2nd Year Chemistry

Fundamental principles of Organic chemistry solved exercise

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Master the core concepts of organic chemistry with this detailed guide to solved exercises from the ‘Fundamental Principles of Organic Chemistry’ chapter. This resource covers key topics such as bonding, hybridization, isomerism, functional groups, and reaction mechanisms. Aligned with the latest syllabus for Lahore Board, Federal Board, and other academic boards, it includes step-by-step solutions, solved MCQs, short questions, and conceptual problems to reinforce learning. Ideal for students aiming to excel in organic chemistry, this guide simplifies complex principles and enhances exam preparation.

Q4. How organic compounds are classified? Give a suitable example of each type.

Organic compounds are classified based on their structure, functional groups, and bonding into the following major categories:

  1. Acyclic or Open-Chain Compounds: These are compounds with straight or branched chains.
  • Example: Butane (C₄H₁₀)
  1. Cyclic Compounds: These compounds have atoms arranged in a ring structure.
  • Example: Cyclohexane (C₆H₁₂)
  1. Aromatic Compounds: Compounds containing one or more benzene rings (arenes).
  • Example: Benzene (C₆H₆)
  1. Heterocyclic Compounds: Cyclic compounds where one or more atoms in the ring are not carbon.
  • Example: Pyridine (C₅H₅N)

Q5. What are homocyclic and heterocyclic compounds? Give one example of each.

  • Homocyclic Compounds: Compounds whose rings are made up entirely of carbon atoms.
  • Example: Benzene (C₆H₆)
  • Heterocyclic Compounds: Compounds that contain at least one atom other than carbon in the ring structure.
  • Example: Pyridine (C₅H₅N) (contains nitrogen in the ring)

Q6. Write the structural formulas of the two possible isomers of C₄H₁₀.

The two isomers of C₄H₁₀ are:

  1. n-Butane (Straight-chain isomer):
    Structure: CH₃-CH₂-CH₂-CH₃
  2. Iso-Butane (Branched-chain isomer):
    Structure: (CH₃)₃CH

Q7. Why is ethene an important industrial chemical?

Ethene (ethylene) is crucial in the chemical industry because:

  1. It is used as a raw material for producing polymers such as polyethylene, the most widely used plastic.
  2. It is involved in the production of other chemicals such as ethanol, ethylene oxide, and ethylene glycol, which are used in manufacturing antifreeze, detergents, and solvents.
  3. Ethene is also used as a plant hormone to stimulate fruit ripening.

Q8. What is meant by a functional group? Name typical functional groups containing oxygen.

A functional group is a specific group of atoms within a molecule responsible for the characteristic chemical reactions of that molecule. Typical oxygen-containing functional groups include:

  1. Hydroxyl group (-OH): Found in alcohols (e.g., ethanol)
  2. Carbonyl group (C=O): Found in aldehydes and ketones (e.g., formaldehyde)
  3. Carboxyl group (-COOH): Found in carboxylic acids (e.g., acetic acid)
  4. Ether group (R-O-R’): Found in ethers (e.g., diethyl ether)

Q9. What is an organic compound? Explain the importance of Wöhler’s work in the development of organic chemistry.

An organic compound is a chemical compound containing carbon atoms, usually bonded to hydrogen, oxygen, and/or other elements. Organic compounds are the basis of life and include molecules such as carbohydrates, proteins, and fats.

Wöhler’s work was groundbreaking because he synthesized urea (an organic compound) from ammonium cyanate (an inorganic compound) in 1828. This demonstrated for the first time that organic compounds could be synthesized from inorganic substances, disproving the belief that organic compounds could only be produced by living organisms, leading to the rise of modern organic chemistry.

Q10. Write a short note on cracking of hydrocarbons.

Cracking is a process in which large hydrocarbon molecules (usually alkanes) are broken down into smaller, more useful molecules, often by applying heat and pressure. This process is crucial in the petroleum industry to convert long-chain hydrocarbons into gasoline, diesel, and other products. There are two main types of cracking:

  1. Thermal Cracking: High temperature and pressure are used to break the bonds.
  2. Catalytic Cracking: A catalyst is used to lower the temperature and pressure needed for the process.

Q11. Explain reforming of petroleum with the help of a suitable example.

Reforming is a chemical process used to convert low-octane hydrocarbons into high-octane gasoline components. This process improves the quality of gasoline by rearranging the molecular structure of hydrocarbons.

  • Example: In naphtha reforming, straight-chain alkanes are converted into branched-chain alkanes, cycloalkanes, and aromatic hydrocarbons. For instance, heptane (C₇H₁₆) can be converted into methylcyclohexane or toluene, which have higher octane ratings, improving fuel efficiency.

Q12. Describe important sources of organic compounds.

Important sources of organic compounds include:

  1. Petroleum: The largest source, used for producing fuels, plastics, and chemicals.
  2. Natural Gas: Contains methane and is used as a source for organic synthesis.
  3. Coal: A source of hydrocarbons, aromatic compounds, and various other organics.
  4. Plants and Animals: Provide carbohydrates, proteins, fats, and other biochemicals used in medicine, food, and textiles.

Q13. What is orbital hybridization? Explain sp³, sp², and sp modes of hybridization of carbon.

Orbital hybridization is the mixing of atomic orbitals in an atom to form new hybrid orbitals that influence molecular geometry and bonding properties.

  1. sp³ Hybridization: Involves the mixing of one s and three p orbitals. The geometry is tetrahedral with bond angles of 109.5°.
  • Example: Methane (CH₄)
  1. sp² Hybridization: Involves the mixing of one s and two p orbitals. The geometry is trigonal planar with bond angles of 120°.
  • Example: Ethene (C₂H₄)
  1. sp Hybridization: Involves the mixing of one s and one p orbital. The geometry is linear with bond angles of 180°.
  • Example: Ethyne (C₂H₂)

Q14. Explain the type of bonds and shapes of the following molecules using hybridization approach.

  • CH₃-CH₂-CH₂-CH₃ (Butane):
  • Hybridization: sp³ for each carbon atom
  • Shape: Tetrahedral around each carbon
  • CH=CH₂ (Ethene):
  • Hybridization: sp² for each carbon
  • Shape: Trigonal planar
  • CHCl (Chloromethane):
  • Hybridization: sp³ for the carbon
  • Shape: Tetrahedral around the carbon
  • HCHO (Formaldehyde):
  • Hybridization: sp² for carbon
  • Shape: Trigonal planar

Q15. Why is there no free rotation around a double bond and free rotation around a single bond? Discuss cis-trans isomerism.

In a double bond, one of the bonds is a pi bond (π) that restricts rotation because breaking this bond requires a significant amount of energy. This is unlike a single bond, which is a sigma bond (σ) that allows free rotation because of the symmetric overlap of orbitals along the bond axis.

Cis-trans isomerism occurs due to the restricted rotation around double bonds, resulting in different spatial arrangements of groups attached to the carbon atoms involved in the double bond. In cis-isomers, similar groups are on the same side of the double bond, while in trans-isomers, they are on opposite sides.

Transition Elements Solved Exercise PTB

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Transition Elements Exercsie

Explore the solved exercise of Transition Elements from Punjab Textbook Board (PTB). Get detailed solutions, explanations, and notes tailored for college students to master key concepts of chemistry effectively.

Enhance your understanding of Transition Elements with this comprehensive solved exercise guide tailored for Punjab Textbook Board (PTB) students. Covering essential topics like electronic configurations, oxidation states, complex compounds, catalytic properties, and magnetic behavior, this guide provides step-by-step solutions to textbook exercises. Perfect for exam preparation, it includes solved MCQs, short questions, long questions, and conceptual problems. Aligned with the PTB syllabus, this resource simplifies the study of transition metals for easy learning and academic success.

(a) Binding energy
The binding energy of transition elements is influenced by the number of unpaired electrons in the d-orbitals. More unpaired electrons lead to stronger metallic bonds, increasing the binding energy.

Q.4 How does the electronic configuration of the valence shell affect the following properties of the transition elements?

(b) Paramagnetism
Paramagnetism in transition metals arises due to the presence of unpaired d-electrons. The more unpaired electrons there are, the stronger the paramagnetic property of the element.

(c) Melting points
The melting points of transition metals generally increase with the number of unpaired d-electrons, as this leads to stronger metallic bonding. However, this trend can vary across the series.

(d) Oxidation states
Transition elements exhibit variable oxidation states due to the similar energy levels of their 3d and 4s electrons. As the number of valence electrons available for bonding increases, the number of possible oxidation states also increases.

Q.5 Explain the following terms giving examples.

(a) Ligands
Ligands are ions or molecules that can donate a pair of electrons to the central metal atom/ion to form a coordination bond. Example: In [Cu(NH₃)₄]²⁺, ammonia (NH₃) acts as a ligand.

(b) Coordination sphere
The coordination sphere consists of the central metal atom/ion and the ligands directly attached to it. For example, in [Fe(CN)₆]⁴⁻, the coordination sphere is Fe and six cyanide ions (CN⁻).

(c) Substitutional alloy
A substitutional alloy forms when atoms of one element replace atoms of another element in a metal’s crystal lattice. Example: Brass is a substitutional alloy of copper and zinc.

(d) Central metal atom
The central metal atom is the atom in a coordination complex to which ligands are bonded. For example, in [Co(NH₃)₆]³⁺, cobalt (Co) is the central metal atom.

Q.6 Describe the rules for naming coordination complexes and give examples.

Answer:

  1. Cation before anion: The name of the cationic part comes before the anionic part.
  2. Ligands named first: Ligands are named before the central metal atom. Neutral ligands use their molecule name, while anionic ligands use the suffix ‘-o’.
  • Example: H₂O becomes aqua, NH₃ becomes ammine, Cl⁻ becomes chloro.
  1. Number of ligands: Prefixes like mono-, di-, tri-, etc., indicate the number of each type of ligand.
  2. Metal name: The metal is named, followed by its oxidation state in Roman numerals.
  • Example: [Cr(H₂O)₆]³⁺ is named as hexaaquachromium(III) ion.
  1. For anionic complexes: The metal’s name ends with the suffix ‘-ate’.
  • Example: [Co(CN)₆]³⁻ is named as hexacyanocobaltate(III).

Q.7 What is the difference between wrought iron and steel? Explain the Bessemer’s process for the manufacture of steel.

Answer:

  • Wrought iron is a nearly pure form of iron with less than 0.08% carbon content, making it soft and malleable. It is used for decorative ironwork.
  • Steel contains more carbon (0.1% to 2%), making it stronger and harder than wrought iron. It is widely used in construction and manufacturing.

Bessemer’s process:
The Bessemer process is a method for making steel by blowing air through molten pig iron to oxidize and remove impurities like carbon, silicon, and manganese. The process helps in producing steel rapidly and at a lower cost.

Q.8 Explain the following giving reasons.

(a) Why does damaged tin-plated iron get rusted quickly?
Answer: When tin-plated iron is damaged, the exposed iron reacts with water and oxygen, forming rust. Since tin is less reactive than iron, the iron oxidizes (rusts) faster when exposed in the presence of tin, acting as a sacrificial element.

(b) Under what conditions does aluminum corrode?
Answer: Aluminum corrodes when exposed to moist environments containing salts or acids. However, aluminum forms a protective layer of aluminum oxide (Al₂O₃) that prevents further corrosion under normal conditions.

(c) How does the process of galvanizing protect iron from rusting?
Answer: Galvanizing involves coating iron with a layer of zinc. Zinc acts as a sacrificial anode, meaning it corrodes in place of the iron. Even if the zinc coating is damaged, the exposed iron remains protected as the zinc continues to corrode preferentially.

Q.9 How chromate ions are converted into dichromate ions?

Answer:
Chromate ions CrO42- are converted into dichromate ions Cr2O7^2- in acidic conditions by the following equilibrium reaction:

2 CrO4^2- + 2 H^+ → Cr2O72- + H2O

This conversion involves the protonation of chromate ions, leading to the formation of dichromate ions.

Q.10 Describe the preparation of KMnO₄ and K₂CrO₄.

Answer:

Preparation of Potassium Permanganate (KMnO₄):

  1. Oxidation of Manganese Dioxide (MnO₂): Manganese dioxide is fused with potassium hydroxide (KOH) in the presence of an oxidizing agent like potassium nitrate (KNO₃):
    2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
  2. Conversion of Potassium Manganate to Potassium Permanganate: Potassium manganate ((K_2MnO_4)) is oxidized in an acidic or neutral medium to form potassium permanganate:
    3 K2MnO4 + 2 H2O → 2 KMnO4 + MnO2 + 4 KOH

Preparation of Potassium Chromate (K₂CrO₄):

  1. Oxidation of Chromite Ore (FeCr₂O₄): Chromite ore is heated with sodium carbonate (Na₂CO₃) in the presence of air or oxygen, yielding sodium chromate ((Na₂CrO₄)):
    4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2
  2. Conversion to Potassium Chromate: Sodium chromate is treated with potassium chloride (KCl), forming potassium chromate:
    Na2CrO4 + 2 KCl → K2CrO4 + 2 NaCl

Q.11 Give systematic names to the following complexes:

(a) [Fe(CO)₅]
Answer: Pentacarbonyliron(0)

(b) [Co(NH₃)₆]Cl₃
Answer: Hexaamminecobalt(III) chloride

(c) [Fe(H₂O)₆]²⁺
Answer: Hexaaquairon(II) ion

(d) Na₃[CoF₆]
Answer: Sodium hexafluorocobaltate(III)

(e) K₃[Cu(CN)₄]
Answer: Potassium tetracyanocuprate(I)

(f) K₂[PtCl₆]
Answer: Potassium hexachloroplatinate(IV)

(g) [Pt(OH)₂(NH₃)₄]SO₄
Answer: Tetraamminehydroxoplatinum(IV) sulfate

(h) [Cr(OH)₃(H₂O)₃]
Answer: Trihydroxotriaquachromium(III)

Group IIIA and IVA elements Complete Exercise Solved

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Q6. Why is aluminium not found as a free element? Explain the chemistry of the borax bead test.

  • Answer: Aluminium is highly reactive and readily combines with oxygen to form aluminium oxide (Al₂O₃), which is a stable compound. This prevents aluminium from being found in a free, metallic state in nature.
    The borax bead test is used for detecting metal ions. Borax (Na₂B₄O₇·10H₂O) melts into a clear glassy bead at high temperatures. When certain metal oxides are heated in the bead, they react with the borax to produce characteristic colors. Aluminium doesn’t form distinct colors in this test.

Q7. How does orthoboric acid react with:

(a) Sodium hydroxide
(b) Ethyl alcohol

  • Answer:
    (a) With Sodium Hydroxide (NaOH): Orthoboric acid reacts with NaOH to form sodium borate (Na₂B₄O₇), as shown in the reaction:
    B(OH)₃ + NaOH → NaBO₂ + 2H₂O
    (b) With Ethyl Alcohol: Orthoboric acid reacts with ethyl alcohol in the presence of concentrated sulfuric acid to form triethyl borate:
    B(OH)₃ + 3C₂H₅OH → B(OC₂H₅)₃ + 3H₂O

Q8. How will you convert boric acid into borax and vice versa?

  • Answer:
  • Boric acid to borax: Boric acid (H₃BO₃) can be converted into borax by heating it with sodium carbonate (Na₂CO₃), which forms sodium tetraborate (borax) and water:
    4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂
  • Borax to boric acid: Borax reacts with a strong acid like hydrochloric acid (HCl) to produce boric acid:
    Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl

Q9. Why are liquid silicones preferred over ordinary organic lubricants?

  • Answer: Liquid silicones are preferred over organic lubricants due to their higher thermal stability, resistance to oxidation, chemical inertness, water repellency, and low-temperature fluidity. They also have low vapor pressure, making them suitable for high-temperature applications.

Q10. Explain:

(a) CO₂ is non-polar in nature:

  • CO₂ has a linear molecular geometry with equal bond dipoles in opposite directions. The symmetry of the molecule causes the dipoles to cancel out, making CO₂ non-polar.
    (b) CO₂ is acidic in character:
  • When CO₂ dissolves in water, it reacts to form carbonic acid (H₂CO₃), which dissociates to release hydrogen ions (H⁺), demonstrating its acidic nature:
    CO₂ + H₂O → H₂CO₃ → H⁺ + HCO₃⁻

Q11. Why is CO₂ a gas at room temperature while SiO₂ is a solid?

  • Answer: CO₂ is a small, simple molecule with weak van der Waals forces between molecules, allowing it to remain in the gaseous state at room temperature. In contrast, SiO₂ forms a giant covalent network structure with strong Si-O bonds, making it a solid with a high melting point.

Q12. Give the names and the formulas of different acids of boron.

  • Answer:
  • Boric acid (H₃BO₃)
  • Tetrahydroxyboric acid (H₄BO₄)
  • Metaboric acid (HBO₂)

Q13. What is the importance of oxides of lead in paints?

  • Answer: Lead oxides, such as lead(II) oxide (PbO) and lead(IV) oxide (Pb₃O₄), are used in paints because they enhance durability, increase resistance to corrosion, and provide brilliant color to the pigments. However, due to toxicity, their use is now limited.

Q14. Give the names, electronic configurations, and occurrence of Group-IIIA elements of the periodic table.

  • Answer: The Group-IIIA elements are:
  • Boron (B): 1s² 2s² 2p¹
  • Aluminium (Al): [Ne] 3s² 3p¹
  • Gallium (Ga): [Ar] 3d¹⁰ 4s² 4p¹
  • Indium (In): [Kr] 4d¹⁰ 5s² 5p¹
  • Thallium (Tl): [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹
    These elements are found in various ores and minerals such as borates, bauxite (Al), and zinc ores (Ga and In).

Q15. Discuss the peculiar behavior of boron with respect to the other members of Group-IIIA elements.

  • Answer: Boron exhibits several unique properties compared to other Group-IIIA elements (Al, Ga, In, Tl):
  • Non-metallic Character: Boron is a metalloid, while the others are metals.
  • Covalency: Boron forms covalent bonds, whereas the other elements can form ionic compounds.
  • High Ionization Energy: Boron has a much higher ionization energy compared to the heavier Group-IIIA elements.
  • Lack of Metallic Conductivity: Unlike the metals in its group, boron is a poor conductor of electricity.
  • Allotropes: Boron exists in various allotropic forms, unlike aluminium and other members.

Q16. (a) What is borax?

(b) Describe its commercial preparation.
(c) Outline the principal uses of borax.
(d) How does borax serve as a water-softening agent?

  • Answer:
    (a) Borax is a sodium salt of boric acid with the formula Na₂B₄O₇·10H₂O.
    (b) Commercial Preparation: Borax is prepared by extracting borate minerals from lake beds and then treating them with soda ash (Na₂CO₃) to form borax.
  • Na₂CO₃ + CaB₄O₇ + H₂O → Na₂B₄O₇ + CaCO₃
    (c) Uses of Borax:
  • Used in glass and ceramic production.
  • As a flux in metallurgy.
  • In detergents and soaps as a water softener.
  • As an antiseptic and preservative.
    (d) Water Softening: Borax helps soften water by binding to calcium and magnesium ions, preventing them from interfering with soap action. This allows soap to lather better in hard water.

Q17. (a) What is boric acid?

(b) How is boric acid prepared in the laboratory?
(c) Give properties and uses of boric acid.

  • Answer:
    (a) Boric Acid (H₃BO₃) is a weak monobasic acid derived from boron.
    (b) Laboratory Preparation: Boric acid can be prepared by adding hydrochloric acid to a hot, concentrated solution of borax:
    Na₂B₄O₇ + 2HCl + 5H₂O → 4H₃BO₃ + 2NaCl
    (c) Properties:
  • It is a white crystalline solid, soluble in water.
  • Boric acid is a weak acid, antiseptic, and has a low vapor pressure.
    Uses:
  • In antiseptic solutions for minor cuts.
  • In the manufacture of glass and ceramics.
  • In insecticides to kill ants and roaches.
  • As a pH buffer in swimming pools.

Q18. (a) Give the names along with the formulas of three important ores of aluminium.

(b) How and under what conditions does aluminium react with the following:
i) Oxygen ii) Hydrogen iii) Halogens iv) Acids v) Alkalies

  • Answer:
    (a) Ores of Aluminium:
  • Bauxite: Al₂O₃·2H₂O
  • Cryolite: Na₃AlF₆
  • Corundum: Al₂O₃
    (b) Reactions of Aluminium:
  • With Oxygen: Aluminium reacts readily with oxygen, forming a protective layer of aluminium oxide:
    4Al + 3O₂ → 2Al₂O₃
  • With Hydrogen: Aluminium does not react with hydrogen directly.
  • With Halogens: Aluminium reacts with halogens to form aluminium halides:
    2Al + 3Cl₂ → 2AlCl₃
  • With Acids: Aluminium dissolves in acids to release hydrogen gas:
    2Al + 6HCl → 2AlCl₃ + 3H₂
  • With Alkalies: Aluminium reacts with alkalis to form aluminates and hydrogen gas:
    2Al + 2NaOH + 6H₂O → 2Na[Al(OH)₄] + 3H₂

Q19. Give the names, electronic configurations, and occurrence of Group-IVA elements of the periodic table.

  • Answer: The Group-IVA elements are:
  • Carbon (C): 1s² 2s² 2p²
  • Silicon (Si): [Ne] 3s² 3p²
  • Germanium (Ge): [Ar] 3d¹⁰ 4s² 4p²
  • Tin (Sn): [Kr] 4d¹⁰ 5s² 5p²
  • Lead (Pb): [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
    These elements are found in nature in various forms, such as carbon in coal, silicon in sand (SiO₂), and lead in galena (PbS).

Q20. Discuss the peculiar behavior of carbon with respect to the other members of Group-IVA of the periodic table.

  • Answer:
  • Catenation: Carbon uniquely forms long chains and rings of atoms due to its strong C-C bonds. Other Group-IVA elements do not show such extensive catenation.
  • Allotropes: Carbon exists in various allotropes like graphite, diamond, and fullerenes, whereas silicon and others have fewer allotropic forms.
  • Multiple Bonds: Carbon forms stable double and triple bonds (C=C and C≡C), unlike heavier elements, which rarely form multiple bonds.
  • Non-metallic Nature: Carbon is a non-metal, whereas the others in the group (like Sn and Pb) exhibit metallic properties.

Q21. (a) What are silicones?

(b) Give a brief summary of the principal properties of silicones.
(c) Outline the uses of silicones.
(d) What are silicates?
(e) Describe the structures of silicates.

  • Answer:
    (a) Silicones are synthetic polymers made up of repeating units of siloxane (Si-O-Si) with organic side groups attached to the silicon atoms.
    (b) Properties of Silicones:
  • High thermal stability
  • Water repellency
  • Electrical insulating properties
  • Flexibility and low toxicity
    (c) Uses of Silicones:
  • Used as sealants and adhesives
  • In medical devices and implants
  • As lubricants and insulating materials
  • In cosmetics and personal care products
    (d) Silicates: Silicates are minerals containing silicon and oxygen, commonly combined with other elements like magnesium, iron, and aluminium. Examples include quartz (SiO₂) and feldspar.
    (e) Structures of Silicates: Silicates have a tetrahedral structure, where one silicon atom is surrounded by four oxygen atoms. These tetrahedra can link in various ways to form different types of silicates, such as isolated, chain, sheet, and framework silicates.

Periodic Classification of Elements

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Periodic classification of elements and periodicity (Solved exercise PTB)

Q4:
What are the improvements made in Mendeleev’s periodic table?

  • Mendeleev left gaps for undiscovered elements and predicted their properties accurately.
  • He arranged elements based on increasing atomic mass, but later corrections were made based on atomic number.
  • Grouping of elements with similar chemical properties was better explained by the periodic table developed by Moseley using atomic number.

Q5:
How does the classification of elements in different blocks help in understanding their chemistry?

  • Elements are classified into s-block, p-block, d-block, and f-block based on their valence electron configurations, which helps in understanding their chemical behavior.
  • s-block elements are typically metals, p-block elements include non-metals, metalloids, and some metals, d-block elements are transition metals, and f-block contains lanthanides and actinides with complex chemistry.

Q6:
How do you justify the position of hydrogen at the top of various groups?

  • Hydrogen is placed at the top of Group 1 due to its ability to form a +1 oxidation state like alkali metals, but it can also form a -1 oxidation state (like halogens in Group 17).
  • Its properties resemble both alkali metals (in forming H⁺ ions) and halogens (in forming H⁻ ions), which justifies its unique position.

Q7:
Why is the ionic radii of negative ions larger than the size of their parent atoms?

  • When an atom gains an electron to form a negative ion (anion), the repulsion between the electrons increases, causing the electron cloud to expand. This results in the anion having a larger radius than the parent atom.

Q8:
Why does ionization energy decrease down the group and increase along a period?

  • Down a group: Ionization energy decreases because the atomic size increases, and the outermost electrons are farther from the nucleus, reducing the attraction.
  • Across a period: Ionization energy increases because atomic size decreases and nuclear charge increases, making it harder to remove an electron.

Q9:
Why is the second value of electron affinity of an element usually shown with a positive sign?

  • The second electron affinity is positive because after the addition of one electron, the atom becomes negatively charged. Adding another electron to a negative ion requires energy to overcome the repulsion between the added electron and the negative ion.

Q10:
Why does metallic character increase from top to bottom in a group of metals?

  • As you move down a group, the atomic size increases, and the outermost electrons are farther from the nucleus, making them easier to lose. This enhances the metallic character, which is associated with the ease of losing electrons.

Q11:
Explain the variation in melting points along the short periods.

  • Melting points tend to increase across a period until reaching Group 14, where they reach a maximum and then decrease. This is due to the increasing strength of metallic bonding or covalent bonding in elements like silicon and then weaker van der Waals forces in non-metals.

Q12:
Why is the oxidation state of noble gases usually zero?

  • Noble gases have a completely filled valence shell, making them highly stable and inert. They do not easily lose or gain electrons, which is why their oxidation state is generally zero.

Q13:
Why is diamond a non-conductor and graphite a good conductor?

  • Diamond: In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms, and all valence electrons are used in bonding. There are no free electrons to conduct electricity.
  • Graphite: In graphite, each carbon atom is bonded to three other carbon atoms, with one delocalized electron per atom. These free electrons can move through the layers, allowing graphite to conduct electricity.

Q14:
Give a brief reason for the following:

(a) d- and f-block elements are called transition elements.

  • They are called transition elements because they exhibit properties that are transitional between s- and p-block elements. They have partially filled d- or f-orbitals, which gives them unique chemical and physical properties.

(b) Lanthanide contraction controls the atomic sizes of elements of the 6th and 7th periods.

  • Lanthanide contraction refers to the steady decrease in atomic and ionic sizes of the lanthanides as atomic number increases. This contraction influences the sizes of elements in the 6th and 7th periods, particularly in d- and f-block elements.

(c) The melting and boiling points of elements increase from left to the middle of the s- and p-block elements and decrease onward.

  • From left to the middle of the s- and p-block, elements exhibit stronger metallic or covalent bonding, resulting in higher melting and boiling points. After the middle, the elements become non-metals with weaker van der Waals forces, leading to lower melting and boiling points.

(d) The oxidation states vary in a period but remain almost constant in a group.

  • Across a period, elements can exhibit multiple oxidation states due to the availability of different numbers of valence electrons for bonding. Within a group, the number of valence electrons remains the same, so the oxidation states are more consistent.

(e) The hydration energies of the ions are in the following order: Al³⁺ > Mg²⁺ > Na⁺.

  • Hydration energy is higher for ions with a greater charge and smaller size. Al³⁺ has the smallest size and highest charge, followed by Mg²⁺, and then Na⁺, leading to the given order.

(f) Ionic character of halides decreases from left to right in a period.

  • As you move across a period, the electronegativity of the elements increases. The greater the electronegativity difference between the metal and the halide, the higher the ionic character. This difference decreases across a period, reducing the ionic character.

(g) Alkali metals give ionic hydrides.

  • Alkali metals react with hydrogen to form ionic hydrides (e.g., NaH) where the hydrogen is present as a hydride ion (H⁻) due to the large difference in electronegativity between alkali metals and hydrogen.

(h) Although sodium and phosphorus are present in the same period of the periodic table, their oxides are different in nature: Na₂O is basic while P₄O₁₀ is acidic.

  • Sodium, being a metal, forms a basic oxide (Na₂O), which reacts with water to produce a strong base (NaOH). Phosphorus, being a non-metal, forms an acidic oxide (P₄O₁₀), which reacts with water to form phosphoric acid (H₃PO₄).

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