Maharashtra Board Class 12 Chemistry Chapter 8 Transition and Inner Transition Elements Solutions

12th Chemistry Chapter 8 Exercise Transition And Inner Transition Elements Solutions Maharashtra Board

Chemistry Class 12 Chapter 8 Exercise Solutions

Choose The Most Correct Option.

Question i. Which one of the following is diamagnetic
(a) Cr³⁺
(b) Fe³⁺
(c) Cu²⁺
(d) Sc³⁺
Answer: (d) Sc³⁺
In simple words: Diamagnetic substances have all paired electrons. Sc³⁺ has a d⁰ configuration ([Ar]), meaning it has no unpaired electrons, making it diamagnetic.

🎯 Exam Tip: Identifying diamagnetic species requires knowing their electronic configuration and checking for unpaired electrons. Remember that d⁰ and d¹⁰ configurations are typically diamagnetic.

Question ii. Most stable oxidation state of Titanium is
(a) +2
(b) +3
(c) +4
(d) +5
Answer: (c) +4
In simple words: Titanium achieves a stable noble gas configuration (like Argon) by losing all four valence electrons (3d²4s²) to form Ti⁴⁺, making +4 its most stable oxidation state.

🎯 Exam Tip: Stability of oxidation states in transition metals often relates to achieving half-filled or completely filled d-orbitals, or a noble gas configuration.

Question iii. Components of Nichrome alloy are
(a) Ni, Cr, Fe
(b) Ni, Cr, Fe, C
(c) Ni, Cr
(d) Cu, Fe
Answer: (c) Ni, Cr
In simple words: Nichrome is primarily an alloy of nickel and chromium, often used in heating elements due to its high resistance and ability to withstand high temperatures.

🎯 Exam Tip: Knowing common alloy compositions is important for material science and application-based questions.

Question iv. Most stable oxidation state of Ruthenium is
(a) +2
(b) +4
(c) +8
(d) +6
Answer: (b) +4
In simple words: Ruthenium exhibits several oxidation states, with +4 being the most common and stable in many of its compounds, although higher states like +8 (in RuO₄) are also known.

🎯 Exam Tip: Ruthenium, like other noble metals, shows a variety of oxidation states, but +4 is a common stable state to remember.

Question v. Stable oxidation states for chromium are
(a) +2, +3
(b) +3, +4
(c) +4, +5
(d) +3, +6
Answer: (d) +3, +6
In simple words: Chromium commonly exhibits +3 and +6 oxidation states, with Cr³⁺ being stable in aqueous solutions and Cr⁶⁺ found in chromates and dichromates.

🎯 Exam Tip: Chromium's +3 and +6 states are particularly important in its chemistry, with +6 being highly oxidizing.

Question vi. Electronic configuration of Cu and Cu⁺¹
(a) 3d¹⁰, 4s⁰; 3d⁹, 4s⁰
(b) 3d⁹, 4s¹; 3d⁸ 4s⁰
(c) 3d¹⁰, 4s¹; 3d¹⁰, 4s⁰
(d) 3d⁸, 4s¹; 3d¹⁰, 4s⁰
Answer: (c) 3d¹⁰, 4s¹; 3d¹⁰, 4s⁰
In simple words: Copper's unusual configuration (3d¹⁰ 4s¹) arises from the stability of a completely filled d-subshell. Cu⁺ loses the 4s electron first, maintaining the stable 3d¹⁰ configuration.

🎯 Exam Tip: Remember the anomalous electronic configurations of chromium and copper due to the stability of half-filled and fully-filled d-orbitals, respectively.

Question vii. Which of the following have d⁰s⁰ configuration
(a) Sc³⁺
(b) Ti⁴⁺
(c) V⁵⁺
(d) all of the above
Answer: (d) All of the above
In simple words: Sc³⁺, Ti⁴⁺, and V⁵⁺ all achieve a noble gas configuration (like Argon) by losing their outer s and d electrons, resulting in a d⁰s⁰ electronic configuration.

🎯 Exam Tip: Ions with d⁰s⁰ configuration are very stable as they attain a noble gas configuration, which means they have no unpaired electrons.

Question viii. Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (d) 5
In simple words: The spin-only magnetic moment formula (\(\mu = \sqrt{n(n+2)}\)) suggests that for a magnetic moment of 5.9 B.M., the number of unpaired electrons (n) must be 5.

🎯 Exam Tip: Be familiar with the spin-only formula for magnetic moment and practice calculating 'n' for given magnetic moment values.

Question ix. In which of the following series all the elements are radioactive in nature
(a) Lanthanoids
(b) Actinoids
(c) d-block elements
(d) s-block elements
Answer: (b) Actinides
In simple words: All elements in the actinoid series are radioactive, meaning their nuclei are unstable and undergo spontaneous decay.

🎯 Exam Tip: Remember that radioactivity is a defining characteristic of the entire actinoid series.

Question x. Which of the following sets of ions contain only paramagnetic ions
(a) Sm³⁺, Ho³⁺, Lu³⁺
(b) La³⁺, Ce³⁺, Sm³⁺
(c) La³⁺, Eu³⁺, Gd³⁺
(d) Ce³⁺, Eu³⁺, Yb³⁺
Answer: (d) Ce³⁺, Eu³⁺, Yb³⁺
In simple words: Paramagnetic ions have unpaired electrons. By examining their electronic configurations, Ce³⁺, Eu³⁺, and Yb³⁺ all possess unpaired f-electrons.

🎯 Exam Tip: To identify paramagnetic ions, determine their electronic configuration and count the number of unpaired electrons. Ions with a completely filled or empty f-subshell (like Lu³⁺ or La³⁺) are diamagnetic.

Question xi. Which actinoid, other than uranium, occur in a significant amount naturally?
(a) Thorium
(b) Actinium
(c) Protactinium
(d) Plutonium
Answer: (a) Thorium
In simple words: Thorium is relatively abundant in the Earth's crust and is found naturally, unlike many other actinoids which are mostly synthetic or occur in trace amounts.

🎯 Exam Tip: While most actinoids are synthetic, thorium and uranium are notable exceptions found in nature.

Question xii. The flux added during extraction of Iron from hematite are its?
(a) Silica
(b) Calcium carbonate
(c) Sodium carbonate
(d) Alumina
Answer: (b) Calcium carbonate
In simple words: Calcium carbonate (limestone) is added as a flux during iron extraction to remove acidic impurities (gangue) like silica, forming slag.

🎯 Exam Tip: Understand the role of flux in metallurgical processes, especially how it helps remove impurities by forming slag.

2. Answer The Following

Question i. What is the oxidation state of Manganese in
(i) MnO₄²⁻
(ii) MnO₄⁻?
Answer:
Oxidation state of Manganese in
(i) MnO₄²⁻ is +6
(ii) MnO₄⁻ is +7
In simple words: In the manganate ion (MnO₄²⁻), manganese is in the +6 oxidation state, while in the permanganate ion (MnO₄⁻), it is in the +7 oxidation state.

🎯 Exam Tip: Be able to calculate oxidation states for elements in polyatomic ions by setting up the appropriate algebraic equation based on the known charges of oxygen and the overall ion charge.

Question ii. Give uses of KMnO₄
Answer:
In simple words: KMnO₄ (potassium permanganate) is a strong oxidizing agent, widely used in various applications such as water purification, disinfectant, and in chemical synthesis.

🎯 Exam Tip: Recall the key properties of common inorganic compounds, especially their uses as oxidizing/reducing agents or in practical applications.

Question iii. Why salts of Sc³⁺, Ti⁴⁺, V⁵⁺ are colorless?
Answer:
(i) Sc³⁺ salts are colourless :
• The electronic configuration of ₂₁Sc [Ar] 3d¹ 4s² and Sc³⁺ [Ar] d⁰.
• Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
• Therefore, Sc³⁺ ions do not absorb the radiations in the visible region. Hence salts of Sc³⁺ are colourless (or white).
(ii) Ti⁴⁺ salts are colourless :
• The electronic configuration of ₂₂Ti [Ar] 3d²4s² and Ti⁴⁺ : [Ar] d⁰
• Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
• Therefore, Ti³⁺ ions do not absorb the radiation in visible region. Hence salts of Ti³⁺ are colourless.
(iii) V⁵⁺ salts are eolourless :
• The electronic configuration of ₂₃V : [Ar] 3d³4s² and V⁵⁺ : [Ar] 3d⁰
• Since there are no unpaired electrons in 3d-subshell, d-d transition is not possible.
• Therefore, V⁵⁺ ions do not absorb the radiations in the visible region. Hence, V⁵⁺ salts are colourless, a
In simple words: The ions Sc³⁺, Ti⁴⁺, and V⁵⁺ all have a d⁰ electronic configuration, meaning they lack unpaired d-electrons. Without unpaired d-electrons, d-d electronic transitions, which are responsible for color in many transition metal complexes, cannot occur, making these ions colorless.

🎯 Exam Tip: Remember that d-d transitions are the primary reason for color in transition metal ions. Ions with d⁰ or d¹⁰ configurations typically appear colorless because these transitions are not possible.

Question iv. Which steps are involved in the manufacture of potassium dichromate from chromite ore?
Answer:
Steps in the manufacture of potassium dichromate from chromite ore are :
• Concentration of chromite ore.
• Conversion of chromite ore into sodium chromate (Na₂CrO₄).
• Conversion of Na₂CrO₄ into sodium dichromate (Na₂Cr₂O₇).
• Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇.
In simple words: The process involves concentrating the chromite ore, converting it to sodium chromate, then to sodium dichromate, and finally exchanging sodium for potassium to get potassium dichromate.

🎯 Exam Tip: Memorize the key reactions and conditions for industrial preparation of important compounds like potassium dichromate; focus on the reagents and the sequence of steps.

Question v. Balance the following equation
(i) KMnO₄ + H₂C₂O₄ + H₂SO₄ → MnSO₄ + K₂SO₄ + H₂O + O₂
(ii) K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + I₂
Answer:
(i) \[ 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ \implies K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂ \]
(ii) Acidified potassium dichromate oxidises potassium iodide (KI) to iodine (I₂). Potassium dichromate is reduced to chromic sulphate. Liberated iodine turns the solution brown
\[ K₂Cr₂O₇ + 6KI + 7H₂SO₄ \implies 4K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂ \]
[Oxidation state of iodine increases from -1 to zero]
In simple words: Balancing chemical equations ensures that the number of atoms of each element is conserved on both sides of the reaction, reflecting the law of conservation of mass.

🎯 Exam Tip: Practice balancing redox reactions using both oxidation state and ion-electron methods, as these are fundamental skills in chemistry examinations.

Question vi. What are the stable oxidation states of plutonium, cerium, manganese, Europium?
Answer:
Stable oxidation states :
Plutonium +3 to +7
Cerium +3, +4
Manganese +2, +4, +6, +7
Europium +2, +3
In simple words: Each element exhibits specific stable oxidation states based on its electronic configuration and chemical environment; for instance, manganese shows a wide range due to its many d-electrons.

🎯 Exam Tip: Familiarize yourself with the common and stable oxidation states of important transition and inner transition elements, as this helps predict their chemical behavior.

Question vii. Write the electronic configuration of chromium and copper.
Answer:
Chromium (₂₄Cr) has electronic configuration,
₂₄Cr (Expected) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(Observed) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Explanation :
• The energy difference between 3d- and 4s-orbitals is very low.
• d-orbitals being degenerate, they acquire more stability when they are half-filled (3d⁵).
• Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².
Copper (₂₉Cu) has electronic configuration,
₂₉Cu (Expected) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
(Observed) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
Explanation:
• The energy difference between 3d- and 4s-orbitals is very low.
• d-orbitals being degenerate, they acquire more stability when they are completely filled.
• Therefore, there arises a transfer of one electron from 4s-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.
Hence, the configuration of Cu is [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁰ 4s².
In simple words: Chromium and copper exhibit anomalous electronic configurations due to the enhanced stability gained by achieving half-filled (3d⁵) or completely filled (3d¹⁰) d-orbitals, respectively. This involves promoting an electron from the 4s orbital to the 3d orbital.

🎯 Exam Tip: Always remember the exceptions in electronic configuration for Chromium and Copper, and be able to explain them in terms of orbital stability (half-filled and completely filled d-orbitals).

Question viii. Why nobelium is the only actinoid with +2 oxidation state?
Answer:
• Nobelium has the electronic configuration ₁₀₂No : [Rn] 5f¹⁴6d⁰7s²
• No²⁺ : [Rn] 5f¹⁴6d⁰
• Since the 4f subshell is completely filled and 6d⁰ empty, +2 oxidation state is the stable oxidation state.
• Other actinoids in +2 oxidation state are not as stable due to incomplete 4f subshell.
In simple words: Nobelium's +2 oxidation state is exceptionally stable because it results in a fully filled 5f¹⁴ subshell and an empty 6d⁰ subshell, providing a very stable electronic configuration.

🎯 Exam Tip: Stability in f-block elements is often related to the half-filled (f⁷) or completely filled (f¹⁴) subshells. Look for these configurations when determining stable oxidation states.

Question ix. Explain with the help of balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.
Answer:
Ce⁴⁺ undergoes hydrolysis as,
\[ Ce⁴⁺ + 2H₂O \implies Ce(OH)₄ + 4H⁺ \]
Due to the presence of H⁺ in the solution, the solution is acidic.
In simple words: Ce(IV) solutions are prepared in acidic media to suppress the hydrolysis of Ce⁴⁺ ions, which would otherwise form insoluble Ce(OH)₄, ensuring the stability and solubility of the Ce⁴⁺ ions in solution.

🎯 Exam Tip: Always consider the hydrolysis of highly charged metal ions. Acidic conditions are often used to prevent precipitation of metal hydroxides and maintain the metal ion in solution.

Question x. What is meant by 'shielding of electrons' in an atom?
Answer:
The inner shell electrons in an atom screen or shield the outermost electron from the nuclear attraction. This effect is called the shielding effect.
The magnitude of the shielding effect depends upon the number of inner electrons.
In simple words: Shielding effect describes how inner shell electrons reduce the effective nuclear charge felt by outer electrons, effectively "shielding" them from the full attractive force of the nucleus.

🎯 Exam Tip: Shielding effect is crucial for understanding periodic trends like atomic size, ionization enthalpy, and electron affinity. Remember that d and f electrons are poor shielders.

Question xi. The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?
Answer:
The 90th element thorium has an electronic configuration, [Rn] 6d²7s². Since it has 2 unpaired electrons it is paramagnetic.
In simple words: Thorium (atomic number 90) has an electronic configuration with unpaired electrons in its 6d subshell, which makes it paramagnetic.

🎯 Exam Tip: To determine if an element is paramagnetic or diamagnetic, write its electronic configuration and identify if there are any unpaired electrons. Paramagnetic species have unpaired electrons, while diamagnetic species have all paired electrons.

3. Answer The Following

Question i. Explain the trends in atomic radii of d-block elements
Answer:
1. The atomic or ionic radii of 3d-series transition elements are smaller than those of representative elements, with the same oxidation states.
2. For the same oxidation state, there is an increase in nuclear charge and a gradual decrease in ionic radii of 3d-series elements is observed. Thus ionic radii of ions with oxidation state +2 decreases with increase in atomic number.
3. There is slight increase is observed in Zn²⁺ ions. With the higher oxidation states, effective nuclear charge increases. Therefore ionic radii decrease with increase in oxidation state of the same element. For example, Fe²⁺ ion has ionic radius 77 pm whereas Fe³⁺ has 65 pm.
In simple words: Across a d-block series, atomic radii generally decrease due to increasing nuclear charge, but this decrease is less pronounced towards the end of the series due to electron-electron repulsion and poor shielding by d-electrons.

🎯 Exam Tip: When discussing trends, always mention the key factors affecting the property (e.g., nuclear charge, shielding effect, electron-electron repulsion) and highlight any anomalies or specific cases like the lanthanoid contraction.

Question ii. Name different zones in the Blast furnace. Write the reactions taking place in them.
Answer:
(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
\[ C + 1/2 O₂ \implies CO \quad ΔH= -220kJ \]
Some part of CO dissociates to give finely divided carbon and O₂.
\[ 2CO \implies 2C + O₂ \]
The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.
(ii) Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
\[ Fe₂O₃ + 3CO \implies 2Fe + 3CO₂ \]
Carbon also reduces partially Fe₂O₃ to Fe.
\[ Fe₂O₃ + 3C \implies 2Fe + 3CO \]
(iii) Zone of slag formation : At 1200 K limestone, CaCO₃ in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO₂, Al₂O₃) and forms a slag of CaSiO₃ and Ca₃AlO₃.
\[ CaCO₃ \implies CaO + CO₂ \]
\[ CaO + SiO₂ \implies CaSiO₃ \]
\[ 12CaO + 2Al₂O₃ \implies 4Ca₃AlO₃ + 3O₂ \]
The slag is removed from the bottom of the furnace through an outlet.
(iv) Zone of fusion: The impurities in ore like MnO₂ and Ca₃(PO₄)₂ are reduced to Mn and P while SiO₂ is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.
The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.
In simple words: A blast furnace operates in distinct zones – combustion, reduction, slag formation, and fusion – each facilitating specific chemical reactions (like oxidation of coke, reduction of iron oxides, and removal of impurities) critical for extracting iron from its ore.

🎯 Exam Tip: Understand the role of each zone in the blast furnace and the key chemical reactions occurring within them, including the purpose of flux and the nature of the reducing agents.

Question iii. What are the differences between cast iron, wrought iron and steel?
Answer:

🎯 Exam Tip: Be able to compare and contrast the properties and uses of different forms of iron based on their carbon content, which significantly influences their mechanical characteristics.

Question iv. Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?
Answer:
The electronic configuration of Fe²⁺ and Fe³⁺ :
Fe²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
Fe³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Due to loss of two electrons from the 4s-orbital and one electron from the 3d-orbital, iron attains 3+ oxidation state. Since in Fe³⁺, the 3d-orbital is half-filled, it gets extra stability, hence Fe³⁺ is more stable than Fe²⁺.
In simple words: Fe³⁺ is more stable than Fe²⁺ because Fe³⁺ achieves a half-filled 3d⁵ electronic configuration, which is energetically more stable due to exchange energy and symmetry.

🎯 Exam Tip: Always relate stability to electronic configurations, particularly half-filled (d⁵, f⁷) or fully-filled (d¹⁰, f¹⁴) subshells, as these configurations impart extra stability.

Question v. Give the similarities and differences in elements of 3d, 4d and 5d series.
Answer:
Similarity:
• They are placed between s-block and p-block of the periodic table.
• All elements are metals showing metallic characters.
• Some are paramagnetic.
• Most of them give coloured compounds.
• They have catalytic properties.
• They form complexes.
• They have variable oxidation states.
Differences:
• In 4d and 5d series lanthanide and actinoid contraction is observed. In 3d series atomic size changes are less marked.
• 4d and 5d elements have high coordination numbers compared to 3d elements.
• 4d and 5d series have similar properties whereas 3d series have different properties.
In simple words: Similarities among 3d, 4d, and 5d series elements include metallic character, variable oxidation states, and complex formation, while differences arise mainly from the lanthanide and actinide contractions in 4d and 5d series, affecting atomic size and coordination numbers.

🎯 Exam Tip: When comparing periodic series, focus on both general trends (like metallic nature, variable oxidation states) and specific effects (like lanthanide contraction) that differentiate them.

Question vi. Explain trends in ionisation enthalpies of d-block elements.
Answer:
1. The ionisation enthalpies of transition elements are quite high and lie between those of s-block and p-block elements. This is because the nuclear charge and atomic radii of transition elements lie between those of s-block and p-block elements.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख संक्रमण तत्वों की प्रथम आयनीकरण एन्थैल्पी (किलोजूल प्रति मोल में) को दर्शाता है। यह दर्शाता है कि आवर्त में परमाणु संख्या बढ़ने पर आयनीकरण एन्थैल्पी बढ़ती है, लेकिन यह वृद्धि अनियमित है, जिसमें कुछ तत्वों के लिए विशिष्ट चोटियाँ और घाटियाँ हैं, जो इलेक्ट्रॉनिक विन्यास की स्थिरता से प्रभावित होती हैं।
2. As atomic number of transition elements increases along the period and along the group, first ionisation enthalpy increases even though the increase is not regular.
3. If IE₁, IE₂ and IE₃ are the first, second and third ionisation enthalpies of the transition elements, then IE₁ < IE₂ < IE₃.
4. In the transition elements, the added last differentiating electron enters into (n-1) d-orbital and shields the valence electrons from the nuclear attraction. This gives rise to the screening effect of (n-1) d-electrons.
5. Due to this screening effect of (n-1) d electrons, the ionisation enthalpy increases slowly and the increase is not very regular.
In simple words: Ionization enthalpies of d-block elements generally increase across a period due to increasing effective nuclear charge, but the trend is irregular because of the varying shielding effects of d-electrons and the stability of specific electronic configurations.

🎯 Exam Tip: When explaining ionization enthalpy trends, discuss the interplay between nuclear charge, atomic size, and shielding effect, and highlight the reasons for any observed irregularities.

Question vii. What is meant by diamagnetic and paramagnetic metal? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.
Answer:
1. Paramagnetic substances : When a magnetic field is applied, substances which are attracted towards the applied magnetic field are called paramagnetic substances. Example : Ni²⁺, Pr⁴⁺
2. Diamagnetic substances : When a magnetic field is applied, substances which are repelled by the magnetic fields are called diamagnetic substances. Example : Zn²⁺, La³⁺
3. Ferromagnetic substances : When a magnetic field is applied, substances which are attracted very strongly are called ferromagnetic substances. These substances can be magnetised. For example, Fe, Co, Ni are ferromagnetic.
In simple words: Paramagnetic substances are attracted by a magnetic field due to unpaired electrons, while diamagnetic substances are repelled due to all paired electrons. Ferromagnetic substances exhibit a strong, permanent magnetism.

🎯 Exam Tip: The key differentiator between paramagnetic and diamagnetic substances is the presence or absence of unpaired electrons, respectively. Ferromagnetism is a more intense form of paramagnetism with permanent magnetic domains.

Question viii. Why the ground-state electronic configurations of gadolinium and lawrencium are different than expected?
Answer:
In simple words: Gadolinium and Lawrencium have unexpected ground-state electronic configurations to achieve the greater stability associated with a half-filled (f⁷) or completely filled (f¹⁴) f-subshell, often by promoting an electron from a d or s orbital.

🎯 Exam Tip: Be aware of the exceptions in electronic configurations for f-block elements like Gadolinium and Lawrencium, which aim for half-filled or fully-filled f-orbitals for stability.

Question ix. Write steps involved in the metallurgical process
Answer:
The various steps and principles involved in the extraction of pure metals from their ores are as follows.:
• Concentration of ores in which impurities (gangue) are removed.
• Conversion of ores into oxides or other reducible compounds of metals.
• Reduction of ores to obtain crude metals.
• Refining of metals giving pure metals.
In simple words: Metallurgical processes typically involve concentrating the ore, converting the metal compound to an oxide (if not already), reducing the oxide to crude metal, and finally refining it to achieve desired purity.

🎯 Exam Tip: Understand the four main stages of metallurgy: concentration, extraction (calcination/roasting then reduction), and refining, along with the purpose of each step.

Question x. Cerium and Terbium behaves as good oxidising agents in +4 oxidation state. Explain.
Answer:
• The most stable oxidation state of lanthanoids is +3.
• Hence, Ce⁴⁺ (cerium) and Tb⁴⁺ (terbium) tend to get +3 oxidation state which is more stable.
• Since they get reduced by accepting electron, they are good oxidising agents in +4 oxidation state.
In simple words: Ce⁴⁺ and Tb⁴⁺ are good oxidizing agents because their most stable oxidation state is +3. They readily accept an electron to transition from +4 to +3, thus causing the oxidation of another substance.

🎯 Exam Tip: Remember that elements in higher, less stable oxidation states tend to be good oxidizing agents, as they readily gain electrons to achieve a more stable, lower oxidation state.

Question xi. Europium and Ytterbium behave as good reducing agents in +2 oxidation state explain.
Answer:
• The most stable oxidation state of lanthanoids is +3.
• Hence, Eu²⁺ and Yb²⁺ tend to get +3 oxidation states by losing one electron.
• Since they get oxidised, they are good reducing agents in +2 oxidation state.
In simple words: Eu²⁺ and Yb²⁺ are good reducing agents because their most stable oxidation state is +3. They readily lose an electron to transition from +2 to +3, thereby reducing another substance.

🎯 Exam Tip: Elements in lower, less stable oxidation states tend to be good reducing agents, as they readily lose electrons to achieve a more stable, higher oxidation state.

12th Chemistry Digest Chapter 8 Transition And Inner Transition Elements Intext Questions And Answers

Do You Know? (Textbook Page No 165)

Question 1. In which block of the modern periodic table are the transition and inner transition elements placed?
Answer:
The transition elements are placed in d-block and inner transition elements are placed in f-block of the modern periodic table.
In simple words: Transition elements reside in the d-block, while inner transition elements (lanthanoids and actinoids) are found in the f-block of the periodic table.

🎯 Exam Tip: A fundamental concept to remember is the classification of elements into s, p, d, and f-blocks based on their differentiating electron's orbital.

Use Your Brain Power! (Textbook Page No 167)

Question 1. Fill in the blanks with correct outer electronic configurations.
Answer:
In simple words: The outer electronic configuration describes the arrangement of electrons in the outermost shells, which dictates an element's chemical properties.

🎯 Exam Tip: Mastering electronic configurations, especially for transition and inner transition elements, is crucial for understanding their properties and periodic trends.

Try This..... (Textbook Page No 168)

Question 1. Write the electronic configuration of Cr and Cu.
Answer: \(^{24}\text{Cr}: [\text{Ar}] 3d^54s^1\) \(^{30}\text{Cu} : [\text{Ar}] 3d^{10}4s^1\)
In simple words: Chromium has a half-filled d-orbital and copper has a completely filled d-orbital for extra stability, leading to these specific electron configurations.

🎯 Exam Tip: Remember the anomalous electron configurations of Chromium and Copper due to the stability of half-filled and completely filled d-orbitals. This is a common exam question.

Can You Tell? (Textbook Page No 168)

Question 1. Which of the first transition series element shows the maximum number of oxidation states and why?
Answer:
• \(^{25}\text{Mn}\) shows the maximum number of oxidation states, \(+2\) to \(+7\).
• \(^{25}\text{Mn}: [\text{Ar}] 3d^54s^2\)
• Mn has incompletely filled d-subshell.
• Due to small difference in energy between 3d and 4s -orbitals, Mn can lose (or share) electrons from both the orbitals.
• Hence Mn shows oxidation states from \(+2\) to \(+7\).
In simple words: Manganese shows the most oxidation states in the first transition series because its 3d and 4s orbitals have very similar energies, allowing many electrons to be lost or shared.

🎯 Exam Tip: Focus on the electron configuration and the energy difference between orbitals when explaining variable oxidation states. Manganese is a key example for this concept.

Question 2. Which elements in the 4d and 5d-series will show maximum number of oxidation states?
Answer: In 4d-series maximum number of oxidation states are for Ruthenium Ru (\(+2\), \(+3\), \(+4\), \(+6\), \(+7\), \(+8\)). In 5d-series, maximum number of oxidation states are for Osmium, Os (\(+2\) to \(+8\)).
In simple words: Ruthenium in the 4d series and Osmium in the 5d series exhibit the widest range of oxidation states.

🎯 Exam Tip: Remember the specific elements Ruthenium and Osmium for maximum oxidation states in their respective series. This demonstrates the trend across periods.

Try This (Textbook Page No 168)

Question 1. Write the electronic configuration of Mn\(^{6+}\), Mn\(^{4+}\), Fe\(^{4+}\), Co\(^{5+}\), Ni\(^{2+}\).
Answer:

🎯 Exam Tip: When writing ion electronic configurations, always remove electrons from the outermost s-orbital first, then from the d-orbital. This is a common mistake for students.

Try This (Textbook Page No 171)

Question 1. Pick up the paramagnetic species from the following : Cu\(^{1+}\), Fe\(^{3+}\), Ni\(^{2+}\), Zn\(^{2+}\), Cd\(^{2+}\), Pd\(^{2+}\).
Answer: The following ions are paramagnetic : Fe\(^{3+}\), Ni\(^{2+}\), Pd\(^{2+}\)
In simple words: Paramagnetic species are those with unpaired electrons in their electron configuration, which cause them to be attracted to a magnetic field.

🎯 Exam Tip: To identify paramagnetic species, write their electron configurations and look for unpaired electrons. Remember that completely filled or empty orbitals usually indicate diamagnetism.

Try This (Textbook Page No 171)

Question 1. What will be the magnetic moment of transition metal having 3 unpaired electrons?
(a) equal to 1.73 B.M.?
(b) less than 1.73 BM.
(c) more than 1.73 B.M.?
Answer: By spin-only formula, \( \mu = \sqrt{n(n + 2)} \) where n is number of unpaired electrons.
\( \mu = \sqrt{3(3 + 2)} = \sqrt{3(5)} = 3.87 \) B. M
Thus the value is more than 1.73 B.M.
In simple words: With 3 unpaired electrons, the magnetic moment calculated by the spin-only formula is 3.87 B.M., which is greater than 1.73 B.M.

🎯 Exam Tip: Understand the spin-only formula \( \mu = \sqrt{n(n + 2)} \) for calculating magnetic moments. Practice calculating for different numbers of unpaired electrons. This formula is crucial for determining paramagnetism.

Use Your Brain Power! (Textbook Page No 171)

Question 1. A metal ion from the first transition series has two unpaired electrons. Calculate the magnetic moment.
Answer:\[ \mu = \sqrt{n(n + 2)} \] \[ \mu = \sqrt{2(2 + 2)} \] \[ \mu = \sqrt{8} \] \[ \mu = 2.84 \text{ B.M.} \]
In simple words: For a metal ion with two unpaired electrons, its magnetic moment is calculated to be 2.84 B.M. using the spin-only formula.

🎯 Exam Tip: Always state the formula clearly and substitute the value of 'n' correctly. Precision in calculation and units (B.M.) is important for scoring marks.

Problem (Textbook Page No 172)

Question 1. Calculate the spin-only magnetic moment of divalent cation of a transition metal with atomic number 25.
Answer: For element with atomic number 25. electronic configuration of its divalent cation will be : [Ar] \(3d^5\).
\[\begin{tabular}{|c|c|c|c|c|} \hline \multicolumn{5}{|c|}{\(3d\)} \\ \hline \( \uparrow \) & \( \uparrow \) & \( \uparrow \) & \( \uparrow \) & \( \uparrow \) \\ \hline \end{tabular} \quad \begin{tabular}{|c|} \hline \multicolumn{1}{|c|}{\(4s\)} \\ \hline \\ \hline \end{tabular} \]
There are 5 unpaired electrons.
\( \therefore n = 5\).
\( \therefore \mu = \sqrt{5 (5+2)} = \sqrt{5(7)} = \sqrt{35} = 5.92 \text{ B.M.} \)
In simple words: A divalent cation of an element with atomic number 25 (Manganese) has 5 unpaired electrons in its \(3d^5\) configuration, resulting in a spin-only magnetic moment of 5.92 B.M.

🎯 Exam Tip: First, identify the element, then its electron configuration. Next, determine the ion's configuration, count unpaired electrons, and apply the spin-only formula. Show all steps clearly.

Try This..... (Textbook Page No 172)

Question 1. Calculate the spin-only magnetic moment of a divalent cation of element Slaving atomic number 27.
Answer: Electronic configuration of divalent ion of an element with atomic number 27 : [Ar] \(3d^7\);
\[\begin{tabular}{|c|c|c|c|c|} \hline \multicolumn{5}{|c|}{[Ar] \(3d^7\)} \\ \hline \( \uparrow\downarrow \) & \( \uparrow\downarrow \) & \( \uparrow\downarrow \) & \( \uparrow \) & \( \uparrow \) \\ \hline \end{tabular}\]
There are 3 unpaired electrons.
\( \therefore n = 3\).
\( \therefore \mu = \sqrt{3 (3+2)} = \sqrt{3(5)} = \sqrt{15} = 3.87 \text{ B.M.} \)
In simple words: For a divalent cation of an element with atomic number 27 (Cobalt), its \(3d^7\) configuration has 3 unpaired electrons, yielding a magnetic moment of 3.87 B.M.

🎯 Exam Tip: Accurately filling the d-orbitals according to Hund's rule is essential to correctly count unpaired electrons. Small errors in electron counting lead to incorrect magnetic moments.

Can You Tell? (Textbook Page No 172)

Question 1. Compounds of s and p-block elements are almost white. What could be the absorbed radiation? (uv or visible)?
Answer: The white colour of a compound indicates the absorption of uv radiation.
In simple words: S and p-block element compounds appear white because they absorb radiation in the ultraviolet (UV) region, not the visible light spectrum.

🎯 Exam Tip: Relate the color of a substance to the type of light it absorbs. Absorption in the visible region results in a complementary observed color, while UV absorption leads to colorless or white appearance.

Can You Tell? (Textbook Page No 181)

Question 1. Why f-block elements are called inner transition metals?
Answer: f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.
In simple words: F-block elements are called inner transition metals because their f-orbitals are deeply buried within the atom, making them "inner" compared to the d-orbitals of transition metals.

🎯 Exam Tip: Emphasize the deep-seated nature of the f-orbitals compared to d-orbitals when explaining why f-block elements are "inner" transition metals. This structural difference is key.

Question 2. Are there an similarities between transition and inner transition metals?
Answer: There are some properties similarity between transition and inner transition metals.
• They are placed between s and p-block elements.
• They are metals with filling of inner subshells in their electronic configuration.
• They show variable oxidation states.
• They show magnetism.
• They form coloured compounds.
• They have catalytic property.
In simple words: Yes, both transition and inner transition metals are located between s and p blocks, are metallic, exhibit variable oxidation states, often show magnetism, form colored compounds, and have catalytic properties due to their partially filled d or f orbitals.

🎯 Exam Tip: Focus on shared characteristics like variable oxidation states, metallic nature, color, and magnetism. These similarities stem from their partially filled d or f orbitals.

Problem (Textbook Page No 184)

Question 1. Which of the following will have highest fourth ionisation enthalpy, La\(^{4+}\), Gd\(^{4+}\), Lu\(^{4+}\).
Answer:
La: \(4f^05d^16s^2\)
Gd: \(4f^75d^16s^2\)
Lu : \(4f^{14}5d^16s^2\)
Lu will have the highest fourth ionisation enthalpy since Lu\(^{3+}\) has the most stable configuration of \(4f^{14}\).
In simple words: Lutetium will have the highest fourth ionization enthalpy because after losing three electrons (forming Lu\(^{3+}\)), it achieves a very stable completely filled \(4f^{14}\) configuration, making it difficult to remove a fourth electron.

🎯 Exam Tip: Stability of half-filled or completely filled subshells plays a crucial role in ionization enthalpies. Removing an electron from a stable configuration requires significantly more energy.

Use Your Brain Power! (Textbook Page No 185)

Question 1. Do you think that lanthanoid complex would show magnetism?
Answer: Lanthanoid complexes may show magnetism.
In simple words: Yes, lanthanoid complexes can show magnetism because many lanthanoid ions have unpaired f-electrons.

🎯 Exam Tip: Remember that magnetism in lanthanoids primarily arises from unpaired f-electrons. The shielding of these f-electrons makes their magnetic properties less influenced by the ligand field.

Question 2. Can you calculate the spin only magnetic moment of lanthanoid complexes using the same formula that you used for transition metal complexes?
Answer: You cannot calculate magnetic moment of lanthanoid complexes using spin only formula as you have to consider orbital momentum also.
In simple words: No, the spin-only formula is insufficient for lanthanoid complexes because their f-electrons are well-shielded, allowing orbital angular momentum contributions to the magnetic moment to be significant and not quenched by the ligand field.

🎯 Exam Tip: Differentiate between transition metals and lanthanoids regarding magnetic moment calculations. For lanthanoids, orbital contribution is significant, unlike most transition metals where it's often quenched.

Question 3. Calculate the spin only magnetic moment of La\(^{3+}\). Compare the value with that given in the table.
Answer: La\(^{3+}\) ion has no unpaired electron.
\( \therefore \mu = \sqrt{n(n+2)} \)
\( = \sqrt{0 (0+2)} \)
\( = 0 \)
La\(^{3+}\) ion has zero value of magnetic moment same as given in the table.
In simple words: La\(^{3+}\) has no unpaired electrons, so its spin-only magnetic moment is 0, which matches the table's value.

🎯 Exam Tip: For ions with zero unpaired electrons, the magnetic moment is always zero. This is a quick check to ensure your electron counting is correct for diamagnetic species.