Step 1: In the complex , the titanium ion is in the +3 oxidation state with electronic configuration: Step 2: The presence of one unpaired electron in the 3d orbital allows for electronic transitions between and levels when light is absorbed. Step 3: This transition is responsible for the colour observed in the complex.
02
PYQ 2023
medium
chemistryID: cbse-cla
Assertion (A): Zn, Cd and Hg are not regarded as transition elements. Reason (R): Zn, Cd and Hg do not have partially filled d-orbitals in their ground state or in any one of their common oxidation states.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is \textit{not} the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(1)
Step 1: Zn, Cd, and Hg belong to the d-block but are not considered true transition elements because transition elements must have at least one ion with an incomplete d-subshell. Step 2: In their ground state and common oxidation states (+2), Zn, Cd, and Hg have completely filled d-orbitals (d configuration). Step 3: Therefore, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.
03
PYQ 2023
medium
chemistryID: cbse-cla
The most common oxidation state for all lanthanoids is:
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Lanthanoids belong to the f-block elements, specifically the 4f series. They are characterized by the filling of the 4f orbitals. Step 2: The most stable and common oxidation state for almost all lanthanoids is . This is due to the stable removal of two 6s electrons and one 5d or 4f electron. Step 3: Some lanthanoids can also show and oxidation states, but these are less common and occur only in certain cases (e.g., Eu , Ce ). The oxidation state is very rare and not typical for lanthanoids.
04
PYQ 2023
medium
chemistryID: cbse-cla
Why do transition metals form large numbers of complex compounds?
Official Solution
Correct Option:
(1)
Step 1: Transition metals have small atomic/ionic radii and high nuclear charge. Step 2: Their partially filled (n-1)d orbitals are available to accept lone pairs from ligands. Step 3: Hence, they readily form coordinate bonds with ligands, resulting in stable complex compounds.
05
PYQ 2023
medium
chemistryID: cbse-cla
How does acidified potassium permanganate solution react with ions? Write ionic equation.
Official Solution
Correct Option:
(1)
Step 1: KMnO4 in acidic medium acts as a strong oxidizing agent. Step 2: Fe2+ is oxidized to Fe3+, and MnO4– is reduced to Mn2+. Ionic equation:
06
PYQ 2023
medium
chemistryID: cbse-cla
Calculate the ‘spin only’ magnetic moment of a divalent ion of a metal M in aqueous solution. The atomic number of the metal M is 25.
Official Solution
Correct Option:
(1)
Step 1: Atomic number of M = 25 → Mn Mn: [Ar] 3d5 4s2 Step 2: In divalent state (Mn2+): loses two electrons → 3d5 configuration Step 3: Number of unpaired electrons Step 4: Magnetic moment (spin only):
07
PYQ 2023
medium
chemistryID: cbse-cla
The chemistry of the actinoids is more complex as compared to the lanthanoids. Why?
Official Solution
Correct Option:
(1)
Step 1: In actinoids, 5f orbitals participate in bonding, while in lanthanoids, 4f orbitals are deeply buried and less involved. Step 2: Actinoids show variable oxidation states ranging from +3 to +6 (sometimes +7), whereas lanthanoids predominantly show +3 oxidation state. Step 3: The 5f electrons in actinoids are more shielded and more easily influenced by ligands.
08
PYQ 2023
medium
chemistryID: cbse-cla
Why is E for Mn3+/Mn2+ redox couple more positive?
Official Solution
Correct Option:
(1)
Step 1: Mn3+ has a high charge density and causes strong hydration, making it less stable. Step 2: Due to instability, Mn3+ has a high tendency to gain an electron (i.e., get reduced). Step 3: Hence, the redox potential (E°) becomes more positive for the Mn3+/Mn2+ couple.
09
PYQ 2024
medium
chemistryID: cbse-cla
Define lanthanoid contraction. How does it affect the atomic radii of the third transition series and the second transition series?
Official Solution
Correct Option:
(1)
Lanthanoid Contraction refers to the gradual decrease in atomic size across the lanthanoid series, due to the ineffective shielding of the 4f-electrons. This contraction affects the atomic radii of the third transition series, which are smaller than expected compared to the second transition series. The increased nuclear charge in the lanthanoids results in a stronger pull on the electrons, reducing their size.
10
PYQ 2024
medium
chemistryID: cbse-cla
What is the cause of contraction in the atomic size of lanthanoids?
Official Solution
Correct Option:
(1)
The contraction in the atomic size of lanthanoids is due to the poor shielding effect of the 4f-orbitals. As we move across the lanthanoid series, the 4f-electrons are added, but they do not shield the increasing nuclear charge effectively, leading to a gradual decrease in atomic size.
11
PYQ 2024
medium
chemistryID: cbse-cla
Why do transition metals and their compounds act as good catalysts?
Official Solution
Correct Option:
(1)
Transition metals and their compounds act as good catalysts because they have multiple oxidation states, which allows them to participate in electron transfer during reactions. Additionally, their ability to form complex intermediates and adsorb reactants on their surface reduces the activation energy, facilitating the reaction.
12
PYQ 2024
medium
chemistryID: cbse-cla
Which of the following elements of the 3d series of transition elements has the lowest ?
1
Sc
2
Cr
3
Cu
4
Zn
Official Solution
Correct Option:
(4)
refers to the enthalpy change or heat of formation for a particular element in its standard state. For transition elements, the enthalpy change is affected by the electron configuration, the stability of the element, and its ionization energy. - Scandium (Sc) and chromium (Cr) have relatively high ionization energies and enthalpy changes.
- Copper (Cu) has a relatively stable electron configuration due to its completely filled d-subshell, making it less reactive and having a moderate .
- Zinc (Zn), on the other hand, has a completely filled 3d subshell and a stable electron configuration, making it the least reactive and having the lowest enthalpy change among the 3d transition elements. This is why Zn has the lowest . Thus, the correct answer is Zn.
13
PYQ 2025
medium
chemistryID: cbse-cla
Which of the following transition metal ion is not coloured?
1
Cu
2
Ni
3
Co
4
V
Official Solution
Correct Option:
(1)
The colour of a transition metal ion arises due to the presence of unpaired electrons in the d-orbitals which allow d-d transitions. Cu has an electronic configuration of [Ar] 3d , which has all paired electrons and no possibility of d-d transitions. Therefore, it is colourless.
14
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): Actinoids show irregularities in their electronic configurations. Reason (R): Actinoids are radioactive in nature.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(2)
Actinoids do indeed show irregularities in their electronic configurations, mainly due to the f-orbitals and their greater number of electrons. However, their radioactivity is not the cause of this irregularity. The reason is unrelated to their electronic configuration.
15
PYQ 2025
easy
chemistryID: cbse-cla
Answer the following questions: Name two elements of 3d series for which the third ionisation enthalpies are quite high. Out of KMnO and K MnO , which one is paramagnetic and why? Write any one consequence of lanthanoid contraction. How do you prepare potassium manganate from pyrolusite ore? Why is the ability of oxygen more than fluorine to stabilise higher oxidation states of transition metals?
Official Solution
Correct Option:
(1)
(i) Two elements of the 3d series for which the third ionization enthalpies are quite high are Zinc (Zn) and Copper (Cu). Both have completely filled d-orbitals, making it more difficult to remove the third electron. (ii) Out of KMnO and K MnO , KMnO is paramagnetic. This is because in KMnO , manganese exists in the +7 oxidation state, which leaves it with unpaired electrons in the d-orbitals, making it paramagnetic. K MnO , on the other hand, has manganese in the +6 oxidation state, which does not result in unpaired electrons. (iii) One consequence of lanthanoid contraction is the decrease in ionic radii as we move across the lanthanide series. This causes a similar size among elements, leading to similarities in chemical behavior. (iv) Potassium manganate can be prepared from pyrolusite ore by reacting it with a suitable oxidizing agent like potassium hydroxide (KOH) under heating. The reaction is as follows:
(v) Oxygen stabilizes higher oxidation states of transition metals better than fluorine because oxygen has a smaller size and a higher electronegativity, which helps in accepting electrons from the transition metal. In contrast, fluorine, being more electronegative, tends to form more stable bonds in lower oxidation states.
16
PYQ 2025
medium
chemistryID: cbse-cla
Why is the ability of oxygen greater than fluorine to stabilise higher oxidation states of transition metals?
Official Solution
Correct Option:
(1)
Oxygen can form multiple bonds ( -bonding) with transition metals. This effective -bond overlap stabilises higher oxidation states through oxo ligands. Fluorine, although very electronegative, forms only single -bonds and does not form -bonds with metals. Therefore, oxo ions like CrO stabilise Cr(+6) more than fluorides can.
17
PYQ 2025
medium
chemistryID: cbse-cla
How do you prepare potassium manganate from pyrolusite ore?
Official Solution
Correct Option:
(1)
MnO (pyrolusite) fused with KOH + O : 2MnO + 4KOH + O → 2K MnO + 2H O.
18
PYQ 2025
medium
chemistryID: cbse-cla
Out of KMnO and K MnO , which one is paramagnetic and why?
Official Solution
Correct Option:
(1)
KMnO : Mn in +7 → d → diamagnetic. K MnO : Mn in +6 → d → 1 unpaired → paramagnetic.
19
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): First ionisation enthalpy of Cr is lower than that of Zn. Reason (R): Zinc is a non-transition element.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(2)
Cr has lower I.E. due to stable half-filled d configuration, which is easier to ionise. Zn has stable filled d → higher I.E. While Zn is not a transition element (due to no variable oxidation states), this fact does not explain Cr’s lower I.E.
20
PYQ 2025
medium
chemistryID: cbse-cla
With increase in atomic number, the melting point of transition metals :
1
first increases and then decreases
2
increases continuously
3
first decreases and then increases
4
remains constant
Official Solution
Correct Option:
(1)
In transition metals, melting point first rises due to stronger metallic bonding as more unpaired d-electrons contribute to bonding. After mid-series, pairing of d-electrons reduces bonding strength, so melting point decreases again.
21
PYQ 2025
medium
chemistryID: cbse-cla
Answer the following question: Name two elements of 3d series for which the third ionisation enthalpies are quite high.
Official Solution
Correct Option:
(1)
Mn and Zn. Mn: stable half-filled d . Zn: stable completely filled d .
22
PYQ 2025
medium
chemistryID: cbse-cla
Complete the following ionic equations : 5SO + 2MnO + 6H ⟶ ?
Official Solution
Correct Option:
(1)
5SO + 2MnO + 6H ⟶ 5SO + 2Mn + 3H O.
23
PYQ 2025
medium
chemistryID: cbse-cla
Account for the following : (I) The E value for manganese is highly negative, whereas E is highly positive.
Official Solution
Correct Option:
(1)
Manganese has stable half-filled d configuration in Mn (3d ). Removing an electron to form Mn breaks this stable configuration → requires more energy. Thus, Mn to Mn is difficult (highly negative E ) but Mn to Mn is favourable (high E ).
24
PYQ 2025
hard
chemistryID: cbse-cla
Among the following outermost electronic configurations of transition metals, which one shows the highest oxidation state?
1
2
3
4
Official Solution
Correct Option:
(2)
The configuration corresponds to Chromium, which can achieve its highest oxidation state of +6 (as in or chromates). This is higher than the oxidation states possible with the other given configurations.
25
PYQ 2025
medium
chemistryID: cbse-cla
Give explanation for each of the following observations:
(a) Mn3+ is an oxidizing agent, whereas Cr2+ is a reducing agent (both d4 configuration).
(b) Actinoid contraction is greater than lanthanoid contraction.
(c) Transition metals form interstitial compounds with H, B, C, N.
Official Solution
Correct Option:
(1)
(a) Mn3+ is an oxidizing agent, whereas Cr2+ is a reducing agent (both d4 configuration).
Mn3+ readily gains an electron to become the more stable Mn2+, hence acts as an oxidizing agent. Cr2+ loses an electron to form Cr3+, which is more stable due to half-filled t2g orbitals (d3), thus acts as a reducing agent.
(b) Actinoid contraction is greater than lanthanoid contraction.
Actinoids have 5f orbitals that are more diffused and poorly shielded compared to 4f orbitals in lanthanoids. This leads to a greater effective nuclear charge and stronger contraction across the series.
(c) Transition metals form interstitial compounds with H, B, C, N.
Transition metals have vacant d-orbitals and lattice voids where small atoms like H, B, C, N can fit without disturbing the structure, forming interstitial compounds which are hard and have high melting points.
26
PYQ 2025
medium
chemistryID: cbse-cla
Assertion (A): Actinoid contraction is greater from element to element than lanthanoid contraction. Reason (R): Actinoids show wide range of oxidation states.
1
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
2
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
3
Assertion (A) is true, but Reason (R) is false.
4
Assertion (A) is false, but Reason (R) is true.
Official Solution
Correct Option:
(1)
The actinoid contraction is greater than the lanthanoid contraction because of the greater ef- fective nuclear charge in actinoids. This increased nuclear attraction leads to a smaller size and greater variation in oxidation states, making their chemistry more complex.
27
PYQ 2025
medium
chemistryID: cbse-cla
Complete the following ionic equations:
Official Solution
Correct Option:
(1)
For equation (I), the oxidation states of manganese change from +6 to +7 in the permanganate ion, indicating an oxidation. In acidic medium, sulfur dioxide ( ) is oxidized to sulfate ( ). The balanced ionic equation for this reaction is: . For equation (II), the iodide ion is oxidized to iodine by permanganate in an acidic medium. The balanced ionic equation is: .
28
PYQ 2025
medium
chemistryID: cbse-cla
Account for the following :
The value for manganese is highly negative, whereas is highly positive.
Official Solution
Correct Option:
(1)
(I) The highly negative value for manganese indicates that the reduction of to is not favored, as manganese prefers a lower oxidation state. On the other hand, the positive value suggests that the reduction of to is favorable. This is due to the greater stability of higher oxidation states in manganese compared to its lower oxidation states. (II) Actinoids show a wide range of oxidation states due to the similar energies of their 5f, 6d, and 7s orbitals. This allows for greater flexibility in bonding and more accessible electron transitions, enabling actinoids to exhibit multiple oxidation states. (III) Transition metals have high melting points due to the strong metallic bonding, which arises from the delocalization of electrons in the d-orbitals. These delocalized electrons form a strong electrostatic attraction between the metal cations and the electron cloud, contributing to high melting points.
29
PYQ 2026
medium
chemistryID: cbse-cla
Which of the following is most basic?
1
2
3
4
Official Solution
Correct Option:
(3)
Step 1: Understanding the Concept: For oxides of transition metals, the basicity is inversely proportional to the oxidation state of the metal. Lower oxidation states lead to ionic character and basic nature, while higher oxidation states lead to covalent character and acidic nature. Step 2: Detailed Explanation: Metal oxides in low oxidation states (like ) are basic because the metal can easily donate electron pairs. As the oxidation state increases to or higher, the metal-oxygen bond becomes more covalent, making the oxide amphoteric or acidic. In , Manganese is in its lowest oxidation state ( ) among the options, making it the most basic oxide. Conversely, (+7) is strongly acidic. Step 3: Final Answer: The most basic oxide is .
30
PYQ 2026
medium
chemistryID: cbse-cla
Why transition elements form interstitial compounds?
Official Solution
Correct Option:
(1)
Concept: Transition metals form metallic crystal lattices (FCC, BCC, or HCP) made of closely packed metal atoms. These lattices naturally contain small empty spaces known as interstitial sites. Step 1: Understand interstitial sites. When metal atoms arrange in a crystal lattice, small voids are left between them. These voids are called interstitial spaces. Step 2: Incorporation of small atoms. Small non-metal atoms such as: can easily occupy these interstitial sites without significantly disturbing the metal lattice. Step 3: Formation of interstitial compounds. This leads to the formation of interstitial compounds such as: These compounds are often non-stoichiometric and retain metallic characteristics. Step 4: Final conclusion. Transition metals form interstitial compounds because their crystal lattices contain empty interstitial spaces where small atoms can be accommodated.
31
PYQ 2026
medium
chemistryID: cbse-cla
Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?
Official Solution
Correct Option:
(1)
Concept: Ionization enthalpy is the energy required to remove an electron from an atom or ion. It depends strongly on the stability of the electronic configuration, especially the presence of half-filled or fully-filled subshells. Step 1: Write electronic configurations. Step 2: Remove the first electron (formation of ). The first ionization removes the electron: Step 3: Analyze second ionization enthalpy ( ). The second electron must now be removed from the -subshell:
In Cr : removal from stable half-filled
In Cu : removal from highly stable fully-filled
Both configurations are exceptionally stable due to symmetry and high exchange energy. Step 4: Conclusion. Breaking these stable half-filled and fully-filled configurations requires extra energy. Hence, the second ionization enthalpy is unusually high for both Cr and Cu.
32
PYQ 2026
medium
chemistryID: cbse-cla
Actinoids show larger number of oxidation states:
1
because they are electropositive in nature
2
because they have large atomic numbers
3
because they have large atomic size
4
due to comparable energies of 5f, 6d and 7s orbitals
Official Solution
Correct Option:
(4)
Concept:
Actinoids exhibit a wide range of oxidation states due to the involvement of multiple orbitals in bonding. Unlike lanthanoids, actinoids have:
Partially filled 5f orbitals
Accessible 6d orbitals
Outer 7s electrons
Because the energies of these orbitals are very close, electrons from all three can participate in bonding. Step 1: Analyze the reason. When orbitals of similar energy levels are available, multiple electrons can be lost or shared, leading to variable oxidation states. Step 2: Evaluate options.
Electropositivity — not the main reason.
Large atomic number — not directly related to oxidation states.
Large atomic size — minor factor.
Comparable energies of 5f, 6d, 7s orbitals — correct reason.
Step 3: Conclusion.
33
PYQ 2026
medium
chemistryID: cbse-cla
Complete and balance the following equations:(1)
Official Solution
Correct Option:
(1)
In alkaline medium, permanganate reduces to manganate:
34
PYQ 2026
medium
chemistryID: cbse-cla
Which is the strongest oxidising agent among:
Official Solution
Correct Option:
(1)
An oxidising agent readily gains electrons (gets reduced). Among given ions:
Mn easily reduces to Mn
Mn has stable half-filled configuration
Due to stability of half-filled configuration, Mn strongly accepts electrons.
35
PYQ 2026
medium
chemistryID: cbse-cla
Out of Mn and Zn , which ion will be more paramagnetic and why? (Atomic numbers: Mn = 25, Zn = 30)
Official Solution
Correct Option:
(1)
Electronic configurations: Mn (Z = 25):
Zn (Z = 30):
Magnetic nature:
Mn : 5 unpaired electrons → highly paramagnetic
Zn : No unpaired electrons → diamagnetic
36
PYQ 2026
medium
chemistryID: cbse-cla
Why do transition metals show variable oxidation states?
Official Solution
Correct Option:
(1)
Transition metals have partially filled orbitals along with outer electrons. Both and electrons can participate in bonding. Reason:
Small energy difference between and orbitals
Electrons from both orbitals are used in bond formation
Hence, different numbers of electrons can be lost.
37
PYQ 2026
medium
chemistryID: cbse-cla
Assertion (A) : Actinoids show irregularities in their electronic configurations. Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The f-block elements (lanthanoids and actinoids) have complex electronic configurations because the energy levels of their outer shells are very close. Step 2: Detailed Explanation: In the actinoid series, the 5f, 6d, and 7s subshells are very close in energy (comparable energies). This leads to easy shifting of electrons between these orbitals, resulting in multiple possible configurations and irregularities in the sequence of filling. For example, the transition from Th to Pa or U doesn't follow a simple 5f-filling pattern. Hence, the assertion is true and the reason accurately explains the cause of these irregularities. Step 3: Final Answer: Code (A) is correct.
38
PYQ 2026
medium
chemistryID: cbse-cla
Which is the most stable oxidation state of Co and why?
Official Solution
Correct Option:
(1)
Cobalt commonly shows +2 and +3 oxidation states. Most stable state: +2 Reason:
Removal of two 4s electrons is easier
Co has relatively stable configuration
39
PYQ 2026
medium
chemistryID: cbse-cla
Why are E values for Mn and Zn more negative than expected?
Official Solution
Correct Option:
(1)
For Mn:
Mn has stable half-filled configuration
Reduction potential becomes more negative
For Zn:
Zn has fully filled configuration
Greater stability → more negative reduction potential
40
PYQ 2026
medium
chemistryID: cbse-cla
Why do transition metals form coloured compounds?
Official Solution
Correct Option:
(1)
Transition metal ions have partially filled d-orbitals.
d-orbitals split in ligand field
d–d electronic transitions absorb visible light
Complementary colour observed
41
PYQ 2026
medium
chemistryID: cbse-cla
What is meant by lanthanoid contraction?
Official Solution
Correct Option:
(1)
Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of lanthanoids with increasing atomic number. Cause:
Poor shielding by 4f electrons
Increased effective nuclear charge
42
PYQ 2026
medium
chemistryID: cbse-cla
Official Solution
Correct Option:
(1)
This is a redox reaction in acidic medium:
43
PYQ 2026
easy
chemistryID: cbse-cla
What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Acidic permanganate solutions are thermodynamically unstable and tend to decompose over time. Step 2: Detailed Explanation: When left to stand, acidic undergoes a slow decomposition reaction, liberating oxygen gas. This process is often catalyzed by sunlight or the presence of manganese(II) ions ( ) produced during the reaction itself (autocatalysis). Equation:
This reaction is a redox decomposition reaction. In laboratory settings, it is called the self-decomposition of permanganate. Step 3: Final Answer: The solution decomposes to liberate oxygen. It is a decomposition reaction.
44
PYQ 2026
medium
chemistryID: cbse-cla
Why Cr has higher melting point than Mn ?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Melting point in transition metals depends on the strength of metallic bonds, which is directly related to the number of unpaired d-electrons available for delocalization. Step 2: Detailed Explanation: 1. Chromium ( ): It has a configuration. It has 6 unpaired electrons that participate effectively in metallic bonding, leading to strong interatomic forces and a high melting point. 2. Manganese ( ): It has a configuration. Although it has 5 unpaired electrons in the 3d subshell, these electrons are held very tightly in the stable, symmetrical half-filled configuration. As a result, they participate less in metallic bonding, leading to weaker interatomic forces and a lower melting point compared to Chromium. Step 3: Final Answer: has more effectively involved delocalized electrons than , leading to higher melting point.
45
PYQ 2026
medium
chemistryID: cbse-cla
Why transition metals act as good catalyst ?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Catalysts work by providing an alternative reaction pathway with lower activation energy. Step 2: Detailed Explanation: Transition metals are excellent catalysts due to: 1. Variable Oxidation States: They can lose or gain electrons easily, allowing them to form unstable intermediate compounds with reactants, which later decompose to yield products and regenerate the catalyst. 2. Adsorption: They provide a large surface area for reactant molecules to adsorb. Adsorption on the metal surface increases the local concentration of reactants and weakens the internal bonds of the reactants, facilitating the reaction. 3. Complex Formation: Their ability to form complex compounds also helps in bringing reactant molecules together. Step 3: Final Answer: The catalytic property is primarily due to variable valency and effective surface area.
46
PYQ 2026
medium
chemistryID: cbse-cla
Name a member of the lanthanoid series (I) which exhibits +4 oxidation state (II) which exhibits +2 oxidation state.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The most common oxidation state for lanthanoids is +3. However, some elements exhibit +2 or +4 states to achieve stable configurations like , , or . Step 2: Detailed Explanation: (I) +4 Oxidation State: Cerium ( , ) has the configuration . By losing 4 electrons, it achieves the stable noble gas configuration of Xenon ( ). Thus, is well-known. (II) +2 Oxidation State: Europium ( , ) has the configuration . By losing 2 electrons, it achieves a stable half-filled configuration. Thus, is a common ion. Step 3: Final Answer: (I) Cerium exhibits +4. (II) Europium exhibits +2.
47
PYQ 2026
medium
chemistryID: cbse-cla
Write the ionic equations for the oxidising action of potassium permanganate for its reaction with in both acidic and alkaline solutions.
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: Potassium permanganate ( ) is a powerful oxidizing agent. In acidic medium, it is reduced to (+2), while in neutral or alkaline medium, it is reduced to (+4). Step 2: Detailed Explanation: 1. In Acidic Solution: Permanganate ion ( ) oxidizes iodide ( ) to free iodine ( ). Reduction half-reaction: Oxidation half-reaction: Overall balanced ionic equation:
2. In Alkaline Solution: Permanganate ion oxidizes iodide ( ) specifically to iodate ion ( ). Reduction half-reaction: Oxidation half-reaction: Overall balanced ionic equation:
Step 3: Final Answer: The reaction products of iodide oxidation depend heavily on the pH: in acid and in alkali.
48
PYQ 2026
medium
chemistryID: cbse-cla
From the given data of values, answer the following questions : (I) Why show irregular trend in the above values ? (II) Why is value exceptionally positive ? (III) Why value is highly negative ?
Official Solution
Correct Option:
(1)
Step 1: Understanding the Concept: The standard electrode potential ( ) of a metal ion is a measure of its tendency to be reduced. It is determined by three main energy factors: enthalpy of sublimation ( ), ionization enthalpy ( ), and enthalpy of hydration ( ). Step 2: Detailed Explanation: (I) Irregular Trend: The irregularity in the values across the 3d series is due to the non-regular variation in the sublimation enthalpies and the sum of the first and second ionization enthalpies. These properties do not change smoothly across the series, causing the observed fluctuations in electrode potentials. (II) Copper's Positive Value: Copper is the only metal in the 3d series with a positive value ( V). This is because the high energy required to transform to (high sublimation and high sum of and ) is not compensated by the relatively low hydration enthalpy of the ion. (III) Manganese's Highly Negative Value: The for is highly negative because of the extra stability associated with the half-filled electronic configuration in the ion. This stability makes the formation of from energetically favourable. Step 3: Final Answer: The trends are explained by the balance of thermodynamic parameters (sublimation, ionization, and hydration) and electronic stability ( configuration).
About D And F Block Elements - CBSE-CLASS-XII
D And F Block Elements is a vital chapter for CBSE-CLASS-XII aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
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Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within D And F Block Elements carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
