CUET UG Chemistry Booster Test - 3 General Properties
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QUESTION 1 OF 20
Based on the passage, which of the following sets entirely consists of exceptions to possessing typical metallic structures at normal temperatures?
QUESTION 2 OF 20
The passage attributes "high thermal and electrical conductivity" to transition elements. In alloys like brass (copper-zinc) and bronze (copper-tin), how is this metallic characteristic generally maintained structurally?
QUESTION 3 OF 20
The hardness of inner transition elements (lanthanoids) generally increases with increasing atomic number. Which specific lanthanoid is distinctly described as being "steel hard"?
QUESTION 4 OF 20
Transition metals are generally highly malleable. However, Zinc is an exception in some physical properties. What is the fundamental atomic reason for Zinc's anomalous metallic behaviour compared to typical hard transition metals?
QUESTION 5 OF 20
Arrange the following transition metals in decreasing order of their melting points according to the general graphical trend observed across the transition series.
1. Zinc (Zn)
2. Tungsten (W)
3. Chromium (Cr)
4. Molybdenum (Mo)
QUESTION 6 OF 20
Elements with very high boiling points (indicating very high enthalpies of atomisation) tend to behave in a specific way regarding their standard electrode potentials and reactions. How do they typically behave?
QUESTION 7 OF 20
The atomic radii of the elements from the 3d series decrease slightly and then remain almost constant towards the end of the series. What governs this constancy?
QUESTION 8 OF 20
Match the 3d transition metal M³⁺ ion with its ionic radius.
| List I | List II |
|---|---|
| 1. Ti³⁺ | a. 64 pm |
| 2. V³⁺ | b. 65 pm |
| 3. Cr³⁺ | c. 67 pm |
| 4. Fe³⁺ | d. 62 pm |
QUESTION 9 OF 20
Across the actinoid series, there is a gradual decrease in the size of atoms or M³⁺ ions. Compared to the lanthanoid contraction, the actinoid contraction is:
QUESTION 10 OF 20
Identify the correct statements regarding the shielding effect in transition and inner transition metals.
Statements:
1. One d electron shields another d electron perfectly.
2. Imperfect shielding coupled with increasing nuclear charge leads to a net decrease in atomic size.
3. The shielding of one 4f electron by another is less effective than that of one d electron by another.
4. 5f electrons provide poorer shielding than 4f electrons.
QUESTION 11 OF 20
The imperfect shielding responsible for lanthanoid contraction occurs largely because:
QUESTION 12 OF 20
Which physical property of the third transition series is heavily influenced by the lanthanoid contraction, leading to values much higher than expected based on normal group trends?
QUESTION 13 OF 20
From Titanium to Copper, the atomic mass steadily increases while the atomic radius decreases slightly. How does this combination specifically affect the physical properties of later elements like Cu compared to early elements like Sc?
QUESTION 14 OF 20
Arrange the following elements in increasing order of their density (g cm⁻³).
1. Iron (Fe)
2. Scandium (Sc)
3. Nickel (Ni)
4. Vanadium (V)
QUESTION 15 OF 20
Identify the element of the 3d series that possesses the highest enthalpy of atomisation (515 kJ mol⁻¹) according to the standard thermochemical tables provided.
QUESTION 16 OF 20
The very high enthalpies of atomisation observed in transition metals directly facilitate which structural phenomenon in compounds of heavy transition metals?
QUESTION 17 OF 20
When evaluating the reduction of M²⁺ to M, we use Standard Electrode Potentials. What is the standard unit of the associated Standard Electrode Potential (E°) that ties into Enthalpy of Atomisation and Ionisation Energy?
QUESTION 18 OF 20
When predicting the stability of oxidation states from ionisation enthalpies, the unusually high second ionisation enthalpy for Cu (1958 kJ mol⁻¹) indicates that:
QUESTION 19 OF 20
Mn⁺ has a 3d⁵4s¹ configuration, and Cr⁺ has a 3d⁵ configuration. Why is the ionisation enthalpy of Mn⁺ lower than that of Cr⁺?
QUESTION 20 OF 20
Although thermodynamic ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, a unique behavior is seen with Copper (Cu) having a positive E° value (+0.34 V). This positive value indicates that:
Test Complete!
Answer Review
1
Based on the passage, which of the following sets entirely consists of exceptions to possessing typical metallic structures at normal temperatures?
�� Most transition metals possess typical metallic structures. �� Four elements are specifically listed as exceptions. �� These exceptions are Zn, Cd, Hg and Mn.
The passage explicitly states that transition metals generally possess one or more typical metallic structures at normal temperatures. However, four elements are identified as exceptions: Zinc, Cadmium, Mercury and Manganese. These elements exhibit deviations from the structural behaviour commonly observed in transition metals. Therefore, the only option containing all four exceptions is Zn, Cd, Hg and Mn. Since the question directly asks for the complete set of exceptions, Option B is the correct answer.
- �� Option A → Sc, Y and La exhibit normal metallic structures.
- �� Option C → Fe, Co, Ni and Cu are typical transition metals.
- �� Option D → Cr, Mo and W are not exceptions.
Passage Analysis
- Application
- Identify the exception list directly from the passage.
- Final Logic
- Zn, Cd, Hg and Mn are specifically mentioned as exceptions.
"Zn-Cd-Hg-Mn = Structural Exceptions"
2
The passage attributes "high thermal and electrical conductivity" to transition elements. In alloys like brass (copper-zinc) and bronze (copper-tin), how is this metallic characteristic generally maintained structurally?
�� Alloys retain metallic bonding. �� Electrons remain delocalised. �� Conductivity is largely preserved.
Brass and bronze are examples of substitutional alloys in which atoms of one metal replace some atoms of another within the metallic lattice. The alloy generally forms a homogeneous solid solution where atoms are randomly distributed while maintaining metallic bonding. Because the sea of delocalised electrons remains present, these alloys continue to exhibit metallic properties such as electrical and thermal conductivity. Therefore, the conductivity is maintained through the formation of homogeneous metallic solid solutions.
- �� Option A → Alloys do not generally become insulators.
- �� Option B → d-orbitals do not become completely empty.
- �� Option C → They do not form covalent network structures.
Concept Application
- Application
- Relate alloy structure to metallic bonding.
- Final Logic
- Metallic conductivity persists because metallic bonding is retained.
"Alloy ≠ Destroy Metal"
3 The hardness of inner transition elements (lanthanoids) generally increases with increasing atomic number. Which specific lanthanoid is distinctly described as being "steel hard"?
�� Hardness increases across the lanthanoid series. �� Samarium is specifically described as steel hard. �� It is harder than many earlier lanthanoids.
Among the lanthanoids, hardness generally increases with increasing atomic number due to stronger metallic bonding and decreasing atomic size. NCERT specifically mentions Samarium as being "steel hard," highlighting its exceptional hardness compared with several other lanthanoids. This property illustrates the gradual strengthening of metallic interactions across the series. Therefore, Samarium is the lanthanoid identified with this distinctive description.
- �� Option B → Cerium is not described as steel hard.
- �� Option C → Lutetium is hard but not specifically named in this context.
- �� Option D → Europium is relatively softer among lanthanoids.
NCERT Recall
- Application
- Recall the special physical-property descriptions of lanthanoids.
- Final Logic
- NCERT specifically associates "steel hard" with Samarium.
"Samarium = Steel Hard"
4 Transition metals are generally highly malleable. However, Zinc is an exception in some physical properties. What is the fundamental atomic reason for Zinc's anomalous metallic behaviour compared to typical hard transition metals?
�� Zinc has a filled 3d¹⁰ configuration. �� d-electrons contribute little to metallic bonding. �� Metallic bonding becomes relatively weaker.
Zinc possesses the electronic configuration [Ar]3d¹⁰4s². Because the 3d subshell is completely filled, the d-electrons contribute minimally to metallic bonding. In most transition metals, partially filled d-orbitals participate significantly in metallic bonding, increasing hardness and strength. The lack of such participation in Zinc results in weaker metallic bonding and causes several physical properties to differ from those of typical transition metals. Therefore, the filled d¹⁰ configuration is responsible for Zinc's anomalous behaviour.
- �� Option A → Zinc has a filled 4s² configuration.
- �� Option B → Zinc does not possess incomplete p-orbitals.
- �� Option D → 4f electrons are not involved in Zinc.
Concept Application
- Application
- Relate electronic configuration to metallic bonding.
- Final Logic
- Filled d¹⁰ orbitals reduce d-electron participation in bonding.
"Zn d¹⁰ = Weak d-Bonding"
5 Arrange the following transition metals in decreasing order of their melting points according to the general graphical trend observed across the transition series.
1. Zinc (Zn)
2. Tungsten (W)
3. Chromium (Cr)
4. Molybdenum (Mo)
�� Strong metallic bonding increases melting point. �� Tungsten exhibits exceptionally high melting point. �� Zinc has one of the lowest melting points among the listed metals.
Melting points of transition metals depend primarily on the strength of metallic bonding, which in turn is related to the number of unpaired d-electrons participating in bonding and the enthalpy of atomisation. Tungsten possesses one of the highest melting points among all metals because of extremely strong metallic bonding. Molybdenum also exhibits very strong metallic interactions and therefore has a very high melting point, though slightly lower than Tungsten. Chromium follows next with a high melting point due to significant d-electron participation in bonding. Zinc, having a completely filled d¹⁰ configuration, contributes fewer electrons to metallic bonding and therefore exhibits a much lower melting point. Hence, the correct decreasing order is W > Mo > Cr > Zn.
- �� Option A → Chromium does not possess a higher melting point than Molybdenum.
- �� Option B → Represents nearly the reverse trend.
- �� Option C → Places Molybdenum above Tungsten incorrectly.
NCERT Recall
- Application
- Recall melting-point trends and enthalpies of atomisation across transition metals.
- Final Logic
- W > Mo > Cr > Zn.
"W-Mo-Cr-Zn → Heat Falls Down"
6 Elements with very high boiling points (indicating very high enthalpies of atomisation) tend to behave in a specific way regarding their standard electrode potentials and reactions. How do they typically behave?
�� High atomisation enthalpy indicates strong metallic bonding. �� Such metals resist oxidation. �� They often exhibit noble behaviour.
Transition metals possessing very high enthalpies of atomisation have exceptionally strong metallic bonding. A large amount of energy is required to separate atoms from the metallic lattice and convert them into ions. Consequently, these metals are less likely to undergo oxidation and often exhibit low chemical reactivity. This tendency is reflected in their electrode potentials and their resistance to corrosion and chemical attack. Therefore, metals with very high boiling points and atomisation enthalpies generally behave in a noble manner in chemical reactions.
- �� Option A → Noble metals are not highly reactive.
- �� Option C → Many noble metals resist dissolution in non-oxidising acids.
- �� Option D → Noble metals are generally poor reducing agents.
Concept Application
- Application
- Relate atomisation enthalpy and metallic bonding to chemical reactivity.
- Final Logic
- Stronger metallic bonding generally leads to more noble behaviour.
"High ΔₐH° = Noble Metal"
7 The atomic radii of the elements from the 3d series decrease slightly and then remain almost constant towards the end of the series. What governs this constancy?
�� Nuclear charge increases across the series. �� d-electrons provide partial shielding. �� Opposing effects produce nearly constant radii.
Across the first transition series, the nuclear charge increases steadily, which tends to decrease atomic size. However, additional electrons are simultaneously added to the 3d orbitals. These electrons contribute to shielding and increase electron-electron repulsion within the d-subshell. Towards the end of the series, these effects increasingly counterbalance the attraction caused by the growing nuclear charge. As a result, the decrease in atomic radius becomes very small and the radii remain nearly constant. Therefore, the observed constancy arises from the balance between increasing nuclear attraction and increasing shielding/repulsion effects.
- �� Option A → Shielding is reduced but not completely lost.
- �� Option B → Expansion of the 4s orbital is not the governing factor.
- �� Option D → Core-electron pairing does not explain the trend.
Concept Application
- Application
- Balance the effects of nuclear charge and shielding.
- Final Logic
- Increasing attraction and increasing repulsion nearly cancel each other.
"More Charge, More Shielding → Size Nearly Constant"
8 Match the 3d transition metal M³⁺ ion with its ionic radius.
| List I | List II |
|---|---|
| 1. Ti³⁺ | a. 64 pm |
| 2. V³⁺ | b. 65 pm |
| 3. Cr³⁺ | c. 67 pm |
| 4. Fe³⁺ | d. 62 pm |
�� Ionic radii decrease with increasing atomic number. �� Greater nuclear charge contracts the ion. �� Cr³⁺ possesses the smallest radius among the listed ions.
The ionic radii of these M³⁺ ions decrease progressively across the transition series because the nuclear charge increases while shielding does not increase proportionally. The NCERT values are approximately Ti³⁺ = 67 pm, V³⁺ = 64 pm, Cr³⁺ = 62 pm and Fe³⁺ = 65 pm. Therefore, Ti³⁺ matches 67 pm, V³⁺ matches 64 pm, Cr³⁺ matches 62 pm and Fe³⁺ matches 65 pm. Hence, the correct matching is 1-c, 2-a, 3-d and 4-b.
- �� Option B → Assigns incorrect radii to multiple ions.
- �� Option C → Reverses several correct pairings.
- �� Option D → Does not follow NCERT tabulated values.
NCERT Recall
- Application
- Recall ionic-radius values of important transition-metal ions.
- Final Logic
- Ti³⁺–67, V³⁺–64, Cr³⁺–62 and Fe³⁺–65 pm.
"Ti Largest, Cr Smallest"
9 Across the actinoid series, there is a gradual decrease in the size of atoms or M³⁺ ions. Compared to the lanthanoid contraction, the actinoid contraction is:
�� 5f electrons shield poorly. �� Effective nuclear charge increases significantly. �� Actinoid contraction is more pronounced.
The actinoid contraction refers to the gradual decrease in atomic and ionic radii across the actinoid series. The 5f electrons are even less effective at shielding the increasing nuclear charge than the 4f electrons of the lanthanoids. Consequently, the effective nuclear attraction experienced by outer electrons increases more strongly across the series. This causes a greater decrease in size from one element to the next. Therefore, actinoid contraction is generally greater than lanthanoid contraction.
- �� Option A → 5f shielding is poor, not strong.
- �� Option C → The contractions are not exactly equal.
- �� Option D → Actinoid contraction is a real and significant phenomenon.
Concept Application
- Application
- Compare shielding effectiveness of 4f and 5f electrons.
- Final Logic
- Poorer 5f shielding causes a larger contraction.
"5f Shields Worse → Greater Contraction"
10 Identify the correct statements regarding the shielding effect in transition and inner transition metals.
Statements:
1. One d electron shields another d electron perfectly.
2. Imperfect shielding coupled with increasing nuclear charge leads to a net decrease in atomic size.
3. The shielding of one 4f electron by another is less effective than that of one d electron by another.
4. 5f electrons provide poorer shielding than 4f electrons.
�� d and f electrons do not shield perfectly. �� 4f shielding is weaker than d-electron shielding. �� Poor shielding causes contraction effects.
Statement 2 is correct because imperfect shielding combined with increasing nuclear charge produces a gradual decrease in atomic and ionic sizes across transition and inner-transition series. Statement 3 is correct because one 4f electron shields another less effectively than d-electrons shield each other. Statement 4 is also correct because 5f electrons provide even poorer shielding than 4f electrons, contributing to the actinoid contraction. Statement 1 is incorrect because d-electrons do not provide perfect shielding. Therefore, Statements 2, 3 and 4 are correct.
- �� Option B → Statement 1 is incorrect.
- �� Option C → Statement 2 is also correct and cannot be omitted.
- �� Option D → Statement 1 is incorrect and Statements 3 and 4 are omitted.
NCERT Recall
- Application
- Recall shielding trends among d- and f-electrons.
- Final Logic
- Statements 2, 3 and 4 are correct; Statement 1 is incorrect.
"d < Perfect, 4f < d, 5f < 4f"
11 The imperfect shielding responsible for lanthanoid contraction occurs largely because:
�� 4f electrons shield poorly. �� Nuclear charge increases steadily. �� Atomic size decreases gradually.
Lanthanoid contraction arises because the 4f electrons do not effectively shield one another from the increasing nuclear charge across the lanthanoid series. Although 4f orbitals are inner orbitals, their complex shapes and poor penetration result in weak shielding. As additional protons are added to the nucleus, the effective nuclear charge experienced by outer electrons increases continuously. Consequently, the electrons are pulled closer to the nucleus, producing a gradual decrease in atomic and ionic radii. This phenomenon is known as lanthanoid contraction and is one of the most important trends associated with f-block chemistry.
- �� Option B → 4f electrons do not completely screen outer electrons.
- �� Option C → Repulsion from 5d electrons is not the primary cause.
- �� Option D → Nuclear charge actually increases across the series.
NCERT Recall
- Application
- Recall the origin of lanthanoid contraction.
- Final Logic
- Poor 4f shielding + increasing nuclear charge = contraction.
"4f Shields Poorly, Size Shrinks Slowly"
12 Which physical property of the third transition series is heavily influenced by the lanthanoid contraction, leading to values much higher than expected based on normal group trends?
�� Atomic size decreases unexpectedly. �� Atomic mass increases significantly. �� Density becomes unusually high.
Because of lanthanoid contraction, the atomic radii of third-transition-series elements are much smaller than expected. At the same time, these elements possess considerably larger atomic masses than their second-series counterparts. The combination of high mass and relatively small atomic volume results in exceptionally high densities. Elements such as Osmium and Iridium are among the densest known metals because of this effect. Therefore, density is one of the physical properties most strongly influenced by lanthanoid contraction.
- �� Option A → Boiling points are influenced by several factors, not primarily lanthanoid contraction.
- �� Option C → Paramagnetism depends mainly on unpaired electrons.
- �� Option D → Oxidation states are not the principal effect discussed.
Concept Application
- Application
- Relate density to atomic mass and atomic volume.
- Final Logic
- High mass + smaller size = higher density.
"Contraction → Compact → Dense"
13 From Titanium to Copper, the atomic mass steadily increases while the atomic radius decreases slightly. How does this combination specifically affect the physical properties of later elements like Cu compared to early elements like Sc?
�� Atomic mass increases across the series. �� Atomic radius decreases slightly. �� Density therefore increases significantly.
Density is determined by mass per unit volume. Moving from Titanium toward Copper, atomic masses increase substantially while atomic radii decrease slightly because of increasing effective nuclear charge. The increased mass and reduced volume combine to produce a significant rise in density. Consequently, Copper possesses a much higher density than early transition metals such as Scandium. This trend is a characteristic feature of the first transition series and is clearly reflected in NCERT data.
- �� Option A → Density trends do not directly determine oxidation potentials.
- �� Option B → Copper remains a solid at room temperature.
- �� Option D → Scandium is not completely diamagnetic in this context.
Concept Application
- Application
- Apply the relationship Density = Mass/Volume.
- Final Logic
- Higher mass and smaller volume increase density.
"Mass Up + Size Down = Density Up"
14 Arrange the following elements in increasing order of their density (g cm⁻³).
1. Iron (Fe)
2. Scandium (Sc)
3. Nickel (Ni)
4. Vanadium (V)
�� Atomic mass increases across the series. �� Density generally rises from Sc to Ni. �� Scandium has the lowest density among the given elements.
The densities of the listed elements approximately follow the order: Scandium (3.0 g cm⁻³) < Vanadium (6.0 g cm⁻³) < Iron (7.9 g cm⁻³) < Nickel (8.9 g cm⁻³). As we move across the first transition series, atomic mass increases significantly while atomic volume does not increase proportionately. This causes a gradual increase in density. Therefore, when the elements are rearranged as Iron, Scandium, Nickel and Vanadium, the increasing order of density becomes Scandium → Vanadium → Iron → Nickel, corresponding to 2, 4, 1 and 3.
- �� Option A → Represents decreasing density trend.
- �� Option C → Places Vanadium before Scandium incorrectly.
- �� Option D → Iron is denser than Vanadium.
NCERT Recall
- Application
- Recall the density values of important first-transition-series elements.
- Final Logic
- Sc < V < Fe < Ni ⇒ 2 < 4 < 1 < 3.
"Sc → V → Fe → Ni = Density Climbs"
15 Identify the element of the 3d series that possesses the highest enthalpy of atomisation (515 kJ mol⁻¹) according to the standard thermochemical tables provided.
�� Vanadium exhibits exceptionally strong metallic bonding. �� It has one of the highest atomisation enthalpies in the 3d series. �� More bonding electrons strengthen the metallic lattice.
Enthalpy of atomisation measures the energy required to convert one mole of a metallic element into gaseous atoms. A higher value indicates stronger metallic bonding. Among the listed 3d-series elements, Vanadium possesses the highest enthalpy of atomisation, approximately 515 kJ mol⁻¹. This reflects the strong contribution of both ns and d-electrons to metallic bonding. The high atomisation enthalpy also explains Vanadium's relatively high melting point and strong metallic character. Therefore, Vanadium is the correct answer.
- �� Option B → Chromium has a lower atomisation enthalpy.
- �� Option C → Iron does not possess the highest value.
- �� Option D → Cobalt's value is lower than Vanadium's.
NCERT Recall
- Application
- Recall tabulated atomisation-enthalpy values.
- Final Logic
- Vanadium shows the highest ΔₐH° among the listed elements.
"Vanadium = Very High ΔₐH°"
16 The very high enthalpies of atomisation observed in transition metals directly facilitate which structural phenomenon in compounds of heavy transition metals?
�� High atomisation enthalpy indicates strong metallic bonding. �� Strong interatomic interactions favour metal-metal bonds. �� Heavy transition metals show this phenomenon frequently.
The enthalpy of atomisation measures the strength of metallic bonding within a metal. Heavy transition metals of the second and third series possess exceptionally high enthalpies of atomisation because many d-electrons participate in bonding. These strong interactions often extend into their compounds, leading to the formation of direct metal-metal bonds. Such bonding is considerably more common among heavier transition metals than among first-series transition metals. Therefore, high atomisation enthalpies are directly associated with the frequent occurrence of metal-metal bonding in their compounds.
- �� Option A → Ionic lattice formation is not the direct consequence of high atomisation enthalpy.
- �� Option C → Transition metals generally form solid oxides rather than purely gaseous oxides.
- �� Option D → Disproportionation is unrelated to atomisation enthalpy.
Concept Application
- Application
- Relate metallic bond strength to structural features in compounds.
- Final Logic
- Strong metallic interactions favour direct metal-metal bonding.
"High ΔₐH° → Metal-Metal Bonds"
17 When evaluating the reduction of M²⁺ to M, we use Standard Electrode Potentials. What is the standard unit of the associated Standard Electrode Potential (E°) that ties into Enthalpy of Atomisation and Ionisation Energy?
�� Standard electrode potential measures electrical potential. �� The SI-derived unit is volt. �� It is represented by V.
Standard electrode potential (E°) is a measure of the tendency of a species to undergo reduction or oxidation under standard conditions. Since it represents an electrical potential difference, its unit is the volt (V). Although enthalpy of atomisation and ionisation enthalpy are measured in kJ mol⁻¹, electrode potentials are measured separately in volts. These thermodynamic quantities together help explain the stability and reactivity of transition metals and their ions.
- �� Option A → Unit of entropy, not electrode potential.
- �� Option B → Unit of conductivity.
- �� Option D → Unit of enthalpy.
NCERT Recall
- Application
- Recall the standard unit used for electrode potentials.
- Final Logic
- Electrode potential is measured in volts.
"E° = Electrical Potential = Volt"
18 When predicting the stability of oxidation states from ionisation enthalpies, the unusually high second ionisation enthalpy for Cu (1958 kJ mol⁻¹) indicates that:
�� Cu⁺ possesses a stable d¹⁰ configuration. �� Removing another electron disrupts this stability. �� Therefore, the second ionisation enthalpy is high.
Copper has the electronic configuration [Ar]3d¹⁰4s¹. When one electron is removed, Cu⁺ acquires the highly stable configuration [Ar]3d¹⁰. To form Cu²⁺, an electron must then be removed from this completely filled d-subshell. Since d¹⁰ configurations possess exceptional stability, a large amount of energy is required. Consequently, Copper exhibits an unusually high second ionisation enthalpy. This observation indicates the extra stability associated with the Cu⁺ ion.
- �� Option A → The high ionisation enthalpy indicates the opposite.
- �� Option B → Hydration enthalpy is not the reason for the high second ionisation enthalpy.
- �� Option C → The statement specifically concerns the stability of Cu⁺.
Concept Application
- Application
- Examine the stability of electronic configurations after ionisation.
- Final Logic
- Cu⁺ = stable d¹⁰; removing another electron is difficult.
"Cu⁺ = d¹⁰ = Hard to Remove"
19 Mn⁺ has a 3d⁵4s¹ configuration, and Cr⁺ has a 3d⁵ configuration. Why is the ionisation enthalpy of Mn⁺ lower than that of Cr⁺?
�� Cr⁺ possesses a stable d⁵ configuration. �� d⁵ has maximum exchange-energy stabilization. �� Removing an electron is therefore difficult.
The Cr⁺ ion possesses the exceptionally stable 3d⁵ configuration. This arrangement contains the maximum number of parallel spins in degenerate orbitals and therefore benefits from significant exchange-energy stabilization. Removal of an electron from Cr⁺ destroys this highly stable arrangement and requires a large amount of energy. In contrast, Mn⁺ still contains a 4s electron that can be removed more easily. Consequently, the ionisation enthalpy of Mn⁺ is lower than that of Cr⁺.
- �� Option A → 3d electrons do not perfectly shield 4s electrons.
- �� Option B → Cr⁺ actually benefits more strongly from exchange-energy stabilization.
- �� Option C → Atomic radius is not the primary factor.
Concept Application
- Application
- Apply the concept of exchange-energy stabilization.
- Final Logic
- Breaking a stable d⁵ configuration requires extra energy.
"d⁵ = Maximum Stability"
20 Although thermodynamic ionisation enthalpies give some guidance concerning the relative stabilities of oxidation states, a unique behavior is seen with Copper (Cu) having a positive E° value (+0.34 V). This positive value indicates that:
�� Copper has high atomisation and ionisation energies. �� Hydration energy cannot fully compensate. �� This produces a positive E° value.
Copper possesses relatively high enthalpies of atomisation and ionisation. Considerable energy is required to convert metallic copper into Cu²⁺ ions. Although hydration of Cu²⁺ releases energy, this hydration enthalpy is insufficient to completely compensate for the energy consumed during atomisation and ionisation. As a result, the overall reduction potential of Cu²⁺/Cu becomes positive (+0.34 V). This explains why Copper behaves more nobly than many other transition metals and does not readily liberate hydrogen from dilute non-oxidising acids.
- �� Option A → Copper does not readily liberate hydrogen from dilute acids.
- �� Option C → Copper actually has a relatively high enthalpy of atomisation.
- �� Option D → Positive E° does not imply strong disproportionation.
Concept Application
- Application
- Balance atomisation energy, ionisation energy and hydration enthalpy.
- Final Logic
- Hydration energy cannot fully offset the high energy required to form Cu²⁺.
"Cu: High Cost, Low Compensation"
