent. This is an addition reaction where NaH acts as a Lewis base and B₂H₆ as a Lewis acid.
*   **Balanced Equation:** 2NaH + B₂H₆ → 2NaBH₄

**(d) H₃BO₃ →**
*   **Reaction:** Orthoboric acid (H₃BO₃) undergoes thermal decomposition upon heating. Depending on the temperature, it loses water molecules in stages to form metaboric acid (HBO₂) and then boron trioxide (B₂O₃).
*   **Balanced Equation (at ~370 K):** H₃BO₃ → HBO₂ + H₂O
*   **Balanced Equation (at >410 K):** HBO₂ → H₂B₄O₇ (tetraboric acid)
*   **Balanced Equation (red heat):** 2H₃BO₃ → B₂O₃ + 3H₂O
    *(The question implies complete decomposition to the most stable oxide at high temperature, so the last equation is generally expected.)*

**(e) Al + NaOH →**
*   **Reaction:** Aluminium is an amphoteric metal. It reacts with strong bases like sodium hydroxide to produce sodium meta-aluminate and hydrogen gas.
*   **Balanced Equation:** 2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂
    *(Alternatively, in concentrated hot NaOH: 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂)*

**(f) B₂H₆ + NH₃ →**
*   **Reaction:** Diborane reacts with ammonia. The products depend on the conditions (temperature, ratio of reactants).
    *   **At low temperature (excess NH₃):** It forms an adduct, B₂H₆·2NH₃, which is ionic, [BH₂(NH₃)₂]⁺[BH₄]⁻.
    *   **At high temperature:** It forms borazine (B₃N₃H₆), also known as inorganic benzene.
*   **Balanced Equation (Low temperature, excess NH₃):** B₂H₆ + 2NH₃ → [BH₂(NH₃)₂]⁺[BH₄]⁻
*   **Balanced Equation (High temperature, 1:2 ratio):** 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂
    *(The question doesn't specify conditions, but the formation of borazine is a very characteristic reaction.)*
    *Let's provide the borazine formation as it's more commonly asked in this context.*
*   **Balanced Equation (General, forming borazine):** 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂

Question 11

Q11. Explain the difference in properties of diamond and graphite on the basis of their structures.

Accepted Answer

Diamond and graphite are two important allotropes of carbon, exhibiting vastly different physical properties due to their distinct crystal structures.

**1. Diamond:**
*   **Structure:** In diamond, each carbon atom is sp³ hybridised and is covalently bonded to four other carbon atoms in a tetrahedral fashion. This forms a rigid, three-dimensional network structure extending throughout the crystal. The C-C bond length is 154 pm.
*   **Properties based on structure:**
    *   **Hardness:** The strong covalent bonds throughout the entire crystal make diamond the hardest known natural substance. This makes it useful for cutting tools, abrasives, and drilling.
    *   **Electrical Conductivity:** All valence electrons of carbon are involved in forming strong covalent bonds, leaving no free electrons to conduct electricity. Hence, diamond is an electrical insulator.
    *   **Thermal Conductivity:** Diamond is an excellent thermal conductor, even better than metals, due to its strong, rigid lattice and efficient phonon propagation.
    *   **Transparency:** The large band gap between the valence and conduction bands means diamond does not absorb visible light, making it transparent.
    *   **Density:** Due to its compact 3D network, diamond has a high density (3.51 g/cm³).

**2. Graphite:**
*   **Structure:** In graphite, each carbon atom is sp² hybridised and is covalently bonded to three other carbon atoms in the same plane, forming hexagonal rings. These hexagonal rings are arranged in layers. The fourth valence electron of each carbon atom forms a delocalised π-electron cloud above and below each layer. The layers are held together by weak van der Waals forces. The C-C bond length within a layer is 141.5 pm, and the distance between layers is 340 pm.
*   **Properties based on structure:**
    *   **Softness and Lubricity:** The weak van der Waals forces between the layers allow them to slide past each other easily. This makes graphite soft and a good solid lubricant.
    *   **Electrical Conductivity:** The delocalised π-electrons are free to move within the layers, making graphite a good conductor of electricity along the layers. It is an electrical conductor.
    *   **Thermal Conductivity:** Graphite conducts heat well along the layers due to its delocalised electrons, but poorly perpendicular to the layers.
    *   **Opacity:** The delocalised electrons can absorb light across a wide range of wavelengths, making graphite opaque and black.
    *   **Density:** Due to the larger spacing between layers, graphite has a lower density (2.22 g/cm³) compared to diamond.

**Summary of Differences:**

Question 12

Q12. What are the structural similarities and differences between diamond, graphite and fullerene?

Accepted Answer

Diamond, graphite, and fullerene are three important allotropes of carbon, each possessing unique structural characteristics that lead to distinct properties.

**1. Diamond:**
*   **Structure:** Each carbon atom is sp³ hybridised and covalently bonded to four other carbon atoms in a tetrahedral arrangement, forming a rigid, three-dimensional network lattice. The C-C bond length is 154 pm.

**2. Graphite:**
*   **Structure:** Each carbon atom is sp² hybridised and covalently bonded to three other carbon atoms in the same plane, forming hexagonal rings. These rings form flat layers, which are stacked one above another. The layers are held by weak van der Waals forces. The fourth valence electron of each carbon forms a delocalised π-electron cloud. C-C bond length within a layer is 141.5 pm, and inter-layer distance is 340 pm.

**3. Fullerene (e.g., C₆₀ Buckminsterfullerene):**
*   **Structure:** Fullerenes are cage-like molecules. C₆₀, the most common fullerene, has a spherical shape resembling a football (soccer ball). It consists of 60 carbon atoms arranged in 12 pentagonal and 20 hexagonal rings. Each carbon atom is sp² hybridised and is bonded to three other carbon atoms. The bonds are a mixture of single and double bonds. The carbon atoms at the junction of two hexagons and one pentagon are sp² hybridised. The C-C bond lengths vary (e.g., 143.5 pm for single bonds and 138.3 pm for double bonds in C₆₀).

**Structural Similarities:**
1.  **Elemental Composition:** All three are allotropes of carbon, meaning they are composed solely of carbon atoms.
2.  **Covalent Bonding:** All three involve strong covalent bonds between carbon atoms.
3.  **Hybridisation:** While diamond is purely sp³, graphite and fullerene both involve sp² hybridisation of carbon atoms.
4.  **Ring Structures:** Graphite consists of hexagonal rings, while fullerene contains both hexagonal and pentagonal rings.

**Structural Differences:**

Question 13

Q13. How can you explain the higher stability of BCl₃ as compared to TlCl₃?

Accepted Answer

The higher stability of BCl₃ compared to TlCl₃ can be explained by considering the 'inert pair effect' and the stability of oxidation states down Group 13.

1.  **Oxidation States of Group 13 Elements:** Group 13 elements (B, Al, Ga, In, Tl) typically show +3 oxidation state. However, as we move down the group, the stability of the +1 oxidation state increases, and the stability of the +3 oxidation state decreases.

2.  **Inert Pair Effect:** This trend is attributed to the 'inert pair effect'. As we go down the group, the ns² electrons become increasingly reluctant to participate in bonding. This is due to the poor shielding effect of intervening d and f electrons (in Ga, In, and Tl), which causes the valence ns² electrons to be held more tightly by the nucleus. Consequently, more energy is required to unpair these ns² electrons and promote one to a p-orbital to achieve the +3 oxidation state.

3.  **Stability Comparison:**
    *   **Boron (B):** Being the first element in the group, boron does not exhibit the inert pair effect. Its most stable oxidation state is +3. Therefore, BCl₃ (where B is in +3 oxidation state) is highly stable.
    *   **Thallium (Tl):** Thallium is the last element in Group 13. The inert pair effect is most pronounced for thallium. As a result, the +1 oxidation state (where only the p-electron participates in bonding) is more stable than the +3 oxidation state (where both ns² and np¹ electrons participate). TlCl (Tl in +1 state) is more stable than TlCl₃ (Tl in +3 state).

Therefore, BCl₃, with boron in its most stable +3 oxidation state, is significantly more stable than TlCl₃, where thallium is in its less stable +3 oxidation state.

Question 14

Q14. Account for the following observations:
(a) CCl₄ is not hydrolysed whereas SiCl₄ is easily hydrolysed.
(b) Carbon shows catenation property but lead does not.
(c) BF₃ is an acid

Accepted Answer

**(a) CCl₄ is not hydrolysed whereas SiCl₄ is easily hydrolysed.**

**Reason:** The difference in hydrolysis behavior is due to two main factors: the presence of vacant d-orbitals and the size of the central atom.

1.  **Presence of Vacant d-orbitals:**
    *   **CCl₄:** Carbon belongs to the second period and has no vacant d-orbitals in its valence shell. Therefore, it cannot expand its octet. When water molecules approach CCl₄, there is no empty orbital for the lone pair of electrons from the oxygen atom of water to attack and form an intermediate. Thus, CCl₄ resists hydrolysis.
    *   **SiCl₄:** Silicon belongs to the third period and has vacant 3d-orbitals in its valence shell. When water molecules approach SiCl₄, the oxygen atom's lone pair can readily attack the silicon atom and donate electrons into its vacant 3d-orbitals, forming an unstable pentacoordinate intermediate. This intermediate then undergoes nucleophilic attack by another water molecule or rearrangement, leading to the elimination of HCl and the formation of Si(OH)₄ (silicic acid).

**Reaction:**
    SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl

2.  **Steric Hindrance:**
    *   **CCl₄:** The four bulky chlorine atoms around the small carbon atom create significant steric hindrance, making it difficult for water molecules to approach the carbon atom for nucleophilic attack.
    *   **SiCl₄:** Silicon is larger than carbon, and the Si-Cl bond is longer, reducing steric hindrance and allowing easier approach of water molecules.

**Conclusion:** The absence of vacant d-orbitals and significant steric hindrance around the small carbon atom prevent the hydrolysis of CCl₄, while the presence of vacant d-orbitals and larger size of silicon facilitate the easy hydrolysis of SiCl₄.

---

**(b) Carbon shows catenation property but lead does not.**

**Reason:** Catenation is the ability of an atom to form bonds with other atoms of the same element, forming long chains or rings. This property is strongest for carbon and decreases significantly down Group 14.

1.  **Carbon:**
    *   **Small Size:** Carbon's small size allows for effective overlap of atomic orbitals, leading to strong C-C covalent bonds (bond enthalpy ~348 kJ/mol).
    *   **High Electronegativity:** Carbon's relatively high electronegativity ensures that C-C bonds are strong and stable.
    *   **Absence of d-orbitals:** The absence of vacant d-orbitals prevents expansion of octet and reduces the likelihood of forming weak bonds with other elements that could break the chain.
    *   **Bond Strength:** The C-C bond strength is comparable to C-O or C-H bond strengths, making carbon chains stable even in the presence of other atoms.

2.  **Lead (Pb):**
    *   **Large Size:** Lead is a much larger atom than carbon. As we move down Group 14, the atomic size increases, and the bond length between identical atoms (Pb-Pb) increases. This leads to weaker orbital overlap and consequently weaker Pb-Pb covalent bonds (bond enthalpy ~105 kJ/mol).
    *   **Inert Pair Effect:** The inert pair effect is prominent in lead, making its +2 oxidation state more stable than +4. This reduces its tendency to form multiple strong covalent bonds required for extensive catenation.
    *   **Shielding Effect:** The presence of intervening d and f electrons in heavier elements like lead leads to poor shielding, increasing the effective nuclear charge on valence electrons and making them less available for strong covalent bonding in chains.
    *   **Reactivity:** Pb-Pb bonds are much weaker and more reactive than C-C bonds, making long chains unstable and prone to breaking.

**Conclusion:** Carbon's small size and strong C-C bonds enable extensive catenation, forming a vast array of organic compounds. In contrast, lead's large size, weak Pb-Pb bonds, and the inert pair effect severely limit its ability to catenate.

---

**(c) BF₃ is an acid according to Lewis concept.**

**Reason:** According to the Lewis concept, an acid is a species that can accept a pair of electrons, and a base is a species that can donate a pair of electrons.

1.  **Electronic Configuration of Boron:** Boron (atomic number 5) has an electronic configuration of [He] 2s²2p¹. In BF₃, boron is sp² hybridised and forms three covalent bonds with three fluorine atoms.

2.  **Incomplete Octet:** After forming three covalent bonds, the boron atom in BF₃ has only six valence electrons (3 bond pairs) in its outermost shell. This means it has an incomplete octet.

3.  **Electron-Deficient Nature:** Due to its incomplete octet, the boron atom in BF₃ is electron-deficient. It has a vacant 2p orbital that can accommodate a pair of electrons.

4.  **Lewis Acid Behavior:** Because BF₃ can accept a pair of electrons to complete its octet, it acts as a Lewis acid. For example, it readily reacts with Lewis bases like ammonia (NH₃) or fluoride ion (F⁻) by accepting a lone pair of electrons.

**Reaction with Ammonia:**
    BF₃ + :NH₃ → F₃B ← NH₃ (an adduct is formed)

**Reaction with Fluoride ion:**
    BF₃ + F⁻ → [BF₄]⁻

**Conclusion:** The electron-deficient nature of boron in BF₃, due to its incomplete octet and the presence of a vacant orbital, allows it to accept a pair of electrons, thus classifying it as a Lewis acid.

Question 15

Q15. Explain the difference in properties of diamond and graphite on the basis of their structures.

Accepted Answer

Diamond and graphite are two important allotropes of carbon. They exhibit vastly different physical properties despite being composed solely of carbon atoms. These differences arise directly from their distinct crystal structures and bonding patterns.

### 1. Structure of Diamond
*   **Hybridization:** Each carbon atom in diamond is sp³ hybridized.
*   **Bonding:** Each sp³ hybridized carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement.
*   **Network:** This forms a giant, rigid, three-dimensional network structure where all carbon atoms are linked by strong covalent bonds.
*   **Bond Length:** The C-C bond length is 154 pm.

### 2. Structure of Graphite
*   **Hybridization:** Each carbon atom in graphite is sp² hybridized.
*   **Bonding:** Each sp² hybridized carbon atom is covalently bonded to three other carbon atoms in the same plane, forming hexagonal rings. The fourth valence electron of each carbon atom forms a delocalized π-electron system above and below the plane of the layers.
*   **Layers:** Graphite consists of distinct layers of these hexagonal carbon networks. Within each layer, the C-C bonds are strong covalent bonds.
*   **Inter-layer Forces:** The different layers are held together by weak van der Waals forces.
*   **Bond Length:** The C-C bond length within a layer is 141.5 pm. The distance between layers is 340 pm.

**Explanation of Properties:**
*   **Hardness:** Diamond's extreme hardness is due to its strong, extensive 3D covalent network, requiring a large amount of energy to break the bonds. Graphite's softness and slipperiness result from the weak van der Waals forces between its layers, allowing them to slide over each other easily.
*   **Electrical Conductivity:** Diamond is an insulator because all its valence electrons are localized in strong C-C covalent bonds, with no free electrons to conduct electricity. Graphite is a good conductor because of the presence of delocalized π-electrons, which can move freely within the layers.
*   **Density:** Diamond has a more compact, tightly packed structure, leading to a higher density compared to graphite, where layers are spaced further apart.

These structural differences are fundamental to understanding the diverse applications of diamond (e.g., cutting tools, jewelry) and graphite (e.g., lubricants, electrodes, pencil lead).

Question 16

Q16. What are the structural features of diborane?

Accepted Answer

Diborane (B₂H₆) is the simplest boron hydride and an electron-deficient compound. Its unique structure cannot be explained by conventional two-centre two-electron (2c-2e) bonds alone. It features a combination of conventional and unconventional bonding.

### Structural Features of Diborane (B₂H₆)

1.  **Molecular Formula:** B₂H₆
2.  **Central Atoms:** It contains two boron atoms.
3.  **Hybridization:** Each boron atom in diborane is sp³ hybridized.
4.  **Terminal Hydrogen Atoms:** There are four terminal hydrogen atoms (Hₜ), two attached to each boron atom. These B-Hₜ bonds are conventional 2-centre 2-electron (2c-2e) covalent bonds.
5.  **Bridging Hydrogen Atoms:** There are two bridging hydrogen atoms (H_b), which form a 'bridge' between the two boron atoms. These bridging bonds are unique.
6.  **3-centre 2-electron (3c-2e) Bonds (Banana Bonds):** The two bridging hydrogen atoms form two 3-centre 2-electron (3c-2e) bonds, often referred to as 'banana bonds' or 'tau bonds'. In these bonds, two electrons are shared among three atoms (B-H_b-B). This is why diborane is considered electron-deficient, as it doesn't have enough electrons to form a 2c-2e bond between every adjacent pair of atoms.
7.  **Planar Arrangement:** The two boron atoms and the two bridging hydrogen atoms lie in one plane. This forms a B-H_b-B-H_b ring-like structure in the plane.
8.  **Perpendicular Arrangement:** The four terminal hydrogen atoms lie in a plane perpendicular to the B-H_b-B plane, with two terminal hydrogens on each boron atom.
9.  **Bond Lengths and Angles:**
    *   **B-Hₜ bond length:** Approximately 119 pm (shorter and stronger).
    *   **B-H_b bond length:** Approximately 133 pm (longer and weaker).
    *   **Hₜ-B-Hₜ angle:** Approximately 120°.
    *   **H_b-B-H_b angle:** Approximately 97°.
    *   **B-H_b-B angle:** Approximately 83°.

### Summary of Bonding:
*   **Total Valence Electrons:** 3 (from B) × 2 + 1 (from H) × 6 = 12 valence electrons.
*   **Terminal Bonds:** Four B-Hₜ bonds (2c-2e) consume 4 × 2 = 8 electrons.
*   **Bridging Bonds:** Two B-H_b-B bonds (3c-2e) consume 2 × 2 = 4 electrons.
*   **Total Electrons Used:** 8 + 4 = 12 electrons. All valence electrons are accounted for.

This unique bonding arrangement gives diborane its characteristic structure and reactivity.

Question 17

Q17. What are the uses of silicones?

Accepted Answer

Silicones are organosilicon polymers containing R₂SiO repeating units, where R is an alkyl or aryl group. They possess unique properties such as high thermal stability, water repellency, chemical inertness, and low surface tension, making them useful in a wide range of applications:

1.  **Sealants and Adhesives:** Due to their flexibility, durability, and strong bonding capabilities, silicones are widely used as sealants in construction, automotive, and electronic industries. They are also used as high-performance adhesives.
2.  **Lubricants:** Silicone oils are excellent lubricants, especially at high temperatures and in situations where conventional organic lubricants break down. They have good thermal stability and their viscosity changes little with temperature.
3.  **Waterproofing Agents:** Their water-repellent nature makes them ideal for waterproofing fabrics, paper, leather, and masonry. They form a protective, hydrophobic layer.
4.  **Electrical Insulators:** Silicones are good electrical insulators and are used in various electrical and electronic applications, such as encapsulating electronic components and as dielectric fluids.
5.  **Medical and Surgical Applications:** Silicones are biocompatible and non-toxic, making them suitable for medical implants (e.g., breast implants, catheters), prosthetics, and components in medical devices.
6.  **Cosmetics and Personal Care Products:** They are used in shampoos, conditioners, lotions, and deodorants to provide a smooth feel, enhance spreadability, and offer conditioning properties.
7.  **Antifoaming Agents:** Silicone fluids are effective defoamers and antifoaming agents in various industrial processes, including food processing, textiles, and wastewater treatment.
8.  **Mould Making:** Silicone rubbers are used to make flexible moulds for casting various materials like resins, wax, and concrete due to their non-stick properties and ability to reproduce fine details.
9.  **High-Temperature Applications:** Due to their thermal stability, they are used in high-temperature gaskets, O-rings, and coatings.

Question 18

Q18. What is the basic structural unit of silicates?

Accepted Answer

The basic structural unit of all silicates is the **silicon dioxide tetrahedron (SiO₄⁴⁻)**.

### Description of the SiO₄⁴⁻ Tetrahedron:

1.  **Central Atom:** A single silicon (Si) atom is located at the center of the tetrahedron.
2.  **Surrounding Atoms:** Four oxygen (O) atoms are covalently bonded to the central silicon atom, occupying the four corners of a regular tetrahedron.
3.  **Bonding:** Each silicon atom forms four single covalent bonds with the four oxygen atoms.
4.  **Charge:** Since silicon typically forms four bonds and each oxygen atom (when bonded to only one silicon) carries a formal negative charge, the overall charge of this unit is 4- (SiO₄⁴⁻). This is because silicon is in the +4 oxidation state, and each oxygen atom is considered in the -2 oxidation state, but when it's part of a silicate, it can be shared. In the isolated unit, each oxygen has one bond to Si and three lone pairs, hence a -1 formal charge, leading to 4- overall.

### Formation of Silicates:
These SiO₄⁴⁻ tetrahedral units can link together by sharing one or more of their oxygen atoms at the corners. This sharing leads to the formation of various complex silicate structures, including:
*   **Orthosilicates:** Discrete SiO₄⁴⁻ units (e.g., olivine).
*   **Pyrosilicates:** Two SiO₄⁴⁻ units sharing one oxygen atom (Si₂O₇⁶⁻).
*   **Cyclic Silicates:** Three or more SiO₄⁴⁻ units forming rings (e.g., beryl).
*   **Chain Silicates:** Units linked in single or double chains (e.g., pyroxenes, amphiboles).
*   **Sheet Silicates:** Units forming two-dimensional sheets (e.g., mica, talc).
*   **Three-dimensional Network Silicates:** All four oxygen atoms of each tetrahedron are shared, forming a vast 3D network (e.g., quartz, feldspars).

Question 19

Q19. What is the difference between glass and quartz?

Accepted Answer

Both glass and quartz are primarily composed of silicon dioxide (SiO₂), but they differ significantly in their atomic arrangement, which leads to distinct physical properties. The key difference lies in their crystalline nature.

### 1. Quartz
*   **Nature:** Quartz is a **crystalline solid**. It is a naturally occurring mineral.
*   **Structure:** It has a highly ordered, long-range repeating three-dimensional network structure. Each silicon atom is tetrahedrally bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. This forms a regular, repeating, and highly organized crystal lattice.
*   **Melting Point:** Quartz has a sharp and definite melting point (around 1713 °C) because its ordered structure requires a specific amount of energy to break all the bonds simultaneously.
*   **Properties:**
    *   **Anisotropic:** Its physical properties (like refractive index, electrical conductivity) vary with direction.
    *   **Hardness:** Hard and brittle.
    *   **Transparency:** Transparent.
    *   **Cleavage:** Exhibits definite cleavage planes.

### 2. Glass
*   **Nature:** Glass is an **amorphous solid**, often considered a supercooled liquid. It is typically manufactured by rapidly cooling molten silica or a mixture containing silica.
*   **Structure:** It lacks a long-range ordered structure. While the basic SiO₄ tetrahedra units are present, they are randomly oriented and connected, forming an irregular, disordered network. It possesses only short-range order, meaning the local environment around each silicon atom is tetrahedral, but there is no repeating pattern over long distances.
*   **Melting Point:** Glass does not have a sharp melting point. Instead, it softens gradually over a range of temperatures as it transitions from a rigid solid to a viscous liquid. This is because the disordered structure means bonds break at varying energy levels.
*   **Properties:**
    *   **Isotropic:** Its physical properties are uniform in all directions.
    *   **Hardness:** Hard and brittle.
    *   **Transparency:** Transparent (for common types).
    *   **Cleavage:** Fractures irregularly (conchoidal fracture).

### Summary of Differences:

In essence, the fundamental difference between glass and quartz lies in the degree of order in their atomic arrangement. Quartz has a highly ordered, crystalline structure, while glass has a disordered, amorphous structure.

Question 20

Q20. Explain the following reactions:
(i) Silicon is heated with methyl chloride in the presence of copper powder at high temperature.
(ii) Silicon dioxide is treated with hydrogen fl

Accepted Answer

Here are the explanations for the given reactions:

### (i) Silicon is heated with methyl chloride in the presence of copper powder at high temperature.
This reaction is a key step in the industrial preparation of methyl-substituted chlorosilanes, which are precursors for silicones. When silicon is heated with methyl chloride (CH₃Cl) in the presence of copper powder as a catalyst at approximately 573 K, a mixture of methylchlorosilanes is formed. The main product is dimethyldichlorosilane.

**Reaction:**
2CH₃Cl + Si \xrightarrow{	ext{Cu powder, 573 K}} (CH₃)₂SiCl₂ + other methylchlorosilanes

**Explanation:**
This is a direct synthesis method. The copper catalyst facilitates the reaction by forming intermediate copper-silicon compounds, which then react with methyl chloride. The dimethyldichlorosilane, (CH₃)₂SiCl₂, is then hydrolyzed to form silanols, which subsequently condense to form linear silicones.

### (ii) Silicon dioxide is treated with hydrogen fluoride.
Silicon dioxide (SiO₂), also known as silica, is generally unreactive towards most acids. However, it reacts with hydrogen fluoride (HF) to form silicon tetrafluoride (SiF₄) and water. This reaction is significant because it is used for etching glass (which is primarily SiO₂) and for the removal of sand impurities in certain industrial processes.

**Reaction:**
SiO₂(s) + 4HF(aq) ightarrow SiF₄(g) + 2H₂O(l)

**Explanation:**
Hydrogen fluoride is unique among hydrogen halides in its ability to react with silica. This is due to the high bond energy of the Si-F bond, which makes the formation of SiF₄ energetically favorable. The reaction proceeds by breaking the strong Si-O bonds in the silica network and forming new Si-F bonds. If excess HF is used, SiF₄ can further react to form fluorosilicic acid (H₂SiF₆).

SiF₄(g) + 2HF(aq) ightarrow H₂SiF₆(aq)

### (iii) Carbon monoxide is heated with ZnO.
This reaction demonstrates the reducing property of carbon monoxide. Carbon monoxide (CO) is a good reducing agent, especially at high temperatures. When heated with zinc oxide (ZnO), carbon monoxide reduces ZnO to metallic zinc (Zn), while CO itself gets oxidized to carbon dioxide (CO₂).

**Reaction:**
ZnO(s) + CO(g) \xrightarrow{	ext{Heat}} Zn(s) + CO₂(g)

**Explanation:**
This reaction is an example of a carbothermal reduction. Carbon monoxide removes oxygen from the metal oxide, converting it into the free metal. This principle is widely used in metallurgy, particularly in blast furnaces for the extraction of iron from its oxides, where CO acts as the primary reducing agent.

Question 21

Q21. Give reasons:
(i) Conc. HNO₃ can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as a dry

Accepted Answer

Here are the reasons for the given statements:

### (i) Conc. HNO₃ can be transported in aluminium container.
**Reason:** Aluminium (Al) reacts with concentrated nitric acid (HNO₃) to form a very thin, tough, and impervious layer of aluminium oxide (Al₂O₃) on its surface. This passive oxide layer prevents further reaction of aluminium with the acid, making the aluminium container resistant to corrosion by concentrated nitric acid. This phenomenon is known as passivity.

### (ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
**Reason:** Aluminium is an amphoteric metal, meaning it reacts with both acids and strong bases. When aluminium pieces are mixed with dilute sodium hydroxide (NaOH), a vigorous reaction occurs, producing hydrogen gas (H₂) and sodium tetrahydroxoaluminate(III).

**Reaction:** 2Al(s) + 2NaOH(aq) + 6H₂O(l) ightarrow 2Na[Al(OH)₄](aq) + 3H₂(g)

The hydrogen gas produced during this reaction creates pressure and agitation, which helps to dislodge and push away blockages (like grease, hair, etc.) in drains. The heat generated by the exothermic reaction also aids in melting grease.

### (iii) Graphite is used as a dry lubricant in machines.
**Reason:** Graphite has a layered structure. Each layer consists of hexagonal rings of carbon atoms, where each carbon atom is sp² hybridized and bonded to three other carbon atoms. The layers are held together by weak van der Waals forces. These weak forces allow the layers to slide easily over one another. This property of being soft and slippery makes graphite an excellent dry lubricant, reducing friction between moving parts in machinery, especially at high temperatures where oil lubricants might decompose.

### (iv) Aluminium alloys are used to make aircraft body.
**Reason:** Aluminium itself is a relatively soft metal. However, when alloyed with other metals like copper, magnesium, manganese, and zinc, its mechanical properties are significantly improved. Aluminium alloys possess a unique combination of properties that are ideal for aircraft construction:
*   **High strength-to-weight ratio:** They are strong yet very light, which is crucial for reducing the overall weight of the aircraft and improving fuel efficiency.
*   **Corrosion resistance:** The passive oxide layer provides good resistance to corrosion.
*   **Ductility and Malleability:** They can be easily formed into complex shapes required for aircraft components.

### (v) Aluminium wire is used to make transmission cables.
**Reason:** Aluminium is a good conductor of electricity, although its conductivity is lower than that of copper. However, aluminium has a significantly lower density than copper. This means that for a given electrical conductivity, an aluminium wire will be much lighter than a copper wire. For long-distance overhead transmission cables, lightness is a critical factor as it reduces the structural load on supporting towers and makes installation easier and more cost-effective. While its conductivity is lower, its high abundance and lower cost also contribute to its widespread use.

Question 22

Q22. Explain the following:
(a) CO₂ is a gas while SiO₂ is a solid.
(b) Silicon forms SiF₆²⁻ but no SiCl₆²⁻.
(c) E° for Cl₂/Cl⁻ is +1.36, for I₂/I⁻ is +0.53, for Ag⁺/Ag is +0.80, for

Accepted Answer

Here are the explanations for the given statements:

### (a) CO₂ is a gas while SiO₂ is a solid.
This difference in physical state arises from their fundamental structural differences:

**Carbon Dioxide (CO₂):**
*   **Structure:** CO₂ is a discrete, simple molecular compound. Each carbon atom is double-bonded to two oxygen atoms (O=C=O). The carbon atom is sp hybridized, resulting in a linear molecular geometry.
*   **Intermolecular Forces:** The individual CO₂ molecules are held together by weak van der Waals forces (specifically London dispersion forces). These forces are easily overcome by thermal energy at room temperature.
*   **Physical State:** Due to weak intermolecular forces, CO₂ exists as a gas at room temperature and pressure.

**Silicon Dioxide (SiO₂):**
*   **Structure:** SiO₂ (silica) is a giant covalent network solid. Each silicon atom is tetrahedrally bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms. This forms an extended three-dimensional network structure, similar to diamond.
*   **Interatomic Forces:** The entire structure is held together by strong covalent Si-O bonds throughout the network. There are no discrete SiO₂ molecules.
*   **Physical State:** A large amount of energy is required to break these strong covalent bonds. Therefore, SiO₂ exists as a hard, high-melting point solid at room temperature.

### (b) Silicon forms SiF₆²⁻ but no SiCl₆²⁻.
This difference is primarily due to a combination of steric factors (size of the ligand) and electronic factors (electronegativity and bond strength):

*   **Size of Halogen Atoms:** Fluorine atoms are much smaller than chlorine atoms. Six small fluorine atoms can easily fit around the central silicon atom without significant steric hindrance, allowing the formation of SiF₆²⁻. However, chlorine atoms are larger, and six chlorine atoms would experience considerable steric repulsion around the relatively small silicon atom, making the formation of SiCl₆²⁻ sterically unfavorable.

*   **Si-X Bond Strength:** The Si-F bond is significantly stronger than the Si-Cl bond. The high electronegativity of fluorine allows for effective overlap with silicon's d-orbitals (though less significant than steric factors). The formation of strong Si-F bonds contributes to the stability of SiF₆²⁻.

*   **Electronic Factors (d-orbital participation):** While silicon has vacant 3d orbitals, their energy is relatively high. The smaller, more electronegative fluorine can induce a greater positive charge on silicon, making its 3d orbitals more accessible for bonding. However, the primary reason remains steric hindrance.

### (c) E° for Cl₂/Cl⁻ is +1.36, for I₂/I⁻ is +0.53, for Ag⁺/Ag is +0.80, for Na⁺/Na is -2.71 and for Li⁺/Li is -3.04. Arrange the following in increasing order of reducing power: I⁻, Ag, Cl⁻, Li, Na.

**Understanding Reducing Power:**
A reducing agent is a species that gets oxidized (loses electrons). A stronger reducing agent has a greater tendency to lose electrons. For a redox couple Mⁿ⁺/M, a more negative (or less positive) standard electrode potential (E°) indicates a greater tendency for the species M to get oxidized (i.e., M is a stronger reducing agent). Conversely, for a couple X/X⁻, a more negative (or less positive) E° for the reduction of X to X⁻ means X⁻ is a weaker reducing agent. We are looking at the reducing power of the *reduced form* (I⁻, Ag, Cl⁻, Li, Na).

Let's list the given standard electrode potentials (reduction potentials) and identify the species whose reducing power we need to compare:

*   Cl₂/Cl⁻: E° = +1.36 V (Reducing agent: Cl⁻)
*   I₂/I⁻: E° = +0.53 V (Reducing agent: I⁻)
*   Ag⁺/Ag: E° = +0.80 V (Reducing agent: Ag)
*   Na⁺/Na: E° = -2.71 V (Reducing agent: Na)
*   Li⁺/Li: E° = -3.04 V (Reducing agent: Li)

**Rule:** The more negative the standard reduction potential (E°), the stronger the reducing agent (the species on the right side of the reduction half-reaction).

Let's arrange the E° values in increasing order (most negative to most positive) and then list the corresponding reducing agents:

1.  Li⁺/Li: E° = -3.04 V (Li is the strongest reducing agent)
2.  Na⁺/Na: E° = -2.71 V (Na is a strong reducing agent)
3.  I₂/I⁻: E° = +0.53 V (I⁻ is a reducing agent)
4.  Ag⁺/Ag: E° = +0.80 V (Ag is a reducing agent)
5.  Cl₂/Cl⁻: E° = +1.36 V (Cl⁻ is the weakest reducing agent among these)

Therefore, the increasing order of reducing power is:

**Cl⁻ < Ag < I⁻ < Na < Li**

Question 23

Q23. Explain the structure of diborane and boron trifluoride.

Accepted Answer

### Structure of Diborane (B₂H₆)
Diborane is the simplest boron hydride and is an electron-deficient compound. Its structure cannot be explained by conventional two-centre two-electron (2c-2e) bonds.

**Key Features:**
1.  **Hybridization:** Each boron atom in diborane is sp³ hybridized.
2.  **Geometry:** The molecule has a unique bridge-like structure.
3.  **Bonding:**
    *   There are four terminal B-H bonds, which are conventional 2c-2e covalent bonds. These four hydrogen atoms and the two boron atoms lie in one plane.
    *   There are two bridging B-H-B bonds, often called 'banana bonds' or three-centre two-electron (3c-2e) bonds. In these bonds, one electron pair is shared among three atoms (two boron atoms and one hydrogen atom). The two bridging hydrogen atoms lie in a plane perpendicular to the plane containing the terminal hydrogen atoms and boron atoms.
4.  **Electron Deficiency:** Diborane has 12 valence electrons (3 from each B, 1 from each H, total 2*3 + 6*1 = 12 electrons). If all bonds were 2c-2e, it would require 14 electrons (7 bonds * 2 electrons/bond). Since it has fewer electrons than required for classical bonding, it is electron-deficient.
5.  **Bond Lengths and Angles:** The terminal B-H bond length is shorter (approx. 119 pm) than the bridging B-H bond length (approx. 133 pm). The H-B-H angle for terminal hydrogens is approximately 120°, while the B-H-B angle in the bridge is approximately 83°.

**Visual Representation (Answer Image Prompt):**
*   A central horizontal line representing the two boron atoms (B-B). *   Two terminal hydrogen atoms attached to each boron atom, extending outwards in the same plane as the B-B line. *   Two bridging hydrogen atoms, one above and one below the B-B line, forming a 'banana' shape with the boron atoms. *   Label the terminal B-H bonds as 2c-2e and bridging B-H-B bonds as 3c-2e. *   Indicate bond lengths and angles (e.g., B-H_terminal ~119 pm, B-H_bridge ~133 pm, H_terminal-B-H_terminal ~120°, B-H_bridge-B ~83°).

### Structure of Boron Trifluoride (BF₃)
Boron trifluoride is a simple molecular compound with a central boron atom bonded to three fluorine atoms.

**Key Features:**
1.  **Hybridization:** The central boron atom is sp² hybridized.
2.  **Geometry:** Due to sp² hybridization and the absence of lone pairs on the central boron atom, BF₃ has a trigonal planar geometry.
3.  **Bonding:** Each boron atom forms three sigma (σ) bonds with three fluorine atoms. These are 2c-2e covalent bonds. All B-F bonds are equivalent.
4.  **Bond Angles:** The F-B-F bond angles are all 120°.
5.  **Electron Deficiency (Lewis Acid):** Boron has only six valence electrons in its outermost shell after forming three bonds (3 from B + 3*1 from F = 6 electrons shared in bonds). It does not complete its octet, making it an electron-deficient molecule and a strong Lewis acid.
6.  **Back Bonding (Minor Effect):** Although boron is electron-deficient, there is a possibility of pπ-pπ back bonding from the filled 2p orbitals of fluorine to the vacant 2p orbital of boron. This back bonding provides some double bond character to the B-F bonds, making them slightly shorter and stronger than expected for a pure single bond. However, the primary geometry remains trigonal planar.

**Visual Representation (Answer Image Prompt):**
*   A central boron atom (B).
*   Three fluorine atoms (F) arranged symmetrically around the boron atom in a plane.
*   Lines connecting B to each F, representing covalent bonds.
*   Label the F-B-F bond angles as 120°.
*   Indicate the trigonal planar geometry.

Chapter 11: The p-Block Elements