CUET UG Chemistry Booster Test - 3 Introduction & Classification
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QUESTION 1 OF 20
Arrange the following subshells in decreasing order of the principal quantum number (n) corresponding to the transition series they define:
1. 3d orbitals
2. 6d orbitals
3. 4d orbitals
4. 5d orbitals
QUESTION 2 OF 20
Identify the correct statements regarding the electronic configuration of the d-block.
Statements:
1. Half-filled and completely filled sets of orbitals are relatively more stable.
2. Cr has a 3d⁴ 4s² configuration to minimize repulsion.
3. Cu has a 3d¹⁰ 4s¹ configuration due to the stability of the fully filled d-orbital.
4. The energy gap between (n−1)d and ns orbitals is very large.
QUESTION 3 OF 20
Match the following orbital fillings with their corresponding series.
| List I | List II |
|---|---|
| 1. 3d progressive filling | a. Actinoids |
| 2. 4f progressive filling | b. Third transition series |
| 3. 5f progressive filling | c. First transition series |
| 4. 5d progressive filling | d. Lanthanoids |
QUESTION 4 OF 20
Analytically speaking, why do transition and inner transition metals exhibit drastically different properties such as coloured ions, paramagnetism and multiple oxidation states compared to main group elements?
QUESTION 5 OF 20
Silver atom has a completely filled d orbital (4d¹⁰) in its ground state. Under the IUPAC definition, identify the state that justifies silver's classification as a transition metal.
QUESTION 6 OF 20
Which fundamental orbital interactions account for exceptions like Cr and Cu deviating from the idealized (n−1)d¹⁻¹⁰ ns² configuration?
QUESTION 7 OF 20
Identify the correct statements concerning the inner transition elements.
Statements:
1. They comprise strictly 14 elements in the 4f series and 14 elements in the 5f series.
2. Lanthanum itself possesses 4f electrons in its ground state.
3. The actinoids include Thorium (Th) through Lawrencium (Lr).
4. They are placed in the main block of the periodic table between s and p blocks.
QUESTION 8 OF 20
How many elements are formally classified under the lanthanoid series alone?
QUESTION 9 OF 20
While Scandium (Z = 21) is a transition element, Zinc (Z = 30) is not. What specific reasoning based on electron configuration justifies this?
QUESTION 10 OF 20
Match the following transition elements with their respective series.
| List I | List II |
|---|---|
| 1. Yttrium (Y) | a. 6d series |
| 2. Hafnium (Hf) | b. 3d series |
| 3. Copernicium (Cn) | c. 4d series |
| 4. Scandium (Sc) | d. 5d series |
QUESTION 11 OF 20
Based on the passage, the 5d series shows a discontinuity between Lanthanum (La) and Hafnium (Hf). What causes this discontinuity in the sequence of transition metals?
QUESTION 12 OF 20
Which element signifies the completion of the 6d transition series according to the passage provided?
QUESTION 13 OF 20
Identify the correct statements regarding Zn, Cd, Hg and Cn.
Statements:
1. Their general outer orbital formula is (n−1)d¹⁰ ns².
2. Their d-orbitals are completely filled in the ground state.
3. Their d-orbitals become partially empty in their common oxidation states.
4. They are firmly categorized as true transition elements by IUPAC.
QUESTION 14 OF 20
The exclusion of Group 12 elements from the transition metal definition relies on the stability of the d¹⁰ configuration. Why doesn't ionization change their transition status?
QUESTION 15 OF 20
Arrange the major blocks in order of increasing electronegativity/non-metallic character historically linked by the transition elements.
1. p-block elements
2. s-block elements
3. d-block (transition) elements
QUESTION 16 OF 20
Identify the fundamental concept that allowed the "usual theory of valence," initially developed for s- and p-block elements, to be successfully applied to transition elements.
QUESTION 17 OF 20
Name the d-block metal specifically highlighted as a critical industrial metal from the first transition series (3d), often alloyed into steel.
QUESTION 18 OF 20
Among the following transition metals, which one is explicitly categorized as an industrial metal rather than a precious metal?
QUESTION 19 OF 20
Match the following applications with the corresponding elements.
| List I | List II |
|---|---|
| 1. Precious jewellery / Currency historically | a. Fe, Ti |
| 2. Nuclear energy source | b. Ag, Au, Pt |
| 3. Excluded end-member elements | c. U, Th, Pa |
| 4. Structural / Industrial civilization base | d. Zn, Cd, Hg |
QUESTION 20 OF 20
Assess the following statements regarding the applications of f-block elements.
Statements:
1. Thorium (Th) and Uranium (U) belong to the actinoid series.
2. These elements are excellent sources of nuclear energy.
3. They are historically known for making ancient civilization tools like bronze.
4. They undergo 4f orbital filling.
Test Complete!
Answer Review
1 Arrange the following subshells in decreasing order of the principal quantum number (n) corresponding to the transition series they define:
1. 3d orbitals
2. 6d orbitals
3. 4d orbitals
4. 5d orbitals
�� Principal quantum number increases from 3d to 6d. �� Higher d-series correspond to larger n values. �� 6d has the highest principal quantum number.
The principal quantum number (n) indicates the main energy level in which an orbital exists. For d-orbitals, the notation directly reveals the value of n. Thus, 3d orbitals belong to the third principal shell, 4d orbitals belong to the fourth shell, 5d orbitals belong to the fifth shell and 6d orbitals belong to the sixth shell. When arranging in decreasing order of principal quantum number, the orbital with the highest value of n must be placed first. Therefore, the sequence becomes 6d > 5d > 4d > 3d. These orbitals define the fourth, third, second and first transition series respectively. Understanding the relationship between orbital notation and principal quantum number is essential for classifying transition series in the periodic table.
- �� Option A → Gives increasing order instead of decreasing order.
- �� Option B → Places 5d before 6d incorrectly.
- �� Option C → Places 4d before 5d incorrectly.
NCERT Recall
- Application
- Recall the meaning of the number preceding the orbital symbol.
- Final Logic
- 6d > 5d > 4d > 3d, therefore Option B is correct.
"6-5-4-3: Highest to Lowest d-Series"
2 Identify the correct statements regarding the electronic configuration of the d-block.
Statements:
1. Half-filled and completely filled sets of orbitals are relatively more stable.
2. Cr has a 3d⁴ 4s² configuration to minimize repulsion.
3. Cu has a 3d¹⁰ 4s¹ configuration due to the stability of the fully filled d-orbital.
4. The energy gap between (n−1)d and ns orbitals is very large.
�� Half-filled and fully filled orbitals possess extra stability. �� Chromium and copper are important exceptions. �� The energy difference between d and s orbitals is small.
The electronic configurations of chromium and copper are important exceptions to the expected Aufbau pattern. Half-filled and completely filled subshells possess extra stability because of symmetrical electron distribution and maximum exchange energy. Chromium achieves the stable configuration 3d⁵ 4s¹ instead of 3d⁴ 4s², while copper adopts 3d¹⁰ 4s¹ instead of 3d⁹ 4s². Therefore, Statements 1 and 3 are correct. Statement 2 is incorrect because chromium does not remain in the 3d⁴ 4s² configuration. Statement 4 is also incorrect because the energy difference between (n−1)d and ns orbitals is very small, which allows electron transfer between these orbitals. This small energy difference is responsible for many anomalous configurations observed among transition elements.
- �� Option A → Statement 2 is incorrect because chromium has 3d⁵ 4s¹ configuration.
- �� Option C → Statements 2 and 4 are both incorrect.
- �� Option D → Statement 4 is incorrect because the energy gap is small.
Concept Application
- Application
- Apply the concepts of exchange energy and stability of half-filled and fully filled orbitals.
- Final Logic
- Statements 1 and 3 are correct; Statements 2 and 4 are incorrect.
"Cr = Half Full, Cu = Fully Full"
3 Match the following orbital fillings with their corresponding series.
| List I | List II |
|---|---|
| 1. 3d progressive filling | a. Actinoids |
| 2. 4f progressive filling | b. Third transition series |
| 3. 5f progressive filling | c. First transition series |
| 4. 5d progressive filling | d. Lanthanoids |
�� 3d filling forms the first transition series. �� 4f filling forms lanthanoids. �� 5f filling forms actinoids. �� 5d filling forms the third transition series.
The classification of transition and inner transition elements is based on the orbital being progressively filled. The first transition series corresponds to the filling of 3d orbitals. The lanthanoid series involves progressive filling of 4f orbitals, while the actinoid series involves progressive filling of 5f orbitals. The third transition series corresponds to the filling of 5d orbitals. These classifications arise directly from the electronic configurations of the elements. The arrangement helps chemists understand periodic trends, oxidation states and chemical behaviour. Therefore, the correct matching is 3d → First Transition Series, 4f → Lanthanoids, 5f → Actinoids and 5d → Third Transition Series.
- �� Option B → All major orbital-series relationships are mismatched.
- �� Option C → 4f and 5f series are interchanged incorrectly.
- �� Option D → 3d and 5d assignments are incorrect.
NCERT Recall
- Application
- Recall the orbital associated with each transition and inner transition series.
- Final Logic
- 3d → First Transition, 4f → Lanthanoids, 5f → Actinoids, 5d → Third Transition.
"3d-First, 4f-Lanth, 5f-Act, 5d-Third"
4 Analytically speaking, why do transition and inner transition metals exhibit drastically different properties such as coloured ions, paramagnetism and multiple oxidation states compared to main group elements?
�� Partially filled d and f orbitals influence chemical behaviour. �� These orbitals interact strongly with surrounding species. �� Unique properties arise from these interactions.
Transition and inner transition elements possess partially filled d or f orbitals that extend sufficiently towards the outer region of the atom. Because these orbitals are exposed to surrounding atoms, ligands and external fields, they strongly influence the physical and chemical properties of the elements. This leads to characteristic features such as coloured ions, variable oxidation states, magnetic behaviour and complex formation. The availability of d and f electrons allows multiple electronic transitions and different bonding possibilities. Main group elements generally lack such partially filled d or f subshells and therefore do not exhibit these properties to the same extent. Thus, the unique behaviour of transition and inner transition elements is directly related to the presence and accessibility of their partially filled d or f orbitals.
- �� Option A → This does not explain the characteristic properties of transition elements.
- �� Option B → Their properties are not due to unusually low nuclear charge.
- �� Option C → Many transition elements possess completely filled inner shells.
Concept Application
- Application
- Relate the properties of transition metals to the behaviour of d and f electrons.
- Final Logic
- Partially filled d and f orbitals are responsible for their distinctive chemical properties.
"d and f = Different Features"
5 Silver atom has a completely filled d orbital (4d¹⁰) in its ground state. Under the IUPAC definition, identify the state that justifies silver's classification as a transition metal.
�� IUPAC definition includes atoms or ions. �� Silver atom has a filled 4d subshell. �� Ag²⁺ possesses an incompletely filled d-subshell.
According to the IUPAC definition, a transition element is one whose atom or at least one of its common oxidation states possesses an incompletely filled d-subshell. Silver has the ground-state configuration [Kr] 4d¹⁰ 5s¹. When silver forms Ag⁺, the configuration becomes [Kr] 4d¹⁰, which still contains a completely filled d-subshell. However, in the Ag²⁺ oxidation state, one electron is removed from the 4d subshell, resulting in a 4d⁹ configuration. This incompletely filled d-subshell satisfies the IUPAC criterion for classification as a transition element. Therefore, silver is regarded as a transition metal because it can form an ion with a partially filled d-orbital.
- �� Option A → Neutral silver has a filled d-subshell.
- �� Option B → Ag⁺ also possesses a filled 4d¹⁰ configuration.
- �� Option D → Anionic states are not the basis of IUPAC classification.
Concept Application
- Application
- Apply the IUPAC definition to the electronic configurations of silver ions.
- Final Logic
- Ag²⁺ has an incompletely filled 4d subshell and therefore satisfies the definition.
"Ag²⁺ Makes Silver Transition"
6 Which fundamental orbital interactions account for exceptions like Cr and Cu deviating from the idealized (n−1)d¹⁻¹⁰ ns² configuration?
�� d and s orbitals have very similar energies. �� Half-filled and fully filled subshells possess extra stability. �� Exchange energy contributes to stabilization.
The anomalous electronic configurations of chromium and copper arise because the energy difference between the (n−1)d and ns orbitals is very small. As a result, the transfer of one electron from the ns orbital to the d orbital requires very little energy. This transfer leads to the formation of particularly stable half-filled (d⁵) or completely filled (d¹⁰) subshells. Chromium adopts the configuration 3d⁵4s¹ instead of 3d⁴4s², while copper adopts 3d¹⁰4s¹ instead of 3d⁹4s². The extra stability is attributed to symmetrical electron distribution and increased exchange energy. These factors outweigh the small energy required for electron rearrangement. Thus, the combined effect of a small energy gap and exchange energy stabilization explains the deviations observed in Cr and Cu.
- �� Option A → The energy barrier is not high; it is actually very small.
- �� Option C → Shielding by ns electrons does not explain these exceptions.
- �� Option D → Transition metals are not highly electronegative, and electronegativity is not the cause.
Concept Application
- Application
- Apply the concepts of exchange energy and orbital energy differences to explain anomalous configurations.
- Final Logic
- Small d–s energy difference plus stability of d⁵ and d¹⁰ configurations causes the exceptions.
"Small Gap, Stable Half and Full"
7 Identify the correct statements concerning the inner transition elements.
Statements:
1. They comprise strictly 14 elements in the 4f series and 14 elements in the 5f series.
2. Lanthanum itself possesses 4f electrons in its ground state.
3. The actinoids include Thorium (Th) through Lawrencium (Lr).
4. They are placed in the main block of the periodic table between s and p blocks.
�� Lanthanoids involve 4f filling. �� Actinoids involve 5f filling. �� Inner transition elements are shown separately.
The inner transition elements consist of two series: the lanthanoids and the actinoids. The lanthanoid series involves the progressive filling of 4f orbitals and contains fourteen elements from Cerium (Ce) to Lutetium (Lu). Similarly, the actinoid series involves the progressive filling of 5f orbitals and contains fourteen elements from Thorium (Th) to Lawrencium (Lr). Therefore, Statements 1 and 3 are correct. Statement 2 is incorrect because Lanthanum has the electronic configuration [Xe]5d¹6s² and does not possess a 4f electron in its ground state. Statement 4 is incorrect because inner transition elements are placed separately at the bottom of the periodic table rather than in the main body. This separate placement helps maintain the compact structure of the periodic table.
- �� Option B → Statement 2 is incorrect because La has no 4f electron.
- �� Option C → Statements 2 and 4 are incorrect.
- �� Option D → Statement 4 is incorrect because inner transition elements are not placed in the main block.
NCERT Recall
- Application
- Recall the composition and placement of the lanthanoid and actinoid series.
- Final Logic
- Only Statements 1 and 3 are correct.
"14 in 4f, 14 in 5f"
8 How many elements are formally classified under the lanthanoid series alone?
�� Lanthanoids involve filling of 4f orbitals. �� Fourteen electrons can occupy the 4f subshell. �� Therefore, fourteen elements belong to the series.
The lanthanoid series corresponds to the progressive filling of the 4f subshell. Since an f-subshell contains seven orbitals and each orbital can accommodate two electrons, the maximum number of electrons that can occupy the 4f subshell is fourteen. Consequently, fourteen elements are formally classified as lanthanoids, extending from Cerium (Ce) to Lutetium (Lu). These elements exhibit similar chemical properties due to their closely related electronic configurations. They are placed in a separate row below the main periodic table and collectively constitute the first series of inner transition elements. Therefore, the lanthanoid series formally contains fourteen elements.
- �� Option A → Ten corresponds to the capacity of a d-subshell.
- �� Option C → The formal lanthanoid series contains fourteen elements, not fifteen.
- �� Option D → Eighteen exceeds the capacity of the 4f subshell.
Concept Application
- Application
- Use the electron capacity of the f-subshell to determine the number of lanthanoids.
- Final Logic
- 7 f-orbitals × 2 electrons each = 14 elements.
"f = 14"
9 While Scandium (Z = 21) is a transition element, Zinc (Z = 30) is not. What specific reasoning based on electron configuration justifies this?
�� Sc possesses a partially filled d-subshell. �� Zn possesses a completely filled d-subshell. �� IUPAC definition depends on d-subshell occupancy.
According to the IUPAC definition, a transition element is one whose atom or at least one of its common oxidation states contains an incompletely filled d-subshell. Scandium has the electronic configuration [Ar]3d¹4s² and therefore possesses a partially filled d-orbital in its ground state. This satisfies the definition of a transition element. Zinc, however, has the configuration [Ar]3d¹⁰4s². When it forms its common Zn²⁺ ion, the configuration becomes [Ar]3d¹⁰, which still contains a completely filled d-subshell. Since neither the atom nor its common ion contains an incompletely filled d-subshell, zinc does not qualify as a transition element under the strict IUPAC definition. This difference in electronic configuration explains the contrasting classifications of Sc and Zn.
- �� Option B → Sc commonly forms +3 ions, not primarily +2 ions.
- �� Option C → Zinc does not have electrons in the 4p orbital in its ground state.
- �� Option D → Radioactivity is unrelated to the classification.
Concept Application
- Application
- Apply the IUPAC definition using the electronic configurations of Sc and Zn.
- Final Logic
- Sc contains an incomplete d-subshell, whereas Zn has a complete d¹⁰ configuration.
"Sc Starts d, Zn Finishes d"
10 Match the following transition elements with their respective series.
| List I | List II |
|---|---|
| 1. Yttrium (Y) | a. 6d series |
| 2. Hafnium (Hf) | b. 3d series |
| 3. Copernicium (Cn) | c. 4d series |
| 4. Scandium (Sc) | d. 5d series |
�� Sc begins the 3d series. �� Y begins the 4d series. �� Hf belongs to the 5d series. �� Cn belongs to the 6d series.
Transition elements are grouped into different series according to the d-orbital being filled. Scandium is the first element of the 3d transition series. Yttrium occupies a similar position in the second transition series and belongs to the 4d series. Hafnium is part of the 5d transition series that follows the lanthanoid elements. Copernicium belongs to the 6d transition series, representing one of the heaviest known transition elements. Correctly associating these elements with their respective series requires understanding their positions in the periodic table and the orbital filling pattern that characterizes each transition series. Therefore, the correct matching is Y → 4d, Hf → 5d, Cn → 6d and Sc → 3d.
- �� Option A → Y and Cn are mismatched.
- �� Option B → All series assignments are shifted incorrectly.
- �� Option C → Hf and Cn are assigned to incorrect series.
NCERT Recall
- Application
- Recall the representative elements associated with each transition series.
- Final Logic
- Sc → 3d, Y → 4d, Hf → 5d, Cn → 6d.
"Sc-Y-Hf-Cn = 3d-4d-5d-6d"
11
Based on the passage, the 5d series shows a discontinuity between Lanthanum (La) and Hafnium (Hf). What causes this discontinuity in the sequence of transition metals?
�� The 5d series is interrupted by the lanthanoids. �� Lanthanoids involve progressive filling of 4f orbitals. �� Hf appears after completion of the 4f series.
The third transition series extends from Lanthanum (La) to Mercury (Hg). However, after Lanthanum, the sequence is interrupted by fourteen lanthanoid elements in which the 4f orbitals are progressively filled. These elements occupy atomic numbers 58 to 71 and are collectively known as the lanthanoid series. After completion of the 4f filling, the 5d series resumes with Hafnium (Hf). This interruption creates an apparent discontinuity between La and Hf in the periodic table. The placement of lanthanoids in a separate row helps maintain the compact structure of the periodic table. Therefore, the discontinuity is caused by the insertion of the 4f-series elements between La and Hf.
- �� Option A → s-block elements do not occur between La and Hf.
- �� Option C → Radioactivity is not responsible for the discontinuity.
- �� Option D → The interruption is due to 4f filling, not 6p filling.
NCERT Recall
- Application
- Recall the position of lanthanoids relative to the 5d transition series.
- Final Logic
- La is followed by fourteen lanthanoids before the series continues with Hf.
"La → Lanthanoids → Hf"
12
Which element signifies the completion of the 6d transition series according to the passage provided?
�� The 6d series begins with Ac and continues from Rf. �� Copernicium is the last element of the series. �� It represents completion of the 6d transition row.
The fourth transition series is known as the 6d series. It begins with Actinium (Ac) and, after the interruption caused by the actinoid series, continues from Rutherfordium (Rf) to Copernicium (Cn). Copernicium is the final element of this transition series and therefore signifies its completion. Like the earlier transition series, the 6d series is characterized by the progressive filling of d-orbitals. The identification of Copernicium as the terminal element of the 6d series helps classify the modern periodic table into four transition series. Therefore, Copernicium correctly represents the end of the 6d transition series.
- �� Option A → Mercury is the last element of the 5d series.
- �� Option B → Actinium marks the beginning of the 6d series.
- �� Option D → Rutherfordium is an early member of the 6d series, not the last.
NCERT Recall
- Application
- Recall the boundary elements of the fourth transition series.
- Final Logic
- The 6d series ends with Copernicium (Cn).
"Ac Starts, Cn Completes"
13 Identify the correct statements regarding Zn, Cd, Hg and Cn.
Statements:
1. Their general outer orbital formula is (n−1)d¹⁰ ns².
2. Their d-orbitals are completely filled in the ground state.
3. Their d-orbitals become partially empty in their common oxidation states.
4. They are firmly categorized as true transition elements by IUPAC.
�� Group 12 elements possess d¹⁰ configurations. �� Their d-subshell remains filled in common oxidation states. �� They are excluded from the strict IUPAC definition.
Zinc, Cadmium, Mercury and Copernicium belong to Group 12 of the periodic table. Their general outer electronic configuration is (n−1)d¹⁰ns². In the ground state, all these elements possess completely filled d-orbitals, making Statements 1 and 2 correct. When these elements form their most common oxidation state of +2, the two ns electrons are removed while the d¹⁰ configuration remains intact. Consequently, they do not possess partially filled d-orbitals in their common oxidation states. According to the IUPAC definition, a transition element must have an incomplete d-subshell in either its atom or one of its common oxidation states. Therefore, these elements are generally excluded from the strict definition of transition elements.
- �� Option B → Statements 3 and 4 are incorrect.
- �� Option C → Statement 3 is incorrect because the d-subshell remains filled.
- �� Option D → Statements 3 and 4 are incorrect.
Concept Application
- Application
- Apply the IUPAC definition to Group 12 electronic configurations.
- Final Logic
- Only Statements 1 and 2 are correct.
"Group 12 = d¹⁰ Always"
14 The exclusion of Group 12 elements from the transition metal definition relies on the stability of the d¹⁰ configuration. Why doesn't ionization change their transition status?
�� ns electrons are removed first during ionization. �� The d¹⁰ core remains unchanged. �� Common ions therefore retain filled d-subshells.
When Group 12 elements such as Zn, Cd and Hg undergo ionization, electrons are removed first from the outermost ns orbital rather than from the inner d-subshell. As a result, the common +2 oxidation state is formed by losing the two ns electrons while retaining the completely filled (n−1)d¹⁰ configuration. Because the d-subshell remains fully occupied, these ions do not possess the incompletely filled d-orbitals required by the IUPAC definition of transition elements. Consequently, ionization does not alter their classification. The stability of the d¹⁰ configuration persists in their common oxidation states and prevents them from being considered true transition metals under the strict definition.
- �� Option A → d-electrons are not removed before ns electrons.
- �� Option C → These elements can form ionic compounds.
- �� Option D → The energy required is high but not infinite.
Concept Application
- Application
- Use the order of electron removal during ionization.
- Final Logic
- Removal of ns electrons leaves the d¹⁰ core unchanged.
"ns Leaves, d¹⁰ Stays"
15 Arrange the major blocks in order of increasing electronegativity/non-metallic character historically linked by the transition elements.
1. p-block elements
2. s-block elements
3. d-block (transition) elements
�� s-block elements are most electropositive. �� d-block elements show intermediate properties. �� p-block elements are relatively more electronegative.
Historically, transition elements were named because they provide a transition in properties between the highly electropositive s-block metals and the more electronegative p-block elements. The s-block elements readily lose electrons and therefore exhibit strong metallic character. The d-block elements occupy an intermediate position and display properties between these two extremes. The p-block contains many non-metals and metalloids with comparatively higher electronegativity and non-metallic character. Therefore, when arranged in increasing order of electronegativity or non-metallic character, the correct sequence is s-block, followed by d-block, followed by p-block. This historical interpretation explains the origin of the term "transition elements."
- �� Option B → Reverses the correct trend.
- �� Option C → Places d-block before s-block incorrectly.
- �� Option D → Places p-block before d-block incorrectly.
Logical Analysis
- Application
- Compare the general metallic and non-metallic character of the blocks.
- Final Logic
- s-block < d-block < p-block in electronegativity and non-metallic character.
"s → d → p"
16 Identify the fundamental concept that allowed the "usual theory of valence," initially developed for s- and p-block elements, to be successfully applied to transition elements.
�� d and s orbitals have comparable energies. �� Both orbital types participate in bonding. �� Variable oxidation states can be explained systematically.
The chemistry of transition elements can be understood using valence concepts because the energies of the (n−1)d and ns orbitals are very close to one another. As a result, electrons from both sets of orbitals can participate in bond formation and ionization. This explains the occurrence of multiple oxidation states, which is a characteristic feature of transition elements. Unlike representative elements, where valence is often fixed, transition metals can exhibit several oxidation states due to the availability of both d and s electrons. Nevertheless, these oxidation states follow predictable patterns that can be interpreted using the principles of valence and electronic configuration. Therefore, the comparable energies and overlap of the (n−1)d and ns orbitals provide the fundamental basis for applying valence concepts to transition elements.
- �� Option A → Transition-metal bonding is not explained by exclusively p-orbital hybridization.
- �� Option B → The anti-penultimate shell is not the basis of valence theory.
- �� Option D → Transition metals are chemically active and not inert.
Concept Application
- Application
- Relate oxidation states to the participation of both d and s electrons.
- Final Logic
- Similar energies of (n−1)d and ns orbitals enable predictable valence behaviour.
"d + s = Variable Valence"
17 Name the d-block metal specifically highlighted as a critical industrial metal from the first transition series (3d), often alloyed into steel.
�� Iron belongs to the first transition series. �� It is the primary constituent of steel. �� It has immense industrial importance.
Iron is one of the most important transition metals in human civilization and modern industry. It belongs to the first transition series (3d series) and serves as the principal component of steel, one of the most widely used engineering materials. Iron and its alloys are used extensively in construction, transportation, machinery, infrastructure and manufacturing industries. The strength, durability and versatility of steel have made iron indispensable for industrial development. NCERT specifically highlights iron as a major industrial metal among transition elements. Its abundance and favourable mechanical properties have contributed significantly to technological progress and economic growth worldwide.
- �� Option A → Platinum is a precious metal rather than a major structural industrial metal.
- �� Option C → Cadmium has limited industrial structural applications.
- �� Option D → Gold is primarily valued as a precious metal.
NCERT Recall
- Application
- Recall the industrially important transition metals emphasized in NCERT.
- Final Logic
- Iron is the key transition metal used extensively in steel production.
"Iron = Industry"
18 Among the following transition metals, which one is explicitly categorized as an industrial metal rather than a precious metal?
�� Titanium is widely used in industry. �� Silver, gold and platinum are precious metals. �� Titanium possesses high strength and corrosion resistance.
Titanium is classified as an industrial transition metal because of its extensive technological and engineering applications. It is valued for its high strength-to-weight ratio, excellent corrosion resistance and durability. Titanium is widely used in aerospace structures, marine equipment, chemical processing industries and medical implants. In contrast, silver, gold and platinum are commonly categorized as precious metals because of their rarity, high economic value and use in jewellery, currency and specialized applications. NCERT identifies titanium as an important industrial metal among transition elements. Therefore, Titanium is correctly categorized as an industrial metal rather than a precious metal.
- �� Option A → Silver is categorized as a precious metal.
- �� Option B → Gold is categorized as a precious metal.
- �� Option D → Platinum is categorized as a precious metal.
NCERT Recall
- Application
- Recall the distinction between industrial and precious transition metals.
- Final Logic
- Titanium is the industrial metal among the given options.
"Ti = Technology and Industry"
19 Match the following applications with the corresponding elements.
| List I | List II |
|---|---|
| 1. Precious jewellery / Currency historically | a. Fe, Ti |
| 2. Nuclear energy source | b. Ag, Au, Pt |
| 3. Excluded end-member elements | c. U, Th, Pa |
| 4. Structural / Industrial civilization base | d. Zn, Cd, Hg |
�� Precious metals include Ag, Au and Pt. �� U, Th and Pa are nuclear-energy elements. �� Zn, Cd and Hg are excluded end members. �� Fe and Ti are important industrial metals.
Silver, gold and platinum have historically been used for jewellery, ornaments and currency due to their lustre, rarity and resistance to corrosion. Uranium, Thorium and Protactinium belong to the actinoid series and are important sources of nuclear energy. Zinc, Cadmium and Mercury are end members of their respective transition series and are generally excluded from the strict IUPAC definition of transition elements because of their filled d¹⁰ configurations. Iron and Titanium are major industrial metals that contribute significantly to infrastructure, engineering and technological development. Matching these applications with their corresponding elements gives the sequence 1-b, 2-c, 3-d and 4-a.
- �� Option A → All major application-element pairings are incorrect.
- �� Option C → Nuclear and industrial classifications are mismatched.
- �� Option D → Precious metals and excluded elements are incorrectly assigned.
Concept Application
- Application
- Associate each group of elements with its characteristic application.
- Final Logic
- Precious → Ag/Au/Pt, Nuclear → U/Th/Pa, Excluded → Zn/Cd/Hg, Industrial → Fe/Ti.
"Jewels–Nuclear–Excluded–Industry"
20 Assess the following statements regarding the applications of f-block elements.
Statements:
1. Thorium (Th) and Uranium (U) belong to the actinoid series.
2. These elements are excellent sources of nuclear energy.
3. They are historically known for making ancient civilization tools like bronze.
4. They undergo 4f orbital filling.
�� Th and U are actinoids. �� Actinoids involve 5f filling. �� They are important nuclear fuels.
Thorium and Uranium belong to the actinoid series, which consists of inner transition elements characterized by the progressive filling of 5f orbitals. These elements are well known for their radioactive nature and their ability to release enormous amounts of energy through nuclear reactions. Consequently, they are widely used as fuels in nuclear reactors and are regarded as excellent sources of nuclear energy. Statement 3 is incorrect because bronze was produced primarily from copper and tin rather than actinoid elements. Statement 4 is also incorrect because actinoids involve the filling of 5f orbitals, not 4f orbitals. Therefore, only Statements 1 and 2 are correct.
- �� Option A → Statement 4 is incorrect because actinoids involve 5f filling.
- �� Option B → Statement 3 is incorrect because bronze is not associated with actinoids.
- �� Option C → Statements 3 and 4 are both incorrect.
NCERT Recall
- Application
- Recall the classification and applications of actinoid elements.
- Final Logic
- Only Statements 1 and 2 correctly describe Thorium and Uranium.
"Th-U = Nuclear 5f"
