GSEB Class 12 Chemistry Solutions Chapter 8 d-and f-Block Elements

 

Question 1. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Answer: Silver displays a +2 oxidation state. In this particular state, its 4d-sub-shell contains nine electrons, meaning one of the 4d-orbitals is only partially occupied. Consequently, silver can be categorized as a transition element.

In simple words: A transition element is defined by having partially filled d-orbitals in its stable oxidation states. Although silver has full d-orbitals in its ground state, its +2 oxidation state reveals a partially filled d-orbital, classifying it as a transition metal.

🎯 Exam Tip: Remember that the definition of a transition element includes having partially filled d-orbitals in *any* common oxidation state, not just the ground state. This is a common point of confusion for elements like silver or copper.

 

Question 2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1. Why?
Answer: In the formation of metallic bonds for zinc, no electrons from its 3d-orbitals are involved. This differs from all other metals in the 3d series, where d-orbital electrons consistently participate in forming metallic bonds.

In simple words: Zinc has the lowest enthalpy of atomization because its 3d electrons are not used for metallic bonding, unlike other 3d transition metals where d-electrons contribute to stronger bonds.

🎯 Exam Tip: Lower enthalpy of atomization often correlates with weaker metallic bonding. For zinc, the fully filled d-orbitals mean these electrons are less available for delocalization, leading to weaker metallic bonds compared to other transition elements.

 

Question 3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?
Answer: Manganese (Mn), with an atomic number of 25, shows the broadest range of oxidation states. This is because manganese possesses the maximum number of unpaired electrons among the 3d series, enabling it to exhibit oxidation states from +2 up to +7.

In simple words: Manganese shows the most oxidation states (from +2 to +7) because it has the highest number of unpaired electrons, which can all be involved in bonding.

🎯 Exam Tip: The variability in oxidation states of transition metals is directly linked to the number of available unpaired d-electrons. Elements with d⁵ configuration, like Mn, tend to show a wide range due to the stability of half-filled orbitals.

 

Question 4. The E°(M²⁺/M) value for copper is positive (+ 0.34V). What is possibly the reason for this? (Hint: consider its high ∆H° and low ∆hydH°)
Answer: The positive standard electrode potential (E°) for copper can be attributed to the significant energy required to convert solid copper to its aqueous divalent ion \(Cu^{2+}(aq)\), which is not adequately compensated by its hydration enthalpy.

In simple words: Copper's positive electrode potential means it's harder to oxidize to \(Cu^{2+}\) than hydrogen, primarily because the energy needed for atomization and ionization is not sufficiently offset by the energy released during hydration.

🎯 Exam Tip: A positive E° value for M²⁺/M indicates that the metal is less reactive than hydrogen and does not readily lose electrons to form its ions, often due to high ionization energy or low hydration energy for the ion.

 

Question 5. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Answer: The inconsistent changes in the first and second ionization enthalpies across the first transition series are primarily due to the varying degrees of stability associated with different 3d-electron configurations. Configurations such as \(d^0\), \(d^5\), and \(d^{10}\) exhibit enhanced stability, which typically results in higher ionization enthalpy values for elements possessing these configurations. For instance, chromium (Cr) has a low first ionization enthalpy because an electron is removed from the 4s-orbital, but its second ionization enthalpy is very high as \(Cr^+\) attains a stable \(d^5\) configuration. Similarly, zinc (Zn) shows a very high first ionization enthalpy because an electron must be removed from its highly stable \(3d^{10}4s^2\) configuration.

In simple words: Ionization energies in transition metals vary unevenly because some electron configurations like half-filled (\(d^5\)) or completely filled (\(d^{10}\)) d-orbitals are extra stable, making it harder to remove electrons.

🎯 Exam Tip: Always consider the electron configurations, especially half-filled and completely filled d-orbitals, when explaining trends or irregularities in ionization energies for transition elements, as these impart significant stability.

 

Question 6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer: Metals typically achieve their highest oxidation states when combined with oxygen or fluorine. This is due to the inherently small size and high electronegativity of both oxygen and fluorine, which enables them to effectively oxidize the metal to its maximum possible oxidation state.

In simple words: Oxygen and fluorine are very electronegative and small, allowing them to pull electrons most effectively from metals and thus achieve the highest possible oxidation states in compounds.

🎯 Exam Tip: The ability of an element to stabilize higher oxidation states in a metal is directly related to its electronegativity and size. Oxygen and fluorine excel in this regard, forming strong covalent bonds that facilitate high oxidation states.

 

Question 7. Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Answer: \(Cr^{2+}\) acts as a stronger reducing agent compared to \(Fe^{2+}\). This enhanced reducing capability stems from the electronic configuration change of \(Cr^{2+}\) from \(d^4\) to \(d^3\). The resultant \(d^3\) configuration is particularly stable, corresponding to a half-filled \(t_{2g}\) level in an octahedral ligand field.

In simple words: \(Cr^{2+}\) is a stronger reducing agent than \(Fe^{2+}\) because it can easily lose an electron to form \(Cr^{3+}\), which has a stable \(d^3\) configuration (half-filled \(t_{2g}\) orbitals).

🎯 Exam Tip: The stability of products is a key factor in determining reducing or oxidizing strength. For transition metals, achieving stable \(d^3\) (half-filled \(t_{2g}\)), \(d^5\) (half-filled), or \(d^{10}\) (fully filled) configurations drives redox reactions.

 

Question 8. Calculate the 'spin only' magnetic moment of Mg²⁺(aq) ion (Z = 27)?
Answer: For a divalent ion with atomic number 27 (Cobalt), its electronic configuration will be \(d^7\). The spin-only magnetic moment \(\mu\) is calculated using the formula: \(\mu = \sqrt{n(n+2)}\) BM
In this case, for a \(d^7\) configuration, there are 3 unpaired electrons (n=3).
\(\mu = \sqrt{3(3+2)}\) BM
\(\mu = \sqrt{3 \times 5}\) BM
\(\mu = \sqrt{15}\) BM
\(\mu \approx 3.87\) BM

In simple words: The magnetic moment of an ion depends on the number of unpaired electrons. For an ion with Z=27 (Cobalt) in its +2 state, it has 3 unpaired electrons, resulting in a magnetic moment of approximately 3.87 Bohr Magnetons.

🎯 Exam Tip: To calculate spin-only magnetic moment, first determine the number of unpaired electrons (n) from the electronic configuration of the ion. Then apply the formula \(\mu = \sqrt{n(n+2)}\) BM. Be careful to correctly identify the element from its atomic number (Z).

 

Question 9. Explain why Cu²⁺ ion is not stable in aqueous solutions?
Answer: The provided content for this question states: The electronic configurations of the three transition series are given in the table.

In simple words: The \(Cu^{2+}\) ion is highly stable in aqueous solutions due to its significantly greater hydration enthalpy compared to \(Cu^{+}\), which compensates for the higher second ionization energy.

🎯 Exam Tip: When evaluating ion stability in aqueous solutions, always consider the balance between ionization enthalpies and hydration enthalpies. A higher hydration energy can often stabilize an ion even if its formation requires more ionization energy.

 

Question 10. Explain giving reasons?
(1) Transition metals and many of their compounds show paramagnetic behaviour.
(2) The enthalpies of atomisation of the transition metals are high.
(3) The transition metals generally form coloured compounds.
(4) Transition metals and their many compounds act as good catalysts.
Answer:
(1) Transition metals and many of their compounds exhibit paramagnetic behavior because they possess unpaired electrons.
(2) The high enthalpies of atomization observed in transition metals are attributed to the presence of unpaired electrons in their valence shells, which leads to strong interatomic metallic bonding.
(3) Transition metals often form colored compounds. This is due to the presence of unpaired electrons, which can readily undergo d-d electronic transitions by absorbing specific wavelengths of visible light.
(4) Many transition metals and their compounds function as effective catalysts. This catalytic activity arises from their ability to have vacant orbitals and to exhibit multiple variable oxidation states.

In simple words: Transition metals have unique properties like paramagnetism, high atomization enthalpies, color, and catalytic activity because they possess unpaired d-electrons and can exist in various oxidation states, allowing for strong bonding, d-d transitions, and reaction pathways.

🎯 Exam Tip: A comprehensive understanding of transition metal properties hinges on their electronic configurations, particularly the availability and participation of d-electrons in bonding and energy transitions.

 

Question 11. What are interstitial compounds? Why are such compounds well known for transition metals?
Answer: Interstitial compounds are formed when most transition elements incorporate small non-metallic atoms (such as H, B, C, N, Si) into the interstitial voids within their crystal lattice at elevated temperatures. These small non-metal atoms fit into the spaces between the metal atoms.
Properties:
(1) The properties of these interstitial compounds resemble those of the parent metal, though their physical properties can differ.
(2) These compounds are typically very hard.
(3) Their ductility and malleability are reduced.
(4) Their electrical conductivity decreases.
(5) Examples include TiH and NiH₂.

In simple words: Interstitial compounds are formed when small atoms like hydrogen or carbon lodge in the gaps of a transition metal's structure, making the metal harder but less ductile, while still retaining some metallic properties.

🎯 Exam Tip: Interstitial compounds are a key characteristic of transition metals due to their suitable crystal structures with voids. Focus on the impact of interstitial atoms on properties like hardness and conductivity.

 

Question 12. How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate with examples?
Answer: In transition elements, consecutive oxidation states typically differ by unity. For example, manganese (Mn) can display all oxidation states from +2 to +7. In contrast, non-transition metals often exhibit variable oxidation states that differ by two units. For instance, lead shows \(Pb(II)\) and \(Pb(IV)\), while tin shows \(Sn(II)\) and \(Sn(IV)\).

In simple words: Transition metals show oxidation states that change by one unit (e.g., +2, +3, +4), while non-transition metals often show oxidation states that change by two units (e.g., +2, +4).

🎯 Exam Tip: The difference in oxidation state variability between transition and non-transition metals lies in the involvement of d-electrons. Transition metals can involve varying numbers of d-electrons, leading to single-unit changes, whereas non-transition metals often involve s- and p-electrons in pairs.

 

Question 13. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer: Potassium dichromate (\(K_2Cr_2O_7\)) is prepared through a multi-step process from iron chromite ore (\(FeCr_2O_4\)).
Preparation:
(1) Initial Step: Iron chromite ore is heated with sodium carbonate in the presence of air to yield sodium chromate.
\[4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2\]
(2) Conversion to Dichromate: The soluble sodium chromate (\(Na_2CrO_4\)) is then acidified with sulfuric acid to produce sodium dichromate (\(Na_2Cr_2O_7\)).
\[2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O\]
(3) Crystallization: Finally, sodium dichromate is treated with potassium chloride to obtain crystals of potassium dichromate.
\[Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl\]
Potassium dichromate is an orange-red crystalline substance. When heated, it decomposes into potassium chromate, chromium oxide, and oxygen.
\[4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 2Cr_2O_3 +3O_2\]
Structure and pH Effect:
The chromate ion (\(CrO_4^{2-}\)) and dichromate ion (\(Cr_2O_7^{2-}\)) exist in an equilibrium that is interconvertible by adjusting the pH of the solution.
\[2CrO_4^{2-} \text{ (Yellow, Chromate)} + 2H^+ \xrightleftharpoons{} Cr_2O_7^{2-} \text{ (Orange, Dichromate)} + H_2O\]
If the pH of a potassium dichromate solution is increased (made alkaline by adding \(OH^-\)), the orange-red dichromate ions (\(Cr_2O_7^{2-}\)) are converted into yellow chromate ions (\(CrO_4^{2-}\)). Conversely, acidification reverses this process. The four oxygen atoms around the chromium atom in both ions are typically arranged in a tetrahedral geometry.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्रोमेट आयन (\(CrO_4^{2-}\)) और डाइक्रोमेट आयन (\(Cr_2O_7^{2-}\)) की संरचना को दर्शाता है। क्रोमेट आयन में एक केंद्रीय क्रोमियम परमाणु चार ऑक्सीजन परमाणुओं से घिरा होता है, जो एक टेट्राहेड्रल व्यवस्था बनाते हैं। डाइक्रोमेट आयन में दो क्रोमियम परमाणु एक ऑक्सीजन परमाणु से जुड़े होते हैं, और प्रत्येक क्रोमियम परमाणु शेष ऑक्सीजन परमाणुओं से जुड़ा होता है, जिसमें क्रोमियम-ऑक्सीजन बांड लंबाई और कोण दिखाए जाते हैं।
Oxidising properties:
Potassium dichromate is a potent oxidizing agent in acidic solutions. Its oxidizing action can be represented as follows:
\[Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad [E° = +1.33 V]\]
The overall molecular reaction in acidic medium is:
\[K_2Cr_2O_7 + 4H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O + 3[O]\]
(a) It oxidizes iodides to iodine:
\[Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2\]
This reaction is useful for the quantitative estimation of iodide ions in volumetric analysis.
(b) It oxidizes H₂S to S:
\[K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S\]
The ionic equation is:
\[Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 7H_2O + 3S\]
(c) It oxidizes ferrous salts to ferric salts:
\[K_2Cr_2O_7 + 7H_2SO_4 + 6FeSO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + 7H_2O\]
The ionic equation is:
\[Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O\]
This reaction is used in the volumetric estimation of ferrous ions.
(d) It oxidizes Sn²⁺ ions to Sn⁴⁺ ions:
\[3Sn^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O\]
Uses of Potassium Dichromate:
- It serves as a primary standard in volumetric analysis.
- It is used as an oxidizing agent in inorganic chemistry.
- Industrially, it finds application in chrome tanning, the production of chrome alum, and in calico printing and dyeing processes.
In aqueous solution, the dichromate and chromate ions are in equilibrium. Increasing the pH by adding alkali converts the orange dichromate ions into yellow chromate ions.
\[Cr_2O_7^{2-} \text{ (Orange, Dichromate)} \xrightleftharpoons{OH^- / H^+} 2CrO_4^{2-} \text{ (Yellow, Chromate)}\]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्रोमेट आयन (\(CrO_4^{2-}\)) और डाइक्रोमेट आयन (\(Cr_2O_7^{2-}\)) के बीच pH-निर्भर संतुलन को दर्शाता है। क्षारीय माध्यम में क्रोमेट आयन प्रभावी होता है, जबकि अम्लीय माध्यम में डाइक्रोमेट आयन बनता है। तीर के ऊपर OH^- आयन क्षारीयता को और H⁺ आयन अम्लीयता को दर्शाता है।

In simple words: Potassium dichromate is made from chromite ore by roasting with sodium carbonate, acidifying the resulting chromate, and then crystallizing. Its orange color comes from the dichromate ion, which turns yellow to chromate ion when the solution becomes alkaline. It is a strong oxidizing agent in acidic media.

🎯 Exam Tip: For potassium dichromate, remember its preparation steps, the CrO₄^2- / Cr₂O₇^2- equilibrium and color change with pH, and its strong oxidizing power, especially in acidic conditions, focusing on the balanced ionic equations.

 

Question 14. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
1. Iodide
3. H₂S
Answer: Potassium dichromate acts as a potent oxidizing agent, particularly in acidic environments. Its characteristic oxidizing reactions, especially with iodide and hydrogen sulfide, can be represented by the following ionic equations:
(a) Oxidation of iodides to iodine:
Potassium dichromate oxidizes iodide ions (\(I^-\)) to elemental iodine (\(I_2\)) in acidic solutions.
\[Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2\]
This specific reaction is frequently employed in volumetric analysis for the quantitative determination of iodide ions.
(b) Oxidation of H₂S to sulfur:
In acidic conditions, potassium dichromate oxidizes hydrogen sulfide (\(H_2S\)) to elemental sulfur (S).
\[Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 7H_2O + 3S\]
The general oxidizing action in acidic medium can be summarized as:
\[Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\]
This allows potassium dichromate to oxidize various substances, including ferrous salts to ferric salts and Sn²⁺ ions to Sn⁴⁺ ions, as well as being used as a primary standard in volumetric analysis and for chrome tanning.

In simple words: Potassium dichromate is a strong oxidizer in acidic conditions, turning orange dichromate ions into green \(Cr^{3+}\) as it oxidizes other substances like iodide to iodine or hydrogen sulfide to sulfur.

🎯 Exam Tip: Focus on writing accurate balanced ionic equations for redox reactions involving dichromate. Remember that in acidic medium, dichromate reduces to Cr³⁺, and hydrogen ions (\(H^+\)) and water (\(H_2O\)) are used to balance the equation.

 

Question 15. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with
3. Oxalic acid? Write the ionic equations for the reactions.
2. SO₂ and
Answer: Potassium permanganate (\(KMnO_4\)) is prepared through a two-stage process from pyrolusite ore (\(MnO_2\)).
Preparation:
(1) Conversion to Potassium Manganate: Pyrolusite ore (\(MnO_2\)) is fused with potassium hydroxide (KOH) or potassium carbonate (\(K_2CO_3\)) in the presence of atmospheric oxygen or an oxidizing agent like \(KNO_3\) or \(KClO_3\). This reaction produces a green mass of potassium manganate (\(K_2MnO_4\)).
\[2MnO_2 + 4KOH + O_2 \xrightarrow{Heat} 2K_2MnO_4 + 2H_2O\]
(2) Oxidation to Potassium Permanganate: The green potassium manganate is then extracted with water and oxidized to purple potassium permanganate. This oxidation can be achieved either electrolytically or by bubbling chlorine gas or ozone through the solution.
Electrolytic Oxidation:
At anode: \(MnO_4^{2-} \rightarrow MnO_4^- + e^-\)
At cathode: \(2H^+ + 2e^- \rightarrow H_2\)
Chemical Oxidation (e.g., with chlorine):
\[2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KOH + O_2\]
The purple colored solution is concentrated by evaporation, yielding crystals of \(KMnO_4\) upon cooling.
Potassium permanganate is a purple crystalline solid that decomposes to \(K_2MnO_4\), \(MnO_2\), and \(O_2\) upon strong heating.
\[2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2\]
Structure:
In the permanganate ion (\(MnO_4^-\)), manganese is in the \(sp^3\) hybridization state, and four oxygen atoms are tetrahedrally arranged around the central manganese atom.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र मैंगनेट आयन (\(MnO_4^{2-}\), हरा) और परमैंगनेट आयन (\(MnO_4^{-}\), बैंगनी) दोनों की टेट्राहेड्रल संरचनाओं को दर्शाता है। दोनों आयनों में, एक केंद्रीय मैंगनीज परमाणु चार ऑक्सीजन परमाणुओं से घिरा होता है। रंग में अंतर मैंगनीज की ऑक्सीकरण अवस्था और उनके इलेक्ट्रॉनिक विन्यास में भिन्नता के कारण होता है।
Oxidizing properties of acidified potassium permanganate:
Potassium permanganate is a powerful oxidizing agent. In acidic medium, it readily accepts electrons, typically reducing manganese from its +7 oxidation state in \(MnO_4^-\) to +2 in \(Mn^{2+}\).
The general reaction in acidic medium is:
\[2KMnO_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O]\]
Here are the reactions with the specified substances:
(i) Reaction with Oxalic Acid (Oxalates to carbon dioxide):
Acidified potassium permanganate oxidizes oxalic acid (or oxalate ions) to carbon dioxide.
\[2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2\]
(ii) Reaction with Sulphur Dioxide (SO₂ to sulphates):
Acidified potassium permanganate oxidizes sulphur dioxide to sulphates.
\[2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4\]
Other reactions in acidic medium (as per the question about acidified permanganate):
(a) Ferrous salts (green) to ferric salts (yellow):
\[MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}\]
(b) Iodides to iodine:
\[2MnO_4^- + 16H^+ + 10I^- \rightarrow 2Mn^{2+} + 8H_2O + 5I_2\]
(c) H₂S to sulphur:
\[2KMnO_4 + 3H_2SO_4 + 5 H_2S \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5S\]
For completeness, though not explicitly asked about *acidified* permanganate, it also acts as an oxidizer in neutral or faintly alkaline solutions, for instance:
(a) Iodides to iodates:
\[KI + 2KMnO_4 + H_2O \rightarrow KIO_3 + 2MnO_2 + 2KOH\]
(b) Thiosulphate to sulphate:
\[8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-\]
(c) Manganese(II) salt to \(MnO_2\):
\[2MnO_4^- + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^+\]
The presence of zinc sulfate or zinc oxide can catalyze the oxidation. Permanganate titrations are not satisfactory in the presence of HCl because HCl itself gets oxidized to Cl₂.

In simple words: Potassium permanganate is made from pyrolusite ore by first forming green manganate and then oxidizing it to purple permanganate. In acidic solutions, it's a powerful oxidizer, turning purple \(MnO_4^-\) into colorless \(Mn^{2+}\) while oxidizing substances like oxalic acid to carbon dioxide and sulfur dioxide to sulfates.

🎯 Exam Tip: When preparing for questions on potassium permanganate, understand its synthesis from \(MnO_2\), its strong oxidizing ability in acidic medium (reducing to \(Mn^{2+}\)), and its distinct color changes during redox reactions. Remember the stoichiometry and balancing of ionic equations.

 

Question 16. For M\(^{2+}\)/M and M\(^{3+}\)/M\(^{2+}\) systems, the E° values for some metals are as follows:

Cr\(^{2+}\)/Cr	- 0.9V	Cr\(^{3+}\)/Cr\(^{2+}\)	- 0.4 V
Mn\(^{2+}\)/Mn	-1.2V	Mn\(^{3+}\)/Mn\(^{2+}\)	+1.5 V
Fe\(^{2+}\)/Fe	-0.4V	Fe\(^{3+}\)/Fe\(^{2+}\)	+ 0.8 V

Use this data to comment upon
1. The stability of Fe\(^{3+}\) in acid solution as compared to that of Cr\(^{3+}\) or Mn\(^{3+}\) and
2. The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer:

1. The standard reduction potential of Cr\(^{3+}\)/Cr\(^{2+}\) is negative, indicating that Cr\(^{3+}\) is not easily reduced to Cr\(^{2+}\) and thus Cr\(^{3+}\) possesses high stability. Conversely, Mn\(^{3+}\)/Mn\(^{2+}\) has a large positive E° value, suggesting that Mn\(^{3+}\) is readily reduced to Mn\(^{2+}\), making Mn\(^{3+}\) the least stable among the three. The E° value for Fe\(^{3+}\)/Fe\(^{2+}\) is positive but relatively small, implying that Fe\(^{3+}\) is more stable than Mn\(^{3+}\) but less stable than Cr\(^{3+}\) in acidic solutions.
2. The ease of oxidation for iron compared to chromium and manganese can be assessed by comparing the E° values for M\(^{2+}\)/M systems. A higher negative reduction potential means the metal is more easily oxidized. Based on the E° values: Mn (−1.2 V) is oxidized more easily than Cr (−0.9 V), which in turn is oxidized more easily than Fe (−0.4 V). This order reflects the decreasing ease of oxidation: Mn > Cr > Fe. As the E° values for Mn\(^{2+}\)/Mn become less negative, the oxidising power of the Mn\(^{2+}\) ion also decreases.
In simple words: The stability of Fe\(^{3+}\) is intermediate, while Cr\(^{3+}\) is very stable and Mn\(^{3+}\) is unstable due to their reduction potentials. Iron is harder to oxidize than manganese or chromium.

🎯 Exam Tip: Understanding the relationship between E° values and the stability/oxidizing ability of ions is crucial. Remember that a more negative E°(M\(^{n+}\)/M) indicates easier oxidation of the metal, and a more positive E°(M\(^{n+}\)/M\(^{(n-1)+}\)) indicates easier reduction of the higher oxidation state, making it less stable.

 

Question 17. Predict which of the following will be coloured in aqueous solution? Ti\(^{3+}\), V\(^{3+}\), Cu\(^{+}\), Sc\(^{3+}\), Mn\(^{2+}\), Fe\(^{3+}\) and Co\(^{2+}\). Give reasons for each?

Answer:

In aqueous solutions, ions with partially filled d-orbitals typically exhibit color due to d-d electronic transitions. Among the given ions, Sc\(^{3+}\) (3d\(^0\)) has a completely empty d-subshell and Cu\(^{+}\) (3d\(^{10}\)) has a completely filled d-subshell. Therefore, neither Sc\(^{3+}\) nor Cu\(^{+}\) will show d-d transitions and will appear colorless in aqueous solutions. All other ions-Ti\(^{3+}\) (3d\(^1\)), V\(^{3+}\) (3d\(^2\)), Mn\(^{2+}\) (3d\(^5\)), Fe\(^{3+}\) (3d\(^5\)), and Co\(^{2+}\) (3d\(^7\))-possess incompletely filled d-orbitals, allowing for d-d electronic transitions, which results in them being colored in aqueous solutions.
In simple words: Ions that have some electrons in their d-orbitals but not a full set (like Ti\(^{3+}\), V\(^{3+}\), Mn\(^{2+}\), Fe\(^{3+}\), Co\(^{2+}\)) will be colored because their electrons can jump to different d-orbital energy levels. Ions with empty or full d-orbitals (like Sc\(^{3+}\), Cu\(^{+}\)) are typically colorless.

🎯 Exam Tip: The presence of unpaired electrons in d-orbitals is the key criterion for d-d transitions and thus for an ion to be colored in aqueous solution. Always check the electronic configuration, specifically the d-electron count, to determine if d-d transitions are possible.

 

Question 18. Compare the stability of the +2 oxidation state for the elements of the first transition series.

Answer:

The variable oxidation states are a defining characteristic of transition elements. While all transition elements, except the first and last members of each series, display multiple oxidation states, the +2 oxidation state shows varying stability across the first transition series. This variability primarily stems from the small energy difference between the 'ns' and '(n-1)d' orbitals, allowing both sets of electrons to participate in bond formation. The specific electronic configurations dictate the stability of these states. For instance, considering Manganese (Mn, 3d\(^5\)4s\(^2\)), it exhibits oxidation states ranging from +2 to +7.
• Formation of Mn\(^{2+}\) (e.g., in MnO) primarily involves the loss of two 4s electrons.
• Formation of Mn\(^{3+}\) (e.g., in Mn\(_{2}\)O\(_{3}\)) involves the loss of two 4s and one 3d electron.
• Formation of Mn\(^{4+}\) (e.g., in MnO\(_{2}\)) involves the loss of two 4s and two 3d electrons.
• Formation of Mn\(^{5+}\) (e.g., in Mn\(_{2}\)O\(_{5}\)) involves the loss of two 4s and three 3d electrons.
• Formation of Mn\(^{6+}\) (e.g., in K\(_{2}\)MnO\(_{4}\)) involves the loss of two 4s and four 3d electrons.
• Formation of Mn\(^{7+}\) (e.g., in KMnO\(_{4}\)) involves the loss of two 4s and five 3d electrons.
The wide range of oxidation states in transition elements is attributed to the availability of d-electrons for bonding. The highest oxidation state observed for any transition metal is +8. The following table illustrates the common oxidation states for elements in the first transition series:

Element	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
	+3	+2	+2	+2	+2	+2	+2	+2	+1	+2
		+3	+3	+3	+3	+3	+3	+3	+2
		+4	+4	+4	+4	+4	+4	+4
			+5	+5	+5	+6	+6
				+6	+6	+7
					+7

From this data, several observations can be made regarding the +2 oxidation state:
(i) Generally, the most common oxidation state for the first-row transition elements is +2, with the notable exception of scandium, which preferentially forms +3 ions. This occurs because two 4s electrons are lost. After scandium, the d-orbitals become relatively more stable than the s-orbitals.
(ii) In the +2 and +3 oxidation states, metals typically form ionic bonds. However, when exhibiting higher oxidation states, particularly with highly electronegative elements like fluorine or oxygen, the bonds formed tend to be more covalent. For instance, all bonds between manganese and oxygen in permanganate (MnO\(_{4}\)\(^{-}\)) are covalent.
(iii) Within a particular group, the maximum oxidation number generally increases with increasing atomic number. For example, iron commonly exhibits +2 and +3 oxidation states, but heavier elements in the same group, such as ruthenium and osmium, can achieve higher states like +4, +6, and +8.
(iv) Lower oxidation states, such as +1 and 0, are also observed in some transition elements. Notable examples include [Ni(CO)\(_{4}\)] and [Fe(CO)\(_{5}\)], where nickel and iron are in a zero oxidation state.
In simple words: The +2 oxidation state's stability varies across the first transition series because d-electrons can participate in bonding. Elements like Mn show many states, but overall, +2 is common. Stability is influenced by electron configuration and the nature of bonding, ranging from ionic at lower states to more covalent at higher ones.

🎯 Exam Tip: When discussing oxidation state stability, focus on electron configurations (d\(^0\), d\(^5\), d\(^{10}\) stability) and ionization energies. Also, consider the nature of the bonding (ionic vs. covalent) as it relates to the electronegativity of the bonding partners.

 

Question 19. Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
1. Electronic configuration
2. Atomic and ionic sizes
3. Oxidation state

Answer:

Both lanthanide and actinide elements involve the filling of f-orbitals, leading to some similarities in their chemical properties, although significant differences exist.
Similarities:
(i) Both predominantly exhibit a +3 oxidation state.
(ii) In both series, f-orbitals are progressively filled across the period.
(iii) Both groups of elements are electropositive and highly reactive.
(iv) Both exhibit characteristic magnetic and spectral properties.
(v) Analogous to lanthanide contraction, actinides also show actinide contraction, which is a consequence of the poor shielding effect of 5f electrons.
(vi) Lanthanide ions, except those with f\(^0\) (e.g., La\(^{3+}\) or Ce\(^{4+}\)) and f\(^{14}\) (e.g., Yb\(^{2+}\) and Lu\(^{3+}\)) configurations, display paramagnetism due to the presence of unpaired f electrons, with neodymium showing maximum paramagnetism.
Differences:

Lanthanides	Actinides
1. Primarily exhibit the common oxidation state of +3, but also show +2 and +4 states.	1. Beyond the common +3 oxidation state, they show higher oxidation states of +4, +5, +6, and +7.
2. Have a lower tendency to form complexes.	2. Exhibit a greater tendency to form complexes.
3. Except for promethium, all elements are non-radioactive.	3. All actinides are radioactive.
4. Lanthanide compounds are generally less basic.	4. Actinide compounds are generally more basic.

In Period 7, after atomic number Z = 89, the 5f and 6d-orbitals have very similar energy levels. This leads to the successive filling of 5f-orbitals in 14 actinides. Thorium is an exception, having no 5f electrons, with 5f orbital occupation beginning from Protactinium.
(i) **Electronic Configuration of Actinide elements:**

Name	Symbol	Atomic No.	Electronic configuration
Actinium	Ac	89	\(5f^06d^17s^2\)
Thorium	Th	90	\(5f^06d^27s^2\)
Protactinium	Pa	91	\(5f^26d^17s^2\)
Uranium	U	92	\(5f^36d^17s^2\)
Neptunium	Np	93	\(5f^46d^17s^2\)
Plutonium	Pu	94	\(5f^66d^07s^2\)
Americium	Am	95	\(5f^76d^07s^2\)
Curium	Cm	96	\(5f^76d^17s^2\)
Berkelium	Bk	97	\(5f^86d^17s^2\)
Californium	Cf	98	\(5f^{10}6d^07s^2\)
Einsteinium	Es	99	\(5f^{11}6d^07s^2\)
Fermium	Fm	100	\(5f^{12}6d^07s^2\)
Mendelevium	Md	101	\(5f^{13}6d^07s^2\)
Nobelium	No	102	\(5f^{14}6d^07s^2\)
Lawrencium	Lr	103	\(5f^{14}6d^17s^2\)

(ii) There is a noticeable decrease in atomic and ionic size as atomic number increases, a phenomenon known as actinide contraction. This contraction is more pronounced in actinides compared to lanthanides because the shielding provided by 5f electrons is less effective than that of 4f electrons.
**Other properties:**
Actinides are generally more reactive than lanthanides, partly due to their larger size and the greater exposure of 5f orbitals, which allows them to participate more readily in chemical reactions. While hydrochloric acid attacks most actinides, nitric acid only slightly affects them due to the formation of a protective oxide layer. Alkalies generally have no reaction with actinides.
**Uses:**
Elements like uranium, thorium, and neptunium are vital as fuel in atomic reactors. Thorium salts are used in incandescent gas mantles. Uranium compounds find applications in the glass, textile, ceramic industries, and in medicine.
(iii) All actinides are radioactive, which limits their accessibility for laboratory research. Elements beyond uranium (atomic number 92) are synthetic and produced through nuclear-chemical methods. Actinide chemistry is complex due to their ability to exist in various oxidation states. While +3 is the common oxidation state, they also frequently exhibit +4, +5, and +6 states. This wide range is attributed to two main factors:
1. The minimal energy difference between the 5f and 7s orbitals.
2. The 5f orbitals in actinides are more exposed than 4f orbitals in lanthanides, leading to 5f electrons also participating in chemical reactions.
(iv) **Chemical Reactivity:**
Chemical reactivity describes an atom's propensity to react vigorously with other substances. Generally, elements positioned further to the left and lower down on the periodic table tend to be more reactive, as their electrons are more easily given up or taken, leading to higher reactivity.
In simple words: Lanthanides and actinides both fill f-orbitals and share some properties like the +3 oxidation state, but actinides are more radioactive, show higher oxidation states (up to +7), form more stable complexes, and their contraction is more pronounced due to poorer 5f shielding. Actinides are also generally more reactive.

🎯 Exam Tip: For comparative questions, always structure your answer with clear points of similarity and difference. When discussing stability or reactivity, link it back to electronic configuration, shielding effects, and orbital energy differences.

 

Question 20. How would you account for the following:
1. Of the d\(^4\) species, Cr\(^{2+}\) is strongly reducing while manganese(III) is strongly oxidising.
2. Cobalt(II) is stable in an aqueous solution but in the presence of complexing reagents, it is easily oxidised.
3. The d\(^1\) configuration is very unstable in ions.

Answer:

1. Cr\(^{2+}\) has a d\(^4\) configuration. It acts as a strong reducing agent because it readily loses an electron to achieve a more stable d\(^3\) configuration, which benefits from the stable half-filled t\(_{2g}\) level (t\(_{2g}^3\)) in an octahedral ligand field. In contrast, Mn\(^{3+}\), also with a d\(^4\) configuration, acts as a strong oxidizing agent because it can readily gain an electron to achieve the highly stable d\(^5\) configuration (t\(_{2g}^3e_g^2\) high spin or t\(_{2g}^5\) low spin). The d\(^5\) configuration is exceptionally stable due to its half-filled nature.
2. In an aqueous solution, Cobalt(II) ions (Co\(^{2+}\)) are relatively stable. However, when strong complexing ligands are present, Co\(^{2+}\) (d\(^7\)) can be easily oxidized to Co\(^{3+}\) (d\(^6\)). This oxidation is favored because the crystal field stabilization energy (CFSE) gained by the d\(^6\) configuration in a strong ligand field (often leading to a low-spin configuration) more than compensates for the third ionization energy required to remove an electron from Co\(^{2+}\).
3. Ions with a d\(^1\) configuration are inherently very unstable. The removal of a single electron from an s-orbital (to form d\(^1\) from s\(^2\)d\(^1\)) often results in an ion that has a strong tendency to lose further electrons or gain electrons to achieve a more stable d\(^0\) (empty d-orbital) or d\(^2\), d\(^3\), d\(^5\), or d\(^{10}\) configuration. The hydration or lattice energy released when such an ion forms a solution or a crystal is usually insufficient to compensate for the ionization enthalpy required to form a d\(^1\) ion and keep it stable as a standalone species.
In simple words: Cr\(^{2+}\) is a good reducer because it wants to become a stable d\(^3\). Mn\(^{3+}\) is a good oxidizer because it wants to become a stable d\(^5\). Co\(^{2+}\) is usually stable in water, but with special partners, it can easily lose an electron to become Co\(^{3+}\) because of energy benefits. Ions with only one d-electron are unstable as they readily lose or gain electrons to achieve more stable configurations like empty, half-filled, or full d-orbitals.

🎯 Exam Tip: Stability of oxidation states and redox behavior often relates to achieving stable electronic configurations (d\(^0\), d\(^3\) in octahedral t\(_{2g}\), d\(^5\), d\(^{10}\)) or gaining significant crystal field stabilization energy (CFSE) with strong ligands. Always consider both electronic factors and thermodynamic factors (hydration/lattice energy vs. ionization energy).

 

Question 21. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution?

Answer:

Disproportionation reactions are chemical processes where a single substance undergoes simultaneous oxidation and reduction, meaning its oxidation state both increases and decreases. In such reactions, an element in an intermediate oxidation state converts into two different species, one with a higher and one with a lower oxidation state.
**Examples:**
i. The disproportionation of manganate(VI) ion (\(MnO_4^{2-}\)) in acidic solution:
\[3MnO_4^{2-} + 4H^+ \longrightarrow 2MnO_4^- + MnO_2 + 2H_2O\]
In this reaction, manganese in \(MnO_4^{2-}\) (oxidation state +6) is oxidized to \(MnO_4^-\) (oxidation state +7) and reduced to \(MnO_2\) (oxidation state +4).
ii. The disproportionation of hydrogen peroxide (\(H_2O_2\)) in aqueous solution:
\[2H_2O_2 \longrightarrow 2H_2O + O_2\]
Here, oxygen in \(H_2O_2\) (oxidation state -1) is reduced to \(H_2O\) (oxidation state -2) and oxidized to \(O_2\) (oxidation state 0).
In simple words: Disproportionation is when one chemical simultaneously gives and takes electrons, ending up in two different forms—one more oxidized and one more reduced. Hydrogen peroxide breaking down into water and oxygen is a common example.

🎯 Exam Tip: To identify disproportionation, check the oxidation state of a specific element in the reactant and then in all products. If that element's oxidation state both increases and decreases, it's a disproportionation reaction. Balancing redox reactions, especially in acidic or basic media, is a fundamental skill for such problems.

 

Question 22. Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why?

Answer:

Copper (Cu) is the metal in the first transition series that most frequently exhibits a +1 oxidation state. This behavior is attributed to its electronic configuration, which is [Ar] \(3d^{10}4s^1\). Copper can readily lose its single 4s electron to achieve a highly stable completely filled \(3d^{10}\) configuration. The resulting Cu\(^{+}\) ion, with its full d-subshell, gains significant stability, making the +1 oxidation state a common and favorable one for copper.
In simple words: Copper often shows a +1 charge because it has one electron in its outer shell (4s\(^1\)) and a full inner shell (3d\(^{10}\)). Losing that one outer electron leaves it with a very stable, complete 3d shell.

🎯 Exam Tip: For transition metals, exceptional stability often arises from achieving d\(^0\), d\(^5\), or d\(^{10}\) configurations. Copper's +1 state is a classic example of d\(^{10}\) stability driving a common oxidation state.

 

Question 23. Calculate the number of unpaired electrons in the following gaseous ions: Mn\(^{3+}\), Cr\(^{3+}\), V\(^{3+}\), and Ti\(^{3+}\). Which one of these is the most stable in an aqueous solution?

Answer:

To determine the number of unpaired electrons, we first write the electronic configuration for each gaseous ion:
• Original atom configuration: Ti = [Ar] \(3d^24s^2\), V = [Ar] \(3d^34s^2\), Cr = [Ar] \(3d^54s^1\), Mn = [Ar] \(3d^54s^2\).
• Ti\(^{3+}\): Losing three electrons from Ti (two 4s, one 3d) results in [Ar] \(3d^1\). Number of unpaired electrons = 1.
• V\(^{3+}\): Losing three electrons from V (two 4s, one 3d) results in [Ar] \(3d^2\). Number of unpaired electrons = 2.
• Cr\(^{3+}\): Losing three electrons from Cr (one 4s, two 3d) results in [Ar] \(3d^3\). Number of unpaired electrons = 3.
• Mn\(^{3+}\): Losing three electrons from Mn (two 4s, one 3d) results in [Ar] \(3d^4\). Number of unpaired electrons = 4.

Summarizing the unpaired electrons:
- Mn\(^{3+}\) = 4 unpaired electrons
- Cr\(^{3+}\) = 3 unpaired electrons
- V\(^{^{3+}\)}\) = 2 unpaired electrons
- Ti\(^{3+}\) = 1 unpaired electron

Among these ions, Cr\(^{3+}\) is the most stable in an aqueous solution. This stability arises because Cr\(^{3+}\) has a d\(^3\) configuration. In an octahedral ligand field (which is typically formed in aqueous solutions), the three d-electrons occupy the lower energy t\(_{2g}\) orbitals (t\(_{2g}^3e_g^0\)), leading to significant crystal field stabilization energy (CFSE). This stable configuration makes Cr\(^{3+}\) particularly resistant to oxidation or reduction in water.
In simple words: For Ti\(^{3+}\) (1), V\(^{3+}\) (2), Cr\(^{3+}\) (3), and Mn\(^{3+}\) (4), the numbers in parentheses are their unpaired electrons. Cr\(^{3+}\) is the most stable in water because its three d-electrons fill the lower energy orbitals perfectly, making it very stable.

🎯 Exam Tip: To calculate unpaired electrons, first determine the full electronic configuration of the atom, then remove electrons starting from the outermost 's' orbitals, then 'p', then 'd'. For stability in aqueous solution, always consider crystal field theory for transition metal ions, particularly the stability conferred by d\(^3\) (t\(_{2g}^3\)) and d\(^6\) (low spin t\(_{2g}^6\)) configurations in octahedral fields.

 

Question 24. Give examples and suggest reasons for the following features of the transition metal chemistry:
1. A transition metal exhibits the highest oxidation state in oxides and fluorides.
2. The highest oxidation state is exhibited in oxoanions of a metal.

Answer:

1. Transition metals achieve their highest oxidation states when bonded with oxygen or fluorine. This is due to the extremely high electronegativity and small size of both oxygen and fluorine. Their strong electron-withdrawing nature allows them to stabilize the high positive charges on the metal atoms. For example, in compounds like KMnO\(_{4}\) (manganese in +7 oxidation state) or K\(_{2}\)Cr\(_{2}\)O\(_{7}\) (chromium in +6 oxidation state), oxygen helps achieve these high states. Similarly, in VF\(_{5}\), vanadium exhibits a +5 oxidation state.
2. The highest oxidation state of a metal is frequently observed in its oxoanions (anions containing both oxygen and the metal, such as permanganate or dichromate). This is because oxygen is not only highly electronegative but also a very strong oxidizing agent, capable of drawing out the maximum number of electrons from the metal. The formation of multiple covalent bonds with oxygen further stabilizes these high oxidation states within the anionic complex. Examples include \(MnO_4^-\) (manganese +7), \(Cr_2O_7^{2-}\) (chromium +6), and \(VO_4^{3-}\) (vanadium +5).
In simple words: Transition metals show their highest charges when bonded with oxygen or fluorine because these elements are small and very good at pulling electrons. Especially in compounds with oxygen where the metal forms an anion, like in permanganate, the metal reaches its top oxidation state because oxygen is a powerful electron-grabber.

🎯 Exam Tip: The ability of a transition metal to achieve its highest oxidation state is fundamentally linked to the electronegativity and oxidizing power of the surrounding ligands/elements. Oxygen and fluorine are prime examples due to their strong electron-pulling capabilities and ability to form multiple bonds.

 

Question 25. Indicate the steps in the preparation of:
1. K\(_{2}\)Cr\(_{2}\)O\(_{7}\) from chromite ore.
2. KMnO\(_{4}\) from pyrolusite ore.

Answer:

1. **Preparation of Potassium Dichromate (K\(_{2}\)Cr\(_{2}\)O\(_{7}\)) from Chromite Ore:**
The preparation involves several steps starting from chromite ore (FeCr\(_{2}\)O\(_{4}\)):
(i) **Fusion of Chromite Ore:** Chromite ore is heated (fused) with sodium carbonate (\(Na_2CO_3\)) in the presence of air (or an oxidizing agent like \(NaNO_3\)). This reaction produces sodium chromate (\(Na_2CrO_4\)), ferric oxide (\(Fe_2O_3\)), and carbon dioxide (\(CO_2\)):
\[4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \longrightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2\]
(ii) **Acidification of Sodium Chromate:** The soluble sodium chromate (which is yellow) is then acidified, typically with sulfuric acid (\(H_2SO_4\)). This converts chromate ions into dichromate ions (\(Na_2Cr_2O_7\)):
\[2Na_2CrO_4 + H_2SO_4 \longrightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O\]
(iii) **Conversion to Potassium Dichromate:** The sodium dichromate solution is then treated with potassium chloride (KCl). Due to lower solubility, potassium dichromate (K\(_{2}\)Cr\(_{2}\)O\(_{7}\)) crystallizes out as an orange-red substance:
\[Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7 + 2NaCl\]
Potassium dichromate is an orange-red crystalline compound. Upon strong heating, it decomposes into potassium chromate, chromium(III) oxide, and oxygen:
\[4K_2Cr_2O_7 \longrightarrow 4K_2CrO_4 + 2Cr_2O_3 + 3O_2\]

2. **Preparation of Potassium Permanganate (KMnO\(_{4}\)) from Pyrolusite Ore:**
The process for preparing potassium permanganate from pyrolusite ore (MnO\(_{2}\)) also involves multiple steps:
(i) **Fusion of Pyrolusite Ore:** Powdered pyrolusite ore is fused with potassium hydroxide (KOH) or potassium carbonate (\(K_2CO_3\)) in the presence of air or an oxidizing agent. This fusion forms a green-colored potassium manganate (\(K_2MnO_4\)):
\[2MnO_2 + 4KOH + O_2 \longrightarrow 2K_2MnO_4 + 2H_2O\]
(ii) **Electrolytic Oxidation of Manganate:** The green potassium manganate is then extracted with water. It is oxidized electrolytically between iron electrodes to form purple potassium permanganate (KMnO\(_{4}\)).
• At the anode (oxidation): \(MnO_4^{2-} \longrightarrow MnO_4^- + e^-\)
• At the cathode (reduction): \(2H^+ + 2e^- \longrightarrow H_2\)
• The overall reaction in solution (or by chemical oxidation using \(Cl_2\) or \(O_3\)):
\[2K_2MnO_4 + Cl_2 \longrightarrow 2KMnO_4 + 2KCl\]
Alternatively, in the presence of water and a strong oxidant:
\[2K_2MnO_4 + H_2O + O \longrightarrow 2KMnO_4 + 2KOH\]
The resulting purple solution of potassium permanganate yields crystalline KMnO\(_{4}\) upon evaporation.
In simple words: To make potassium dichromate, chromite ore is first heated with sodium carbonate and air to get sodium chromate, then acidified to sodium dichromate, and finally treated with potassium chloride to crystallize the final product. To make potassium permanganate, pyrolusite ore is fused with potassium hydroxide and air to form potassium manganate, which is then oxidized, usually by electrolysis, to purple potassium permanganate.

🎯 Exam Tip: For preparation questions, remember the key starting materials, intermediate products, and the role of oxidizing/reducing agents. Practice writing balanced chemical equations for each step, paying attention to conditions like heating or acidification.

 

Question 26. What are alloys? Name an important alloy which contains some of the lanthanoid metals? Mention its uses?

Answer:

Alloys are metallic substances formed by blending two or more elements, with at least one being a metal. This blending is typically achieved by melting the constituent elements together.
An important alloy that incorporates lanthanoid metals is **mischmetal**. Mischmetal is an alloy predominantly composed of rare-earth elements (typically 94-95%), with about 5% iron and trace amounts of other elements like sulfur, carbon, silicon, calcium, and aluminum. The primary rare-earth components include cerium (approximately 40%), lanthanum, and neodymium (around 44%).
**Uses of Mischmetal:**
• Mischmetal is pyrophoric, meaning it produces sparks when struck, making it suitable for use in cigarette and gas lighters, and in manufacturing flame-throwing tank components and tracer bullets.
• An alloy of magnesium containing about 3% mischmetal is utilized in the production of jet engine parts, where its properties enhance performance at high temperatures.
• The inclusion of lanthanides in steel alloys significantly improves the workability and strength of steel at elevated temperatures.
In simple words: Alloys are mixtures of metals. Mischmetal is an important alloy containing lanthanides (rare-earth metals) mixed with iron. It's used in lighters because it sparks easily, in jet engines when mixed with magnesium, and improves steel's strength at high temperatures.

🎯 Exam Tip: When defining alloys, emphasize they are mixtures with metallic properties. For specific examples like mischmetal, highlight its key components (rare earths + iron) and its most notable applications, especially those leveraging its pyrophoric nature or high-temperature strength enhancement.

 

Question 27. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

Answer:

Inner transition elements are a special group of elements characterized by the filling of their f-orbitals. These elements are located outside the main body of the periodic table, forming two distinct series: the lanthanides (4f-block elements) and the actinides (5f-block elements). Their electron configurations involve the filling of the (n-2)f subshell.
From the given atomic numbers:
• 29 (Copper) is a d-block transition element.
• 59 (Praseodymium) is a lanthanide element (4f-block).
• 74 (Tungsten) is a d-block transition element.
• 95 (Americium) is an actinide element (5f-block).
• 102 (Nobelium) is an actinide element (5f-block).
• 104 (Rutherfordium) is a d-block transition element.

Therefore, the atomic numbers corresponding to inner transition elements are 59, 95, and 102.
In simple words: Inner transition elements are the lanthanides and actinides, which fill their f-orbitals. From the list, atomic numbers 59 (Praseodymium), 95 (Americium), and 102 (Nobelium) are inner transition elements.

🎯 Exam Tip: Remember that inner transition elements correspond to the f-block. Lanthanides typically range from Z=57-71 and actinides from Z=89-103. Knowing these ranges helps quickly identify f-block elements.

 

Question 28. Which is the last element in the series of actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element?

Answer:

The last element in the series of actinoids is Lawrencium (Lr), with atomic number 103.
Its electronic configuration is \([Rn] 5f^{14}6d^17s^2\).
Lawrencium predominantly exhibits a +3 oxidation state. This is the characteristic oxidation state for most actinoids, resulting from the loss of its two 7s electrons and one 6d electron. While other actinides can show higher oxidation states (up to +7), Lawrencium, with its \(5f^{14}\) core, is most stable in the +3 state, where its f-orbitals are completely filled, similar to the d\(^{10}\) stability in transition metals.
In simple words: The last actinoid is Lawrencium (Lr), atomic number 103, with an electron configuration of \([Rn] 5f^{14}6d^17s^2\). It primarily shows a +3 oxidation state, which is stable because it results in a completely filled 5f-subshell.

🎯 Exam Tip: For f-block elements, remember the last element of each series (Lutetium for lanthanides, Lawrencium for actinides) and their characteristic electronic configurations. The +3 oxidation state is fundamental to understanding their chemistry, often linked to stable f-electron counts.

 

Question 29. Use Hund's rule to derive the electronic configuration of Ce\(^{3+}\) ion, and calculate its magnetic moment on the basis of 'spin-only' formula?

Answer:

First, we determine the electronic configuration of the neutral Cerium (Ce) atom and then the Ce\(^{3+}\) ion.
Cerium (Ce) has an atomic number of 58. Its ground state electronic configuration, following Hund's rule and considering the stability of partially filled orbitals, is:
\(^{58}Ce = [Xe] 4f^15d^16s^2\)
To form the Ce\(^{3+}\) ion, three electrons are removed: first the two 6s electrons, and then one 5d electron. This leaves:
\(Ce^{3+} = [Xe] 4f^1\)
According to Hund's rule, the single electron in the 4f subshell will be unpaired.
Now, we calculate the spin-only magnetic moment (\(\mu\)) using the formula \(\mu = \sqrt{n(n+2)}\) BM, where 'n' is the number of unpaired electrons.
For Ce\(^{3+}\), \(n = 1\) (one unpaired 4f electron).
So, \(\mu = \sqrt{1(1+2)}\) BM
\(\mu = \sqrt{1 \times 3}\) BM
\(\mu = \sqrt{3}\) BM
\(\mu \approx 1.73\) BM
Thus, the magnetic moment of Ce\(^{3+}\) is approximately 1.73 Bohr Magnetons.
In simple words: Cerium (Ce) loses three electrons to become Ce\(^{3+}\), leaving it with one electron in its 4f shell. Since this electron is unpaired, its magnetic moment is calculated as approximately 1.73 Bohr Magnetons.

🎯 Exam Tip: When determining electronic configurations for ions, always remove electrons from the outermost 's' orbitals first, then 'p', and then 'd' (or 'f' for lanthanides/actinides). For magnetic moment calculations, 'n' refers specifically to the number of *unpaired* electrons.

 

Question 30. Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements?

Answer:

The exhibition of +2 and +4 oxidation states in lanthanoids is linked to the stability associated with empty (f\(^0\)), half-filled (f\(^7\)), or completely filled (f\(^{14}\)) f-electron configurations, or configurations that are very close to these stable states.
**Lanthanoids exhibiting a +4 oxidation state:**
Elements that display a +4 oxidation state often achieve, or are close to achieving, a stable f\(^0\), f\(^1\), f\(^2\), f\(^7\), or f\(^8\) configuration. These include:
• Cerium (Ce, \(^{58}Ce\), \(4f^15d^16s^2\)) exhibits +4 (forms \(Ce^{4+}\) with \(4f^0\), a very stable empty f-shell).
• Praseodymium (Pr, \(^{59}Pr\)) exhibits +4 (can form \(Pr^{4+}\) with \(4f^1\)).
• Neodymium (Nd, \(^{60}Nd\)) exhibits +4 (can form \(Nd^{4+}\) with \(4f^2\)).
• Terbium (Tb, \(^{65}Tb\)) exhibits +4 (can form \(Tb^{4+}\) with \(4f^7\), a stable half-filled f-shell).
• Dysprosium (Dy, \(^{66}Dy\)) exhibits +4 (can form \(Dy^{4+}\) with \(4f^8\)).
**Lanthanoids exhibiting a +2 oxidation state:**
Elements that display a +2 oxidation state typically achieve, or are close to achieving, a stable f\(^7\) or f\(^{14}\) configuration, or involve configurations where 5d and 6s electrons are primarily removed, leaving the f-shell relatively untouched. These include:
• Neodymium (Nd, \(^{60}Nd\)) exhibits +2 (can form \(Nd^{2+}\) with \(4f^4\)).
• Samarium (Sm, \(^{62}Sm\)) exhibits +2 (can form \(Sm^{2+}\) with \(4f^6\)).
• Europium (Eu, \(^{63}Eu\), \(4f^76s^2\)) exhibits +2 (forms \(Eu^{2+}\) with \(4f^7\), a very stable half-filled f-shell).
• Thulium (Tm, \(^{69}Tm\)) exhibits +2 (can form \(Tm^{2+}\) with \(4f^{13}\)).
• Ytterbium (Yb, \(^{70}Yb\), \(4f^{14}6s^2\)) exhibits +2 (forms \(Yb^{2+}\) with \(4f^{14}\), a very stable filled f-shell).
In simple words: Some lanthanoids show +4 or +2 oxidation states because doing so leads them to more stable electron arrangements—like having no f-electrons (f\(^0\)), a half-filled f-shell (f\(^7\)), or a completely filled f-shell (f\(^{14}\)). For example, Cerium becomes +4 for an f\(^0\) state, while Europium becomes +2 for an f\(^7\) state.

🎯 Exam Tip: The stability of f\(^0\), f\(^7\), and f\(^{14}\) configurations is a recurring theme in lanthanoid chemistry. When asked about anomalous oxidation states (+2 or +4), always link them back to these particularly stable electron counts in the f-subshell.

 

Question 31. Compare the chemistry of the actinoids with that of lanthanoids with reference to:
1. Electronic configuration
2. Oxidation states and
3. Chemical reactivity

Answer:

A comparison of actinoid and lanthanoid chemistry based on the specified criteria reveals both commonalities and key differences:
1. **Electronic Configuration:**
• **Lanthanoids:** Characterized by the filling of 4f orbitals. The general electronic configuration is \([Xe] 4f^{1-14}5d^{0-1}6s^2\). The 5d orbital is often empty or has only one electron, which can vary.
• **Actinoids:** Characterized by the filling of 5f orbitals. The general electronic configuration is \([Rn] 5f^{0-14}6d^{0-2}7s^2\). The 5f and 6d orbitals have very similar energies, leading to less predictable filling and a more complex chemistry than lanthanoids.
2. **Oxidation States:**
• **Lanthanoids:** Predominantly exhibit a +3 oxidation state. Some elements also show +2 (e.g., Eu, Yb for f\(^7\), f\(^{14}\) stability) and +4 (e.g., Ce, Tb for f\(^0\), f\(^7\) stability) oxidation states, but these are less common.
• **Actinoids:** Also exhibit a common +3 oxidation state. However, they display a much wider range of higher oxidation states, including +4, +5, +6, and +7 (e.g., U, Np, Pu). This is due to the smaller energy gap between 5f, 6d, and 7s orbitals, allowing more electrons to participate in bonding.
3. **Chemical Reactivity:**
• **Lanthanoids:** Generally less reactive than actinides. Their early members are quite reactive, forming strong bases, but reactivity decreases across the series due to lanthanide contraction.
• **Actinoids:** More reactive than lanthanoids. This increased reactivity is partly attributed to their larger atomic sizes and the greater exposure of their 5f orbitals, which makes their electrons more available for chemical bonding. All actinides are radioactive, which also influences their handling and study.
In simple words: Lanthanides fill 4f orbitals and mostly show a +3 charge, with some +2 and +4. Actinides fill 5f orbitals, also favor +3, but can show much higher charges (up to +7) because their 5f, 6d, and 7s electrons are close in energy. Actinides are generally more reactive and all are radioactive.

🎯 Exam Tip: When comparing f-block elements, focus on the principal quantum number of the f-orbital (4f vs. 5f) as it dictates the energy differences between subshells, influencing oxidation states and reactivity. Radioactivity is a unique characteristic differentiating actinides from most lanthanides.

 

Question 32. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109?

Answer:

The electronic configurations for the specified atomic numbers are:
• \(^{61}Pm\) (Promethium): \([Xe] 4f^55d^06s^2\)
• \(^{91}Pa\) (Protactinium): \([Rn] 5f^26d^17s^2\)
• \(^{101}Md\) (Mendelevium): \([Rn] 5f^{13}6d^07s^2\)
• \(^{109}Mt\) (Meitnerium): \([Rn] 5f^{14}6d^77s^2\)
In simple words: The electron arrangements for atomic numbers 61 (Pm), 91 (Pa), 101 (Md), and 109 (Mt) involve filling 4f or 5f orbitals after the noble gas core, followed by d and s electrons in their outermost shells.

🎯 Exam Tip: For elements in the f-block, remember to fill the 's' orbitals before 'd' or 'f' in general, and then consider any exceptions (like d\(^1\) entry for some f-block elements) or stability rules (half-filled/full f-orbitals). Use the nearest noble gas core for shorthand configurations.

 

Question 33. Write down the number of 3d electrons in each of the following ions: Ti\(^{2+}\), V\(^{2+}\), Cr\(^{3+}\), Mn\(^{2+}\), Fe\(^{2+}\); Fe\(^{3+}\), Co\(^{2+}\), Ni\(^{2+}\) and Cu\(^{2+}\). Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral)?

Answer:

For hydrated ions in an octahedral environment, the five 3d orbitals split into two energy levels: three lower energy t\(_{2g}\) orbitals and two higher energy e\(_{g}\) orbitals. Assuming high-spin complexes where applicable (due to water being a weak field ligand), the electron occupations are as follows:

Ion	Number of 3d electrons	Configuration (Octahedral, High Spin)
Ti\(^{2+}\)	2	\(3d^2\) (\(t_{2g}^2e_g^0\))
V\(^{2+}\)	3	\(3d^3\) (\(t_{2g}^3e_g^0\))
Cr\(^{3+}\)	3	\(3d^3\) (\(t_{2g}^3e_g^0\))
Mn\(^{2+}\)	5	\(3d^5\) (\(t_{2g}^3e_g^2\))
Fe\(^{2+}\)	6	\(3d^6\) (\(t_{2g}^4e_g^2\))
Fe\(^{3+}\)	5	\(3d^5\) (\(t_{2g}^3e_g^2\))
Co\(^{2+}\)	7	\(3d^7\) (\(t_{2g}^5e_g^2\))
Ni\(^{2+}\)	8	\(3d^8\) (\(t_{2g}^6e_g^2\))
Cu\(^{2+}\)	9	\(3d^9\) (\(t_{2g}^6e_g^3\))

In simple words: For these ions in water (octahedral setting), their d-electrons fill the lower energy t\(_{2g}\) orbitals first, then the higher energy e\(_{g}\) orbitals, following specific patterns based on the total number of d-electrons for each ion.

🎯 Exam Tip: Always recall the splitting of d-orbitals in an octahedral field (t\(_{2g}\) below e\(_{g}\)). For high-spin complexes, fill each orbital singly before pairing electrons. For low-spin (strong field ligands), pair electrons in t\(_{2g}\) first before moving to e\(_{g}\).

 

Question 34. What can be inferred from the magnetic moment values of the following complex species?

Example	Magnetic Moment (BM)
i. K\(_{4}\)[Mn(CN)\(_{6}\)]	2.2
ii. [Fe(H\(_{2}\)O)\(_{6}\)]\(^{2+}\)	5.3
iii. K\(_{2}\)[MnCl\(_{4}\)]	5.9

Answer:

The magnetic moment (\(\mu\)) of a complex can be calculated using the spin-only formula: \(\mu = \sqrt{n(n+2)}\) BM, where 'n' is the number of unpaired electrons. We can infer the number of unpaired electrons and thus the electronic configuration and hybridization from the given magnetic moment values.
1. **For K\(_{4}\)[Mn(CN)\(_{6}\)]:**
The magnetic moment (\(\mu\)) is 2.2 BM.
Using the formula, \(n(n+2) = (2.2)^2 \approx 4.84\). Solving for 'n', we find \(n \approx 1\).
This indicates the presence of 1 unpaired electron. Manganese in K\(_{4}\)[Mn(CN)\(_{6}\)] is in the +2 oxidation state (Mn\(^{2+}\)), which has a \(3d^5\) configuration. Since CN\(^{-}\) is a strong-field ligand, it causes significant crystal field splitting, leading to electron pairing. For \(3d^5\) with 1 unpaired electron, the configuration must be low spin (\(t_{2g}^5e_g^0\)). The complex is octahedral, and the hybridization is \(d^2sp^3\), characteristic of an inner orbital complex.
2. **For [Fe(H\(_{2}\)O)\(_{6}\)]\(^{2+}\):**
The magnetic moment (\(\mu\)) is 5.3 BM.
Using the formula, \(n(n+2) = (5.3)^2 \approx 28.09\). Solving for 'n', we find \(n \approx 4\).
This indicates the presence of 4 unpaired electrons. Iron in [Fe(H\(_{2}\)O)\(_{6}\)]\(^{2+}\) is in the +2 oxidation state (Fe\(^{2+}\)), which has a \(3d^6\) configuration. Since \(H_2O\) is a weak-field ligand, it causes small crystal field splitting, resulting in electrons remaining unpaired. For \(3d^6\) with 4 unpaired electrons, the configuration is high spin (\(t_{2g}^4e_g^2\)). The complex is octahedral, and the hybridization is \(sp^3d^2\), characteristic of an outer orbital complex.
3. **For K\(_{2}\)[MnCl\(_{4}\)]:**
The magnetic moment (\(\mu\)) is 5.9 BM.
Using the formula, \(n(n+2) = (5.9)^2 \approx 34.81\). Solving for 'n', we find \(n \approx 5\).
This indicates the presence of 5 unpaired electrons. Manganese in K\(_{2}\)[MnCl\(_{4}\)] is in the +2 oxidation state (Mn\(^{2+}\)), which has a \(3d^5\) configuration. Since Cl\(^{-}\) is a weak-field ligand and the complex is tetrahedral, there is no significant pairing of electrons. For \(3d^5\) with 5 unpaired electrons, the configuration is high spin. The complex is tetrahedral, and the hybridization is \(sp^3\).
In simple words: By looking at the magnetic moment, we can tell how many unpaired electrons each complex has. For K\(_{4}\)[Mn(CN)\(_{6}\)], 1 unpaired electron means strong ligand pairing. For [Fe(H\(_{2}\)O)\(_{6}\)]\(^{2+}\), 4 unpaired electrons mean weak ligand, high spin. For K\(_{2}\)[MnCl\(_{4}\)], 5 unpaired electrons mean weak ligand, high spin in a tetrahedral shape.

🎯 Exam Tip: Magnetic moment values are critical for determining the number of unpaired electrons, which in turn helps deduce the spin state (high-spin or low-spin), ligand field strength (weak or strong), geometry, and hybridization of a coordination complex. Always work backward from the magnetic moment to 'n' and then correlate with electronic configuration and ligand properties.

GSEB Class 12 Chemistry d-and f-Block Elements Additional Important Questions and Answers

Question 1. The electronic configuration of copper as depicted by two students is given below.

	4s		3d
Student-I:	↑↓	↑↓	↑↓	↑↓	↑↓	↑↓	↑↓
Student-II:	↑	↑↓	↑↓	↑↓	↑↓	↑↓

(1) Whose representation is correct? Justify your answer.
(2) Copper is considered a transition metal. Provide a reason.
(3) Why are Zinc (Zn), Cadmium (Cd), and Mercury (Hg) generally not categorized as transition metals?

Answer:
(1) Student-II's representation of the electronic configuration for copper is correct. The `4s¹3d¹⁰` configuration provides exceptional stability due to the completely filled d-orbitals and the maximal exchange energy achievable by this symmetrical arrangement.
(2) Copper is classified as a transition metal despite having a fully filled `3d¹⁰` sub-shell in its ground state. This classification stems from its most stable oxidation state, Cu²⁺, in which it possesses a partially filled `3d⁹` orbital.
(3) Elements like Zn, Cd, and Hg are generally not considered transition metals. This is because, in both their atomic form and their most common +2 oxidation state, they feature completely filled d-orbitals, thus lacking the characteristic partially filled d-subshells necessary for transition element properties.In simple words: Student-II is correct about copper's stable electron configuration. Copper is a transition metal because its Cu²⁺ ion has an incomplete d-shell. Zinc, cadmium, and mercury are not transition metals because their d-shells are always full in their common states.

🎯 Exam Tip: Remember the special electronic configuration of copper (`4s¹3d¹⁰`) due to stability. The definition of a transition metal relies on having partially filled d-orbitals in at least one stable oxidation state, not just the ground state.

 

Question 2. How do atomic radii of transition elements in a given series vary with an increase in atomic number? What is your observation on the 5d series of transition elements? Discuss the phenomenon involved, if any.

Answer:Atomic radii of transition elements generally decrease across a series with increasing atomic number due to increasing nuclear charge. However, in the 5d series, the atomic radii are very similar to those of the corresponding 4d series elements. This phenomenon is attributed to lanthanide contraction. Lanthanide contraction is the gradual reduction in atomic and ionic sizes from lanthanum to lutetium. It occurs because as the atomic number increases in the lanthanide series, electrons are added to the inner 4f orbitals. These 4f electrons provide very poor shielding from the nuclear charge. Consequently, the effective nuclear charge experienced by the outermost electrons increases, leading to a stronger attraction to the nucleus and a smaller atomic size.In simple words: Atomic sizes usually shrink across a period. But for 4d and 5d series, they are surprisingly similar due to "lanthanide contraction," where 4f electrons poorly shield the nucleus, pulling outer electrons closer.

🎯 Exam Tip: Lanthanide contraction is a crucial concept. Understand its definition, cause (poor shielding of 4f electrons), and consequences (similar atomic radii and properties for 4d and 5d series elements).

 

Question 3. Transition metals are widely used as catalysts in industrial processes?
(1) Give any two industrial processes in which d-block elements are used as catalysts.
(2) Why do transition metals exhibit catalytic properties?
(3) Why do the transition elements exhibit similarity in properties along the period as well as down the group?

Answer:
(1) Two industrial processes utilizing d-block elements as catalysts are:
(a) The Haber process, where iron (Fe) catalyzes the synthesis of ammonia: `\[ N_2 + 3H_2 \xrightarrow{Fe} 2NH_3 \]`
(b) The Contact process, where vanadium pentoxide (\(V_2O_5\)) (or platinum (Pt)) acts as a catalyst for the production of sulfur trioxide: `\[ 2SO_2 + O_2 \xrightarrow{V_2O_5} 2SO_3 \]`
(2) Transition metals display catalytic properties primarily due to two factors: their ability to exhibit multiple oxidation states and the presence of vacant d-orbitals. Their variable oxidation states allow them to readily form unstable intermediate compounds, facilitating reaction pathways.
(3) Transition elements show similarities in properties both across a period and down a group. This is because the differentiating electron enters the penultimate \((n-1)d\) orbitals, rather than the outermost shell. Consequently, the number of electrons in the outermost energy level remains relatively consistent, leading to similar chemical behavior.In simple words: Transition metals are great catalysts (like iron in ammonia production or vanadium oxide in sulfuric acid production) because they can change their oxidation states and have empty d-orbitals to form temporary bonds. Their properties are similar across and down groups because their outer electron count doesn't change much, as new electrons fill inner d-orbitals.

🎯 Exam Tip: For catalysis, focus on variable oxidation states and empty d-orbitals. For periodic trends, emphasize the filling of \((n-1)d\) orbitals as the reason for similarity, especially compared to s- and p-block elements.

 

Question 4. Give reasons for the following:
(1) Zn²⁺ salts are white while Ni²⁺ salts are blue in color.
(2) Chromium (Cr) is a hard metal while Mercury (Hg) is a liquid.
(3) Mn²⁺ shows maximum paramagnetism among the bivalent ions of the first transition series.
(4) Cu²⁺ is paramagnetic while Cu⁺ is diamagnetic.
(5) First ionization energies of 5d elements are higher than those of 3d and 4d elements.

Answer:
(1) Zinc (Zn²⁺) salts appear white because the Zn²⁺ ion has a completely filled \(3d^{10}\) electronic configuration, which prevents d-d electron transitions. In contrast, Nickel (Ni²⁺) salts are blue because the Ni²⁺ ion possesses incompletely filled d-orbitals, allowing for d-d transitions that absorb certain wavelengths of visible light, resulting in a blue color.
(2) Chromium (Cr) is a hard metal due to strong interatomic bonding, which arises from the presence of five unpaired d-electrons in its \((3d^54s^1)\) configuration. Conversely, Mercury (Hg) is a liquid at room temperature because its atoms have completely filled d-orbitals \((5d^{10}6s^2)\), leading to very weak interatomic metallic bonding.
(3) Paramagnetism results from the presence of unpaired electrons. Among the bivalent ions of the first transition series, the Manganese (Mn²⁺) ion has a \(3d^5\) configuration, meaning it possesses the maximum number of unpaired d-electrons (five). This leads to the highest observed paramagnetism for Mn²⁺ ions in this series.
(4) The Copper (Cu²⁺) ion exhibits paramagnetic properties because it has a \(3d^9\) electronic configuration, which includes one unpaired d-electron. Conversely, the Cuprous (Cu⁺) ion is diamagnetic, as it possesses a completely filled \(3d^{10}\) electronic configuration with no unpaired d-electrons.
(5) The first ionization energies of 5d transition elements are generally higher compared to their 3d and 4d counterparts. This is because the 4f electrons, which precede the 5d elements in the periodic table, provide poor shielding from the nuclear charge. This poor shielding results in a stronger attraction between the nucleus and the outermost electrons, requiring more energy to remove the first electron.In simple words: Zn²⁺ salts are white (full d-shell, no d-d transitions), while Ni²⁺ salts are blue (incomplete d-shell allows d-d transitions). Chromium is hard due to strong bonds from unpaired d-electrons, but mercury is liquid because its full d-shells lead to weak bonds. Mn²⁺ is highly paramagnetic because it has the most unpaired d-electrons (five). Cu²⁺ is paramagnetic (one unpaired d-electron), while Cu⁺ is diamagnetic (all electrons paired). 5d elements have higher ionization energies because poor shielding by 4f electrons makes the outer electrons more tightly bound.

🎯 Exam Tip: When explaining color, paramagnetism, or metallic properties, always refer to the electronic configuration (especially d-orbital filling) and the presence/absence of unpaired electrons. For 5d element ionization energies, highlight the impact of lanthanide contraction via poor 4f shielding.

 

Question 5. The following table shows a partial segment of the d-block elements:

Set A	Si	Ti	V	Cr
Set B	Y	Zr	Nb	Mo
Set C	La	Hf	Ta	W

(1) Compare the atomic sizes of elements in Set A with Set B, and Set B with Set C within the same group.
(2) Identify the phenomenon responsible for the relationship between the atomic sizes of elements in Set B and Set C.
(3) Provide the reason for this phenomenon.

Answer:
(1) The atomic sizes of elements in Set B (4d series) are larger than those in Set A (3d series) because a new electron shell is introduced. However, the atomic sizes of elements in Set B are remarkably comparable to those in Set C (5d series) within the same group.
(2) The phenomenon explaining the similar atomic sizes between elements of Set B and Set C is known as lanthanide contraction.
(3) Lanthanide contraction occurs due to the very poor shielding effect of the highly diffused 4f orbitals. As atomic number increases in the lanthanide series, electrons fill these 4f orbitals, which are ineffective at shielding the outer electrons from the increasing nuclear charge. This leads to a greater effective nuclear charge and a subsequent contraction in atomic and ionic radii.In simple words: Moving down a group, atoms usually get bigger. But between the 4d and 5d series, the size stays almost the same because of "lanthanide contraction," which means inner electrons (4f) don't shield the nucleus well, pulling outer electrons closer.

🎯 Exam Tip: Understanding lanthanide contraction is crucial as it explains the similar properties of 4d and 5d series elements and their separation challenges. Be prepared to define it and explain its causes.

 

Question 6. The f-block elements are known as lanthanides and actinides.
(a) Explain why they exhibit similar physical and chemical properties.
(b) Compare the similarities and differences between lanthanides and actinides.
(c) State their important uses.

Answer:
(a) Both lanthanide and actinide series elements are characterized by the filling of \((n-2)f\) orbitals. This orbital filling pattern, combined with their very similar atomic and ionic radii (especially due to lanthanide and actinide contraction), results in their comparable physical and chemical properties.
(b) The similarities and differences between lanthanides and actinides are summarized below:
Similarities:
(i) Both series predominantly display a +3 oxidation state.
(ii) In both series, the f-orbitals are progressively filled.
(iii) Both groups of elements are highly electropositive and chemically reactive.
(iv) Both exhibit characteristic magnetic and spectral properties.
(v) Analogous to lanthanide contraction, actinides also show actinide contraction, which is caused by the poor shielding effect of their 5f-electrons.
(vi) Lanthanide ions, excluding those with \(f^0\) (like La³⁺ or Ce⁴⁺) and \(f^{14}\) (like Yb²⁺ and Lu³⁺) configurations, demonstrate paramagnetism due to the presence of unpaired f-electrons. Neodymium (Nd) typically shows maximum paramagnetism.
Differences:

Lanthanides	Actinides
1. Beyond the common +3 oxidation state, lanthanides primarily show +2 and +4 states.	1. Beyond the common +3 oxidation state, actinides exhibit higher oxidation states, including +4, +5, +6, and +7.
2. They have a lesser tendency to form complexes.	2. They possess a greater tendency to form complexes.
3. All elements are non-radioactive, except for promethium.	3. All actinides are radioactive.
4. Lanthanide compounds are less basic.	4. Actinide compounds are more basic.

(c) Important uses of lanthanides and actinides include:
(i) Lanthanide oxides are utilized as phosphors in color television screens.
(ii) Uranium and thorium are significant as fuels in atomic reactors.
(iii) Thorium is also used in the manufacturing of incandescent gas mantles.In simple words: Lanthanides and actinides are similar because they fill inner f-orbitals and have similar atomic sizes. They differ in their higher oxidation states, complex-forming ability (actinides are better), and radioactivity (all actinides are radioactive, only promethium among lanthanides). They are used in TVs, nuclear fuels, and gas mantles.

🎯 Exam Tip: When comparing lanthanides and actinides, focus on electronic configuration, common and variable oxidation states, complex formation, and radioactivity. These are key differentiators often tested.

 

Question 7. Ramu observed that during the qualitative analysis of Cu²⁺ ions, adding ammonia results in a deep blue colored solution.
(1) Identify the compound formed.
(2) Explain the chemistry of this reaction.
(3) Why do transition metals readily form complexes?
(4) Explain why \([Ti(H_2O)_6]^{3+}\) is colored, whereas \([Sc(H_2O)_6]^{3+}\) is colorless.
(5) Why do Zirconium (Zr) and Hafnium (Hf) exhibit similar properties?

Answer:
(1) The compound formed is tetraamminecopper(II) sulfate, \([Cu(NH_3)_4]SO_4\).
(2) When aqueous Cu²⁺ ions are exposed to ammonia, a deep blue colored solution develops. This color change is due to the formation of the complex ion, tetraamminecopper(II) \([Cu(NH_3)_4]^{2+}\), through the following reaction: `\[ Cu^{2+} + 4NH_3 \implies [Cu(NH_3)_4]^{2+} \]`
(3) Transition metal ions readily form complexes primarily because they possess vacant and/or partially-filled d-orbitals, which can accept electron pairs from ligands. Additionally, their relatively small atomic and ionic radii allow for close packing of ligands around the central metal ion.
(4) The \([Ti(H_2O)_6]^{3+}\) ion is colored because Ti³⁺ has one electron in its d-orbital (\(3d^1\)). This electron can undergo a d-d transition by absorbing energy, typically corresponding to the yellow-green wavelength region. As a result, the complementary purple color is observed. In contrast, \([Sc(H_2O)_6]^{3+}\) is colorless because Sc³⁺ has a \(d^0\) configuration, meaning it has no d-electrons available for d-d transitions.
(5) Zirconium (Zr, Z = 40) and Hafnium (Hf, Z = 72) display very similar properties due to the lanthanide contraction effect. Lanthanide contraction causes the atomic and ionic radii of Hf to be almost identical to those of Zr, despite the significant difference in their atomic numbers. This size similarity leads to comparable chemical behavior.In simple words: When ammonia reacts with copper ions, a blue complex forms because copper is a transition metal that can accept electron pairs into its d-orbitals. Titanium complexes are colored due to d-d electron transitions, while scandium complexes are colorless because they lack d-electrons. Zirconium and Hafnium have similar sizes and properties due to lanthanide contraction.

🎯 Exam Tip: Complex formation, d-d transitions, and lanthanide contraction are fundamental concepts in d-block and f-block chemistry. Be ready to explain these phenomena with examples and appropriate electronic configurations.