7 Most SAQ’s of P – Block Elements Group -13 Chapter in Inter 1st Year Chemistry (TS/AP)

4 Marks

SAQ-1 : Explain Borax bead test with suitable examples.

Introduction

The Borax bead test is a traditional and effective method used for the identification of certain metal ions. It involves using the unique properties of borax to detect and differentiate metal ions based on the colors they produce.

Borax Bead Test Process

1. Starting with Borax: Borax, when heated, undergoes a series of transformations. It initially becomes a white, opaque mass which is anhydrous sodium tetraborate.
2. Formation of Borax Glass: Upon further heating, borax turns into a transparent borax glass. The borax glass is mainly composed of sodium metaborate and boric oxide (B₂O₃).
3. Interaction with Metal Oxides: When a metal oxide (from the sample being tested) is introduced to this molten borax bead, it reacts with the boric oxide. The resulting product is a metal metaborate.
4. Color Indication: These metal metaborates are often colored, and the specific color produced can indicate the presence of a particular metal ion in the sample.

Example of Borax Bead Test

Cobalt Detection: When cobalt oxide (CoO) is introduced to the molten borax bead, it reacts with the boric oxide to produce a compound called cobalt metaborate (Co(BO₂)₂). This results in a bead that is characteristically colored, indicating the presence of cobalt.

Summary

The Borax bead test serves as a qualitative analysis method for identifying metal ions. By observing the change in color of the borax bead upon introducing a metal oxide, one can infer the presence of specific metals. The different colors produced are distinctive and serve as an identifying feature for various metal ions.

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SAQ-2 : Explain the structure of Diborane (B₂H₆). How is it prepared?

Introduction

Diborane (B₂H₆) is an intriguing compound of boron, notable for its unique structure and bond formation that distinguishes it from other compounds. Its structural peculiarities and preparation methods make it a significant topic in the study of inorganic chemistry.

Structure of Diborane (B₂H₆)

1. Hybridisation:
   • Boron in Diborane undergoes sp³ hybridisation, resulting in four sp³ hybrid orbitals.
   • Three of these orbitals hold one electron each, and the fourth remains empty.
   • Each Boron atom uses two of its sp³ hybrid orbitals to form sigma bonds with two Hydrogen atoms.
   • A unique bond, termed the bridge bond (or Banana bond/Tau bond), forms by the overlap of the vacant orbital of one Boron atom, the 1s orbital of Hydrogen, and the sp³ orbital of the second Boron atom.
2. Spatial Structure:
   • Diborane can be visualized as having two coplanar BH₂ units.
   • Four Hydrogen atoms, termed as terminal hydrogens (H_t), are part of these BH₂ units.
   • There are also two bridge hydrogens (H_b), arranged perpendicular to the plane of the BH₂ units – one positioned above and the other below.

Preparation of Diborane

Industrial Preparation:

1. Diborane can be prepared by reacting Boron trifluoride (BF₃) with Lithium hydride (LiH). When heated to 450K, this reaction yields Diborane and Lithium fluoride.
2. The reaction is represented as: $$2BF₃ + 6LiH \rightarrow (at 450K) B₂H₆ + 6LiF$$

Summary

Diborane’s unique structure, highlighted by its banana-shaped bridge bonds, sets it apart from typical compounds. It can be industrially prepared through the reaction of Boron trifluoride with Lithium hydride, exemplifying the complexities and wonders of chemical bonding.

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SAQ-3 : Write an essay on preparation and chemical activities of Diborane.

Introduction

Diborane (B₂H₆) is a chemical compound known for its unique bonding nature and diverse reactions. The preparation and chemical reactions of Diborane provide valuable insights into the behavior and applications of boron compounds.

Preparation of Diborane

1. Industrial Method:
   • Boron trifluoride (BF₃) is treated with lithium hydride (LiH) at a temperature of 450K.
   • The reaction yields Diborane and lithium fluoride (LiF).
   • Reaction: $$2BF_3 + 6LiH \rightarrow (at 450K) B₂H₆ + 6LiF$$
2. Laboratory Methods:
   • Using Boron Trichloride and Lithium Aluminium Hydride:
     • Boron trichloride (BCl₃) reacts with lithium aluminium hydride in the presence of dry ether.
     • This reaction forms Diborane, lithium chloride, and aluminum chloride.
     • Reaction: $$4BCl_3 + 3LiAlH_4 \rightarrow (in dry ether) 2B₂H₆ + 3LiCl + 3AlCl_3$$
   • Using Boron Trichloride and Hydrogen:
     • When boron trichloride and hydrogen are exposed to a silent electric discharge, they combine to form Diborane and hydrogen chloride.
     • Reaction: $$2BCL_3 + 6H_2 \rightarrow (with electric discharge) B₂H₆ + 6HCl$$

Chemical Activities of Diborane

1. Reaction with Water:
   • Diborane reacts with water to yield boric acid and hydrogen gas.
   • Reaction: $$B₂H₆ + 6H₂O \rightarrow 2H₃BO₃ + 6H₂↑$$
2. Combustion:
   • When burnt in the presence of oxygen, Diborane undergoes combustion, producing boron trioxide and water. This reaction liberates a significant amount of heat.
   • Reaction: $$B₂H₆ + 3O₂ \rightarrow B₂O₃ + 3H₂O$$ with ΔH = −2006.4kJ/mol
3. Reaction with Ammonia:
   • Diborane interacts with ammonia to form the diammoniate of Diborane at 120°C. Upon further heating to 200°C, this complex decomposes to produce borazole (or borazine/inorganic benzene).
   • Reactions: $$B₂H₆ + 2NH₃ \rightarrow (at 120°C) B₂H₆,2NH₃ \rightarrow (at 200°C) B₃N₃H₆$$

Summary

Diborane is an intriguing compound, notable for its bonding structure and diverse chemical reactions. From its industrial and laboratory preparation methods to its reactivity with other compounds, Diborane exemplifies the dynamic nature of inorganic chemistry.

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SAQ-4 : Explain the difference in properties of diamond and graphite on the basis of their structure.

Introduction

Carbon, a versatile element, exists in various forms called allotropes. Diamond and graphite are two such allotropes, and their distinct properties arise from their unique structural arrangements. Understanding these properties helps in grasping the reasons behind their differences.

Properties of Diamond

1. Hardness: Diamond is renowned for being the hardest known natural material. This remarkable hardness is attributed to its crystal structure.
2. Structure: It possesses a three-dimensional tetrahedral structure.
3. Hybridization: In diamond, each carbon atom undergoes sp³ hybridization.
4. Bonding: Each carbon atom forms four sigma bonds with four other carbon atoms, resulting in a three-dimensional network of strong covalent bonds.
5. Bond Parameters: The C-C bond length in diamond is 1.54 Å (angstroms) and the bond angle is 109.5°.
6. Electrical Conductivity: Due to the absence of free electrons (all electrons are engaged in covalent bonding), diamond is an insulator and a poor conductor of electricity.
7. Appearance: Diamonds are usually transparent and exhibit a brilliant sparkle.

Properties of Graphite

1. Hardness: Graphite is soft and slippery, making it useful as a lubricant.
2. Structure: Graphite has a planar hexagonal structure. These layers are stacked over one another.
3. Hybridization: Each carbon atom in graphite undergoes sp² hybridization.
4. Bonding: Three of the four valence electrons in carbon form sigma bonds with three other carbon atoms, resulting in a planar hexagonal structure. The fourth electron of each carbon atom is delocalized and moves freely between the layers, forming a pi bond.
5. Bond Parameters: The C-C bond length within the hexagonal layers is 1.42 Å, with a bond angle of 120°.
6. Electrical Conductivity: These delocalized pi electrons make graphite a good conductor of electricity.
7. Interlayer Forces: The hexagonal layers are held together by weak van der Waals forces, allowing them to slide over each other.

Summary

The stark contrast in the properties of diamond and graphite, despite both being forms of carbon, is a testament to how atomic arrangement and bonding can dramatically influence material properties. While diamond’s tetrahedral arrangement results in its unparalleled hardness and insulating properties, graphite’s layered structure endows it with conductivity and lubricating capabilities. The study of these allotropes enriches our understanding of carbon’s versatility and underscores the profound implications of atomic and molecular structures in determining material properties.

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SAQ-5 : Why is diamond hard?

Introduction

Diamond is often hailed as the paragon of hardness among naturally occurring substances. This characteristic is due to its unique atomic arrangement and the nature of carbon-carbon bonds. Let’s explore the reasons behind diamond’s exceptional hardness.

Reasons Behind Diamond’s Hardness

1. Hybridization: Every carbon atom in a diamond undergoes sp³ hybridization. This form of hybridization leads to the formation of four strong sigma bonds with four other carbon atoms.
2. Tetrahedral Arrangement: The carbon atoms are arranged in a tetrahedral fashion. Each carbon atom is symmetrically bonded to four others, leading to a highly ordered and repetitive three-dimensional structure.
3. Extended Covalent Network: The bonding in diamond extends beyond discrete molecules, creating an extensive network. This makes the entire structure one large molecule, where billions of carbon atoms are interconnected.
4. Strength of Carbon-Carbon Bonds: The carbon-carbon covalent bond is particularly strong, with a bond energy of approximately 348 kJ/mol. The sp³ hybridization and tetrahedral arrangement uniformly distribute these bonds, adding robustness to the structure.
5. Absence of Planes of Weakness: Unlike graphite, diamond does not have planes of weakness due to weak van der Waals forces. This absence of slippage between layers contributes to diamond’s hardness.
6. High Melting Point: Diamond’s melting point of around 3,550°C further attests to the strength of the bonds holding the carbon atoms together.

Summary

Diamond’s unmatched hardness arises from its unique tetrahedral structure, where each carbon atom is covalently bonded to four others, forming a rigid and extensive three-dimensional network. The considerable energy required to break these numerous strong covalent bonds contributes to diamond’s exceptional hardness, making it invaluable in abrasive and cutting applications.

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SAQ-6 : What do you understand by (a) Allotropy (b) inert pair effect (c) Catenation.

Understanding Allotropy

1. Allotropy: Allotropy refers to the phenomenon where an element exists in two or more different forms in the same physical state. These different forms, known as allotropes, exhibit distinct physical properties despite having the same chemical composition. Allotropy arises due to different arrangements of atoms within the element.
2. Examples: Carbon exhibits allotropy, existing as diamond, graphite, and graphene.

Inert Pair Effect

1. Inert Pair Effect: The inert pair effect is a term used in inorganic chemistry to describe the tendency of the outermost s-electrons (inert pair) of the heavier elements of groups 13-16 to remain unshared in the valence shell. This effect becomes more pronounced down the group, affecting the element’s valency and chemical behavior.
2. Significance: It explains why lead(II) compounds are more stable than lead(IV) compounds, contrasting with carbon, where the +4 oxidation state is more stable.

Concept of Catenation

1. Catenation: Catenation is the ability of an element to form bonds with itself, leading to the formation of chains or rings of the same element. It is primarily observed in carbon due to its strong carbon-carbon bond.
2. Characteristic of Carbon: Carbon’s propensity for catenation is the foundation for the vast array of organic compounds, making it unique among the elements.

Summary

These concepts play crucial roles in understanding various aspects of chemistry:

• Allotropy shows how one element can exist in multiple forms with different physical properties.
• The inert pair effect helps in understanding the variations in chemical behavior and stability among the heavier elements in a group.
• Catenation underpins the diversity and complexity of organic compounds, primarily due to carbon’s unique bonding capabilities.

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SAQ-7 : What are electron deficient compounds? Is BCl₃ an electron deficient species? Explain.

Introduction

Electron Deficient Compounds: Electron deficient compounds are those that do not have enough electrons to form a complete octet around every atom through normal two-center two-electron bonds. These compounds are typically found among elements with low electronegativity, such as boron and aluminum.

Is BCl₃ an Electron Deficient Species?

1. BCl₃ as an Electron Deficient Compound: Boron trichloride (BCl₃) is a classic example of an electron deficient species. In BCl₃:
   • Boron’s Electron Count: Boron has only three valence electrons.
   • Formation of Bonds: Boron forms three covalent bonds with three chlorine atoms, using each of these valence electrons.
   • Incomplete Octet: Even after forming these bonds, boron’s central atom lacks a complete octet (it only has six electrons around it).
   • Electron Deficiency: This makes BCl₃ an electron deficient compound as the boron atom does not achieve an octet configuration.

Explanation of BCl₃‘s Electron Deficiency

1. Structure and Bonding: In BCl₃, each chlorine atom achieves an octet, but the boron atom does not. This discrepancy is because boron, having only three valence electrons, can only share three electrons with chlorine atoms, leading to an incomplete octet.
2. Nature of Boron: Boron’s ability to form stable compounds despite having an incomplete octet is characteristic of its nature. This electron deficiency allows BCl₃ to act as a Lewis acid, meaning it can accept electron pairs from other compounds.

Summary

In summary, electron deficient compounds like BCl₃ lack a complete octet in their structure. BCl₃, with boron having only three valence electrons, forms three covalent bonds and remains electron deficient. This characteristic makes BCl₃ an effective Lewis acid, capable of accepting electrons from other species.