CUET UG Chemistry Booster Test - 2 General Properties
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QUESTION 1 OF 20
Which of the following elements has a completely filled d¹⁰ configuration in both its ground state and its common oxidation states, thereby not being regarded as a proper transition metal?
QUESTION 2 OF 20
The transition metals retain their distinct metallic conductivity even when:
QUESTION 3 OF 20
Identify the correct statements regarding the mechanical traits of transition metals.
Statements:
1. Interstitial compounds of transition metals are very soft.
2. Some borides of transition metals approach diamond in hardness.
3. The transition metals (excluding Zn, Cd and Hg) are very hard.
4. Alloys formed by transition metals are generally soft and melt easily.
QUESTION 4 OF 20
Malleability is a characteristic of transition metals. What structural feature allows them to exhibit typical metallic properties like malleability and ductility?
QUESTION 5 OF 20
According to the passage, the maximum melting point in a transition series is generally found when the d-orbital configuration is:
QUESTION 6 OF 20
The passage relates the strong interatomic interaction and high boiling/melting points directly to:
QUESTION 7 OF 20
Match the 3d transition metals with their metallic radii (pm) as per NCERT data.
| List I | List II |
|---|---|
| 1. Sc | a. 126 pm |
| 2. V | b. 128 pm |
| 3. Fe | c. 135 pm |
| 4. Cu | d. 164 pm |
QUESTION 8 OF 20
Arrange the following M²⁺ ions in decreasing order of their ionic radii.
1. V²⁺ (79 pm)
2. Cr²⁺ (82 pm)
3. Ni²⁺ (70 pm)
4. Cu²⁺ (73 pm)
QUESTION 9 OF 20
When moving from Scandium to Zinc in the 3d series, the atomic radii decrease less rapidly than initially expected. This is primarily because:
QUESTION 10 OF 20
Comparing the shielding effects within transition and inner-transition metals, which of the following is correct?
QUESTION 11 OF 20
Identify the specific phenomenon responsible for the nearly identical radii of second- and third-transition-series elements such as Zirconium and Hafnium.
QUESTION 12 OF 20
Because of the chemical and physical similarities resulting from lanthanoid contraction, which procedural consequence heavily impacts metals like Zirconium and Hafnium?
QUESTION 13 OF 20
From Titanium to Copper, why does the density show a significant general increase?
QUESTION 14 OF 20
According to the properties of the 3d series, which two metals share the highest identical density (8.9 g cm⁻³)?
QUESTION 15 OF 20
In the first series of transition elements (Sc to Zn), which element has the exceptionally lowest enthalpy of atomisation (126 kJ mol⁻¹)?
QUESTION 16 OF 20
Heavy transition metals from the second and third series exhibit more frequent metal-metal bonding in compounds compared to the first series. This is an important factor in accounting for their:
QUESTION 17 OF 20
The irregular trend in the first ionisation enthalpy of the 3d metals is accounted for by considering that the removal of one electron alters the relative energies of:
QUESTION 18 OF 20
The sum of the first and second ionisation enthalpies is required to form M²⁺ ions. Which metal has a correspondingly low value for this sum because ionisation removes 4s electrons to yield a highly stable d¹⁰ configuration?
QUESTION 19 OF 20
When evaluating energetic parameters associated with transition metals, what is the standard unit applied to Enthalpy of Hydration (ΔhydH°) and Ionisation Enthalpies?
QUESTION 20 OF 20
The third ionisation enthalpies for Mn²⁺ and Zn²⁺ are exceptionally high compared to their neighbours because:
Test Complete!
Answer Review
1 Which of the following elements has a completely filled d¹⁰ configuration in both its ground state and its common oxidation states, thereby not being regarded as a proper transition metal?
�� Zinc has a completely filled d-subshell. �� Zn²⁺ also retains the d¹⁰ configuration. �� Therefore, it is not a true transition element.
According to the IUPAC definition, a transition element is one whose atom or one of its common oxidation states possesses an incomplete d-subshell. Zinc has the electronic configuration [Ar]3d¹⁰4s². When it forms its most common ion, Zn²⁺, the two 4s electrons are removed, giving the configuration [Ar]3d¹⁰. Since both the atom and the common ion possess a completely filled d-subshell, Zinc does not satisfy the IUPAC definition of a transition element. Consequently, although it belongs to the d-block, it is not regarded as a true transition metal.
- �� Option A → Copper can form ions with incomplete d-subshells.
- �� Option B → Titanium possesses partially filled d-orbitals.
- �� Option C → Scandium has an incompletely filled d-subshell in the atom.
NCERT Recall
- Application
- Apply the IUPAC definition of transition elements.
- Final Logic
- Zn and Zn²⁺ both possess d¹⁰ configurations.
"Zn = d¹⁰ Always"
2 The transition metals retain their distinct metallic conductivity even when:
�� Small atoms occupy interstitial spaces. �� Metallic bonding remains largely intact. �� Electrical conductivity is retained.
Transition metals often form interstitial compounds by trapping small atoms such as hydrogen, carbon or nitrogen within the spaces of their crystal lattice. Although these compounds become harder and often possess higher melting points, the metallic framework remains largely unchanged. As a result, they continue to exhibit metallic properties such as electrical conductivity. This characteristic distinguishes interstitial compounds from many ionic and covalent solids. Therefore, transition metals retain their metallic conductivity even after forming interstitial compounds.
- �� Option A → Conductivity is not specifically defined by heating to the boiling point.
- �� Option C → Covalent gaseous oxides are not metallic conductors.
- �� Option D → Disproportionation is unrelated to metallic conductivity.
Concept Application
- Application
- Relate the structure of interstitial compounds to metallic properties.
- Final Logic
- The metallic lattice remains intact in interstitial compounds.
"Interstitial Yet Conductive"
3 Identify the correct statements regarding the mechanical traits of transition metals.
Statements:
1. Interstitial compounds of transition metals are very soft.
2. Some borides of transition metals approach diamond in hardness.
3. The transition metals (excluding Zn, Cd and Hg) are very hard.
4. Alloys formed by transition metals are generally soft and melt easily.
�� Most transition metals are hard. �� Interstitial compounds are generally hard, not soft. �� Certain borides possess exceptional hardness.
Most transition metals, except Zinc, Cadmium and Mercury, are hard because of strong metallic bonding involving both ns and (n−1)d electrons. Therefore, Statement 3 is correct. Some borides of transition metals possess extremely high hardness values and may approach diamond in hardness, making Statement 2 correct. Statement 1 is incorrect because interstitial compounds are generally very hard rather than soft. Statement 4 is also incorrect because transition-metal alloys are usually strong, hard and possess high melting points. Therefore, only Statements 2 and 3 are correct.
- �� Option B → Statement 1 is incorrect.
- �� Option C → Statements 1 and 4 are incorrect.
- �� Option D → Statement 4 is incorrect.
NCERT Recall
- Application
- Recall the mechanical properties of transition metals and their compounds.
- Final Logic
- Statements 2 and 3 are correct; Statements 1 and 4 are incorrect.
"Borides Hard, Metals Hard"
4 Malleability is a characteristic of transition metals. What structural feature allows them to exhibit typical metallic properties like malleability and ductility?
�� Both ns and d-electrons participate in bonding. �� Strong metallic bonds are formed. �� Layers can slide without breaking the structure.
Transition metals possess strong metallic bonding because both the outer ns electrons and several (n−1)d electrons participate in interatomic bonding. The resulting metallic lattice contains delocalized electrons that hold the metal ions together while still allowing layers of atoms to slide past one another. This characteristic gives rise to malleability and ductility. The extensive participation of d-electrons also contributes to the strength and hardness of transition metals. Therefore, strong metallic bonding involving both ns and d-electrons is responsible for these properties.
- �� Option A → Covalent network solids are generally brittle.
- �� Option C → Filled f-orbitals do not explain malleability.
- �� Option D → p-orbitals are not responsible for metallic bonding in transition metals.
Concept Application
- Application
- Connect metallic bonding with mechanical properties.
- Final Logic
- Strong metallic bonding enables malleability and ductility.
"ns + d = Strong Metal"
5
According to the passage, the maximum melting point in a transition series is generally found when the d-orbital configuration is:
�� Maximum unpaired electrons occur at d⁵. �� Strong interatomic interaction results. �� Melting points reach their maximum near d⁵.
The passage states that the melting points of transition metals generally increase and reach a maximum near the middle of the series where the d⁵ configuration occurs. At d⁵, each d-orbital contains one unpaired electron, maximizing the number of electrons available for strong metallic bonding. This produces particularly strong interatomic interactions and consequently very high enthalpies of atomisation. Since melting point is directly related to the strength of metallic bonding, the maximum melting point is observed near the d⁵ configuration, except for a few anomalous cases such as manganese and technetium.
- �� Option A → d¹ has fewer bonding electrons.
- �� Option B → d⁰ lacks d-electrons for strong metallic interaction.
- �� Option C → d¹⁰ does not provide maximum metallic-bond strength.
Passage Analysis
- Application
- Identify the electronic configuration explicitly linked to maximum melting points.
- Final Logic
- Maximum metallic bonding occurs near d⁵.
"d⁵ = Peak Melting"
6
The passage relates the strong interatomic interaction and high boiling/melting points directly to:
�� Strong metallic bonding requires more energy to break. �� High atomisation enthalpy indicates strong interatomic attraction. �� High melting and boiling points result.
The passage clearly states that transition metals possess high enthalpies of atomisation because a greater number of (n−1)d and ns electrons participate in metallic bonding. Strong metallic bonding requires a large amount of energy to separate atoms completely into the gaseous state. Consequently, elements with high enthalpies of atomisation generally exhibit high melting and boiling points. Thus, the strong interatomic interactions responsible for these thermal properties are directly reflected in the high values of enthalpy of atomisation.
- �� Option A → Low atomisation enthalpy indicates weaker bonding.
- �� Option C → Unpaired electrons contribute significantly to bonding.
- �� Option D → Mn is merely an exception to the trend.
Passage Analysis
- Application
- Relate atomisation enthalpy to metallic bond strength.
- Final Logic
- Strong bonding → High atomisation enthalpy → High melting/boiling points.
"High ΔₐH° = High MP & BP"
7 Match the 3d transition metals with their metallic radii (pm) as per NCERT data.
| List I | List II |
|---|---|
| 1. Sc | a. 126 pm |
| 2. V | b. 128 pm |
| 3. Fe | c. 135 pm |
| 4. Cu | d. 164 pm |
�� Sc has the largest metallic radius among the listed elements. �� Metallic radii decrease across the series. �� Cu possesses one of the smallest radii in the list.
The metallic radii of the given elements are approximately: Sc = 164 pm, V = 135 pm, Fe = 126 pm and Cu = 128 pm. These values illustrate the general decrease in metallic radius across the first transition series due to increasing effective nuclear charge. Therefore, Sc matches 164 pm, V matches 135 pm, Fe matches 126 pm and Cu matches 128 pm. Hence the correct matching is 1-d, 2-c, 3-a and 4-b.
- �� Option B → Assigns incorrect radii to all listed elements.
- �� Option C → V and Cu are mismatched.
- �� Option D → Sc cannot have a radius of 135 pm.
NCERT Recall
- Application
- Recall standard metallic-radius values from NCERT tables.
- Final Logic
- Sc–164, V–135, Fe–126 and Cu–128 pm.
"Sc Largest, Fe Smallest"
8 Arrange the following M²⁺ ions in decreasing order of their ionic radii.
1. V²⁺ (79 pm)
2. Cr²⁺ (82 pm)
3. Ni²⁺ (70 pm)
4. Cu²⁺ (73 pm)
�� Ionic radius decreases with increasing nuclear charge. �� Cr²⁺ has the largest radius among the listed ions. �� Ni²⁺ has the smallest radius.
The ionic radii provided are Cr²⁺ = 82 pm, V²⁺ = 79 pm, Cu²⁺ = 73 pm and Ni²⁺ = 70 pm. To arrange them in decreasing order, the ion with the largest radius is placed first and the smallest last. Therefore, the correct order is Cr²⁺ > V²⁺ > Cu²⁺ > Ni²⁺. Converting this into the numbering scheme gives 2, 1, 4 and 3.
- �� Option B → Places V²⁺ ahead of Cr²⁺ incorrectly.
- �� Option C → Gives nearly the reverse trend.
- �� Option D → Represents increasing rather than decreasing radii.
Data Analysis
- Application
- Compare the numerical ionic-radius values directly.
- Final Logic
- 82 > 79 > 73 > 70 pm.
"Cr > V > Cu > Ni"
9 When moving from Scandium to Zinc in the 3d series, the atomic radii decrease less rapidly than initially expected. This is primarily because:
�� Nuclear charge increases across the series. �� Newly added electrons enter 3d orbitals. �� These electrons partially shield the outer 4s electrons.
As we move across the first transition series, the nuclear charge increases steadily. At the same time, additional electrons enter the 3d orbitals. These 3d electrons provide partial shielding to the outer 4s electrons, reducing the full effect of the increasing nuclear charge. Consequently, atomic radii decrease, but the decrease is much less pronounced than expected. This partial shielding effect is responsible for the characteristic size trend observed across transition-metal series.
- �� Option A → The shielding occurs mainly from 3d to 4s electrons.
- �� Option B → The number of principal energy levels remains the same.
- �� Option C → Nuclear charge increases continuously.
Concept Application
- Application
- Apply shielding and effective nuclear-charge concepts.
- Final Logic
- Partial shielding by 3d electrons slows the decrease in atomic radii.
"3d Shields 4s"
10 Comparing the shielding effects within transition and inner-transition metals, which of the following is correct?
�� f-electrons are poor shielding electrons. �� Their ineffective shielding causes lanthanoid contraction. �� d-electrons shield better than f-electrons.
The shielding effect of f-electrons is significantly weaker than that of d-electrons. Because 4f electrons are diffuse and penetrate poorly, they do not effectively counterbalance the increasing nuclear charge across the lanthanoid series. This poor shielding leads to lanthanoid contraction, a gradual decrease in atomic and ionic radii. Since d-electrons provide somewhat better shielding than f-electrons, the correct statement is that one 4f electron shields less effectively than one d electron.
- �� Option A → Opposite of the observed trend.
- �� Option C → Neither d nor f electrons provide perfect shielding.
- �� Option D → s-electrons are actually the most effective shielding electrons.
NCERT Recall
- Application
- Recall the relative shielding abilities of orbital types.
- Final Logic
- s > p > d > f in shielding effectiveness.
"f = Feeble Shielding"
11 Identify the specific phenomenon responsible for the nearly identical radii of second- and third-transition-series elements such as Zirconium and Hafnium.
�� 4f electrons shield poorly. �� Effective nuclear charge increases across the lanthanoids. �� The sizes of 5d elements become similar to those of 4d elements.
Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii across the lanthanoid series due to the poor shielding effect of 4f electrons. Because the 4f electrons do not effectively shield the increasing nuclear charge, the outer electrons are pulled closer to the nucleus. As a result, the atomic sizes of the 5d transition elements become much smaller than expected. Consequently, pairs of elements such as Zirconium (4d series) and Hafnium (5d series) possess almost identical radii and show very similar chemical properties. Therefore, lanthanoid contraction is the phenomenon responsible for this observation.
- �� Option A → Actinoid contraction occurs in the actinoid series.
- �� Option B → The inert pair effect is mainly observed in heavier p-block elements.
- �� Option D → Screening effect is a general concept, not the specific phenomenon.
NCERT Recall
- Application
- Recall the consequences of poor shielding by 4f electrons.
- Final Logic
- Poor 4f shielding causes lanthanoid contraction and similar radii.
"4f Poor Shielding → Lanthanoid Contraction"
12 Because of the chemical and physical similarities resulting from lanthanoid contraction, which procedural consequence heavily impacts metals like Zirconium and Hafnium?
�� Zr and Hf possess nearly identical radii. �� Their chemical properties become extremely similar. �� Separation becomes difficult.
Lanthanoid contraction causes Hafnium and Zirconium to have almost identical atomic and ionic radii. Since chemical behaviour depends strongly on size and electronic structure, these two elements exhibit very similar physical and chemical properties. Consequently, conventional separation techniques become ineffective because both elements behave nearly identically in chemical reactions. This creates significant industrial difficulty in obtaining pure Zirconium and Hafnium from naturally occurring ores. Therefore, the most important consequence is the difficulty of separating these elements.
- �� Option A → No such characteristic combustion behaviour exists.
- �� Option C → Zirconium is not exclusively radioactive.
- �� Option D → They do not undergo spontaneous disproportionation.
Concept Application
- Application
- Connect lanthanoid contraction with similarities in chemical behaviour.
- Final Logic
- Similar radii lead to difficult separation.
"Zr ≈ Hf → Hard to Separate"
13 From Titanium to Copper, why does the density show a significant general increase?
�� Atomic mass increases across the series. �� Metallic radius decreases slightly. �� Density therefore increases.
Density is defined as mass per unit volume. Across the first transition series from Titanium to Copper, atomic mass increases steadily because additional protons and neutrons are added to the nucleus. Simultaneously, metallic radii decrease slightly due to increasing effective nuclear charge. The decrease in size reduces atomic volume while the increase in mass raises the numerator of the density relationship. As a result, density increases significantly across the series. Therefore, the combined effect of increasing atomic mass and decreasing metallic radius accounts for the observed trend.
- �� Option A → Both trends are opposite to reality.
- �� Option B → Metallic radius decreases rather than increases.
- �� Option D → Density is not directly determined by boiling point.
Concept Application
- Application
- Apply the density relationship involving mass and volume.
- Final Logic
- More mass and less volume produce greater density.
"Mass Up, Size Down = Density Up"
14 According to the properties of the 3d series, which two metals share the highest identical density (8.9 g cm⁻³)?
�� Nickel and Copper possess very high densities. �� Both have values close to 8.9 g cm⁻³. �� They represent the highest identical density pair listed.
The densities of transition metals generally increase from Titanium to Copper because atomic masses increase while metallic radii decrease slightly. Among the listed metals, Nickel and Copper both possess densities of approximately 8.9 g cm⁻³ according to standard NCERT data. These values are among the highest in the first transition series and illustrate the strong metallic bonding and compact atomic packing present in these elements. Therefore, Nickel and Copper form the correct pair.
- �� Option A → Iron and Cobalt have lower densities.
- �� Option C → Zinc has a lower density than Copper.
- �� Option D → Chromium and Manganese possess much lower densities.
NCERT Recall
- Application
- Recall density values of important first-transition-series elements.
- Final Logic
- Ni and Cu each have densities of about 8.9 g cm⁻³.
"Ni = Cu = 8.9"
15 In the first series of transition elements (Sc to Zn), which element has the exceptionally lowest enthalpy of atomisation (126 kJ mol⁻¹)?
�� Zinc possesses a stable d¹⁰ configuration. �� Few electrons participate effectively in metallic bonding. �� Its atomisation enthalpy is unusually low.
Zinc has the electronic configuration [Ar]3d¹⁰4s². The completely filled d-subshell contributes relatively little to metallic bonding compared with partially filled d-subshells found in most transition metals. Consequently, metallic bonding in Zinc is weaker than in neighbouring elements. This leads to a much lower enthalpy of atomisation, approximately 126 kJ mol⁻¹, which is the lowest value in the first transition series. Therefore, Zinc exhibits the minimum enthalpy of atomisation among the given elements.
- �� Option A → Scandium has a considerably higher atomisation enthalpy.
- �� Option B → Manganese is anomalous but not the lowest.
- �� Option D → Titanium exhibits strong metallic bonding and higher values.
NCERT Recall
- Application
- Recall atomisation-enthalpy trends across the first transition series.
- Final Logic
- Filled d¹⁰ configuration gives Zinc the lowest atomisation enthalpy.
"Zn d¹⁰ = Lowest ΔₐH°"
16 Heavy transition metals from the second and third series exhibit more frequent metal-metal bonding in compounds compared to the first series. This is an important factor in accounting for their:
�� Metal-metal bonding strengthens interatomic attraction. �� Stronger bonding requires more energy to break. �� Atomisation enthalpy therefore increases.
The second (4d) and third (5d) transition series exhibit stronger metallic bonding than the first series. One important reason is the greater tendency of these heavier transition metals to form metal-metal bonds. Such bonding increases the overall strength of the metallic lattice and enhances interatomic interactions. Consequently, a larger amount of energy is required to separate the atoms completely into the gaseous state. This is reflected in their higher enthalpies of atomisation. Therefore, the frequent occurrence of metal-metal bonding contributes significantly to the greater enthalpies of atomisation observed in these series.
- �� Option A → Stronger bonding increases, not decreases, atomisation enthalpy.
- �� Option B→ Metal-metal bonding is unrelated to complete diamagnetism.
- �� Option C → Stronger bonding generally increases hardness.
Concept Application
- Application
- Relate metal-metal bonding to metallic bond strength.
- Final Logic
- More metal-metal bonding → stronger bonding → higher ΔₐH°.
"Metal-Metal Bond = More Energy Needed"
17 The irregular trend in the first ionisation enthalpy of the 3d metals is accounted for by considering that the removal of one electron alters the relative energies of:
�� 4s and 3d orbitals possess similar energies. �� Electron removal changes their relative stability. �� This causes irregular ionisation trends.
In transition elements, the energies of the 4s and 3d orbitals are very close to one another. When the first electron is removed during ionisation, the relative energies of these orbitals change. This change affects the stability of the resulting ion and produces irregularities in the trend of first ionisation enthalpies across the series. The effect is particularly noticeable in elements with half-filled and completely filled d-subshells. Therefore, the irregular trend is explained by changes in the relative energies of the 4s and 3d orbitals.
- �� Option A → 3s and 3p orbitals are not involved.
- �� Option B → 4p orbitals are not occupied in the 3d series.
- �� Option D → These orbitals belong to the 4d series.
NCERT Recall
- Application
- Recall the orbital-energy relationships responsible for ionisation trends.
- Final Logic
- Removal of an electron alters the 4s–3d energy relationship.
"IE Trend = 4s ↔ 3d"
18 The sum of the first and second ionisation enthalpies is required to form M²⁺ ions. Which metal has a correspondingly low value for this sum because ionisation removes 4s electrons to yield a highly stable d¹⁰ configuration?
�� Zinc has configuration 3d¹⁰4s². �� Removal of two 4s electrons gives Zn²⁺. �� Zn²⁺ possesses a stable d¹⁰ configuration.
Zinc has the electronic configuration [Ar]3d¹⁰4s². During formation of Zn²⁺, the two outermost 4s electrons are removed, resulting in the highly stable configuration [Ar]3d¹⁰. Since the d¹⁰ configuration is exceptionally stable, the formation of Zn²⁺ is energetically favourable compared with many other transition-metal ions. Consequently, the sum of the first and second ionisation enthalpies is relatively low. This stability of the resulting ion plays a major role in the chemistry of Zinc.
- �� Option B → Cu⁺ rather than Cu²⁺ possesses the stable d¹⁰ configuration.
- �� Option C → Sc²⁺ does not yield a stable d¹⁰ arrangement.
- �� Option D → Chromium is associated with d⁵ stability, not d¹⁰ stability.
Concept Application
- Application
- Determine which ion acquires a particularly stable configuration after losing two electrons.
- Final Logic
- Zn → Zn²⁺ gives stable d¹⁰ configuration.
"Zn²⁺ = Perfect d¹⁰"
19 When evaluating energetic parameters associated with transition metals, what is the standard unit applied to Enthalpy of Hydration (ΔhydH°) and Ionisation Enthalpies?
�� Both quantities measure energy changes. �� Values are expressed per mole. �� The standard unit is kJ mol⁻¹.
Enthalpy of hydration and ionisation enthalpy are thermodynamic quantities that represent energy changes occurring during hydration and ionisation processes, respectively. Since these changes are measured for one mole of species, the standard unit used is kilojoule per mole (kJ mol⁻¹). This unit allows direct comparison of energetic trends among transition metals and their ions. Therefore, kJ mol⁻¹ is the correct unit for both quantities.
- �� Option A → Volt is a unit of electrical potential.
- �� Option B → S cm⁻¹ is a unit of conductivity.
- �� Option D → J m⁻² is not used for enthalpy measurements.
NCERT Recall
- Application
- Recall the standard thermodynamic unit for enthalpy.
- Final Logic
- Energy per mole is expressed in kJ mol⁻¹.
"All Enthalpies → kJ mol⁻¹"
20 The third ionisation enthalpies for Mn²⁺ and Zn²⁺ are exceptionally high compared to their neighbours because:
�� Mn²⁺ has a stable d⁵ configuration. �� Zn²⁺ has a stable d¹⁰ configuration. �� Removing another electron destroys this stability.
Mn²⁺ possesses the highly stable half-filled configuration 3d⁵, while Zn²⁺ possesses the highly stable completely filled configuration 3d¹⁰. The third ionisation process requires removing an electron from these exceptionally stable arrangements. Since such removal destroys the extra stability associated with half-filled and fully filled d-subshells, a large amount of energy is required. Consequently, the third ionisation enthalpies of manganese and zinc are significantly higher than those of neighbouring elements. This behaviour is a classic example of the stabilising effect of special electronic configurations.
- �� Option B → Atomic radius is not the principal factor.
- �� Option C → The stable configurations involved are d⁵ and d¹⁰, not d⁶ and d⁹.
- �� Option D → Diamagnetism does not explain the high ionisation enthalpy.
Concept Application
- Application
- Identify the stability associated with half-filled and fully filled d-subshells.
- Final Logic
- Breaking d⁵ or d¹⁰ configurations requires exceptionally high energy.
"Break d⁵ or d¹⁰ → IE₃ Sky High"
