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:
- Cation before anion: The name of the cationic part comes before the anionic part.
- 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.
- Number of ligands: Prefixes like mono-, di-, tri-, etc., indicate the number of each type of ligand.
- 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.
- 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):
- 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 - 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):
- 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 - 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)
