Why is the density of 5d transition series elements (e.g., Hf, Ta, W) almost double that of their corresponding 4d series elements (e.g., Zr, Nb, Mo)?

Why is the density of 5d transition series elements (e.g., Hf, Ta, W) almost double that of their corresponding 4d series elements (e.g., Zr, Nb, Mo)? +

Correct Answer

Due to lanthanoid contraction, the atomic volumes of 5d elements become very small while their atomic masses are much higher than those of corresponding 4d elements.

Detailed Explanation

Density is given by:

ρ = Mass / Volume

The 5d transition elements such as Hf, Ta and W have considerably larger atomic masses than their corresponding 4d elements Zr, Nb and Mo. At the same time, the lanthanoid contraction causes the atomic and metallic radii of the 5d elements to be unexpectedly close to those of the corresponding 4d elements. For example:

Zr ≈ Hf    |    Nb ≈ Ta    |    Mo ≈ W

Thus, their atomic volumes are relatively similar, but the 5d elements contain much heavier atoms. Therefore:

Higher mass + nearly similar volume → much higher density

This is why the densities of many 5d elements are approximately twice those of their corresponding 4d elements.

Role of Lanthanoid Contraction

Between the 4d and 5d series, the lanthanoids are filled with electrons in the 4f subshell. The 4f-electrons shield the nuclear charge poorly. Consequently, effective nuclear charge increases and pulls the outer electrons inward. This produces lanthanoid contraction.

Poor 4f shielding → Z_eff ↑ → atomic radius ↓

As a result, the size of 5d elements is much closer to that of their 4d counterparts than would normally be expected.

Example: Zr vs Hf

Zr (4d series): atomic mass ≈ 91.2 u
Hf (5d series): atomic mass ≈ 178.5 u

Hf has almost twice the atomic mass of Zr, while their atomic sizes are surprisingly similar because of lanthanoid contraction. Therefore, Hf has a much higher density.

ρ(Hf) ≫ ρ(Zr)

The same general idea applies to pairs such as:

Nb → Ta    and    Mo → W

JEE / NEET Quick Trick

5d density ↑↑ because Mass ↑ greatly, but Volume remains nearly similar due to Lanthanoid Contraction.

Mass ↑ + Volume ≈ constant → Density ↑

30 Related MCQs with Solutions

1. The density of many 5d transition elements is much higher than that of corresponding 4d elements mainly because:

A. Their atomic masses are much higher while atomic volumes are similar

B. Their atomic masses are lower

C. Their atomic radii are much larger

D. They have fewer electrons

Answer: A

2. Which phenomenon makes the sizes of Hf and Zr very similar?

A. Inert pair effect

B. Lanthanoid contraction

C. Diagonal relationship

D. Relativistic effect only

Answer: B. Lanthanoid contraction

3. Density is defined as:

A. Volume/Mass

B. Mass/Volume

C. Mass × Volume

D. Mass + Volume

Answer: B. Mass/Volume

4. Which pair belongs to corresponding 4d and 5d series?

A. Zr and Hf

B. Na and K

C. Ca and Sr

D. Sc and Y

Answer: A. Zr and Hf

5. The atomic mass of Hf is approximately:

A. 91 u

B. 120 u

C. 178.5 u

D. 207 u

Answer: C. 178.5 u

6. The atomic mass of Zr is approximately:

A. 40 u

B. 91 u

C. 178 u

D. 238 u

Answer: B. 91 u

7. Which statement about Zr and Hf is correct?

A. Hf is much larger than Zr

B. Their radii are surprisingly similar

C. Zr is much heavier than Hf

D. They belong to the same period

Answer: B

8. Lanthanoid contraction is primarily caused by:

A. Excellent shielding by 4f-electrons

B. Poor shielding by 4f-electrons

C. Absence of 4f-electrons

D. 5d shielding

Answer: B

9. Which 5d element corresponds to Nb?

A. Ta

B. W

C. Hf

D. Re

Answer: A. Ta

10. Which 5d element corresponds to Mo?

A. Hf

B. Ta

C. W

D. Re

Answer: C. W

11. Which 5d element corresponds to Zr?

A. Hf

B. Ta

C. W

D. Os

Answer: A. Hf

12. The high density of W compared with Mo is primarily related to:

A. Lower mass of W

B. Higher mass of W and relatively small atomic volume

C. Larger volume of W only

D. Absence of d-electrons

Answer: B

13. Which electrons show poor shielding in lanthanoids?

A. 1s

B. 2p

C. 3d

D. 4f

Answer: D. 4f

14. Due to lanthanoid contraction, the 5d elements have:

A. Much larger radii than 4d elements

B. Radii comparable to corresponding 4d elements

C. Zero atomic radius

D. Radii smaller than all 3d elements

Answer: B

15. The density of an element increases when:

A. Mass decreases and volume increases

B. Mass increases while volume remains similar

C. Both mass and volume decrease equally

D. Mass becomes zero

Answer: B

16. Which pair has nearly similar atomic radii because of lanthanoid contraction?

A. Zr and Hf

B. Na and K

C. Li and Cs

D. Mg and Ba

Answer: A

17. The 5d series follows the:

A. 3d series

B. 4d series

C. 6d series

D. 4f series only

Answer: B. 4d series

18. Which of the following has higher density?

A. Zr

B. Hf

C. Both have exactly equal density

D. Cannot be predicted

Answer: B. Hf

19. The unexpectedly similar size of Zr and Hf is due to:

A. Poor shielding by 4f-electrons

B. Excellent shielding by 5d-electrons

C. Inert pair effect

D. Hydrogen bonding

Answer: A

20. Which transition series contains Hf, Ta and W?

A. 3d

B. 4d

C. 5d

D. 6d

Answer: C. 5d

21. Which transition series contains Zr, Nb and Mo?

A. 3d

B. 4d

C. 5d

D. 6d

Answer: B. 4d

22. The main factor increasing density from 4d to 5d counterparts is:

A. Higher atomic mass

B. Lower nuclear charge

C. Larger atomic volume

D. Fewer protons

Answer: A. Higher atomic mass

23. The relatively small atomic volume of 5d elements is associated with:

A. Lanthanoid contraction

B. Alkali contraction

C. Hydrogen contraction

D. Inert gas expansion

Answer: A. Lanthanoid contraction

24. Which quantity is in the denominator of density?

A. Mass

B. Volume

C. Atomic number

D. Nuclear charge

Answer: B. Volume

25. Which of the following is a consequence of lanthanoid contraction?

A. Hf and Zr have very different radii

B. Hf and Zr have similar radii

C. Hf has no d-electrons

D. Zr becomes a lanthanoid

Answer: B

26. Which statement is correct regarding Ta and Nb?

A. Ta has much lower atomic mass

B. Ta has higher atomic mass and much higher density

C. Nb is a 5d element

D. Both are lanthanoids

Answer: B

27. Which element is the densest among the following?

A. Zr

B. Hf

C. Nb

D. Ta

Answer: D. Ta

28. Which element has a density close to that expected from a much heavier atom because of its compact atomic volume?

A. Hf

B. Na

C. K

D. Li

Answer: A. Hf

29. The correct relationship is:

A. ρ = V/M

B. ρ = M/V

C. ρ = M×V

D. ρ = M+V

Answer: B. ρ = M/V

30. The best explanation for the unusually high density of 5d transition metals is:

A. Higher atomic mass combined with lanthanoid-contraction-induced small atomic volume

B. Lower atomic mass and larger volume

C. Absence of f-electrons

D. Low nuclear charge

Answer: A

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What is the primary cause of Lanthanoid Contraction?

What is the primary cause of Lanthanoid Contraction? +

Correct Answer

Poor shielding effect of the 4f-electrons

Detailed Explanation

Lanthanoid contraction is the gradual decrease in atomic and ionic radii of the lanthanoid elements from La to Lu with increasing atomic number. The main reason is the poor shielding ability of the 4f-electrons. As we move from La to Lu:

• The nuclear charge increases because the atomic number increases.
• Electrons are progressively added to the 4f subshell.
• The 4f-electrons shield the nuclear charge ineffectively.
• Therefore, the effective nuclear charge experienced by the outer electrons increases.
• The electron cloud is pulled closer to the nucleus.
• Hence, atomic and ionic radii gradually decrease.

Increasing nuclear charge + poor 4f shielding → increasing Z_eff → decreasing radius

Why is 4f Shielding Poor?

The shielding effectiveness generally follows:

s > p > d > f

Since 4f orbitals have poor penetration and are relatively diffuse, their shielding of the nuclear charge is weak. Therefore, the increasing nuclear charge is not effectively cancelled by the added 4f-electrons.

Major Consequences of Lanthanoid Contraction

1. Similarity between 4d and 5d elements
For example, Zr and Hf have very similar atomic radii and chemical properties.

2. Decrease in basicity of lanthanoid hydroxides
Basicity generally decreases from La(OH)₃ to Lu(OH)₃.

3. Difficulty in separation of lanthanoids
Their chemical properties become increasingly similar.

JEE / NEET Quick Trick

Lanthanoid Contraction = Poor shielding by 4f-electrons

4f shielding ↓ → Z_eff ↑ → Radius ↓

30 Related MCQs with Solutions

1. The primary cause of lanthanoid contraction is:

A. Increase in atomic mass

B. Poor shielding by 4f-electrons

C. Increase in number of shells

D. Strong shielding by 4f-electrons

Answer: B. Poor shielding by 4f-electrons

2. Lanthanoid contraction refers to:

A. Increase in atomic radius across lanthanoids

B. Decrease in atomic/ionic radii across lanthanoids

C. Increase in ionic charge

D. Decrease in atomic number

Answer: B

3. Which electrons are mainly responsible for lanthanoid contraction?

A. 2p

B. 3d

C. 4f

D. 5d

Answer: C. 4f

4. The shielding efficiency of 4f-electrons is:

A. Very high

B. Moderate

C. Poor

D. Perfect

Answer: C. Poor

5. Across the lanthanoid series, effective nuclear charge generally:

A. Decreases

B. Remains constant

C. Increases

D. Becomes zero

Answer: C. Increases

6. The gradual decrease in ionic radii from La³⁺ to Lu³⁺ is called:

A. Inert pair effect

B. Lanthanoid contraction

C. Diagonal relationship

D. Shielding effect

Answer: B

7. Which orbital has the poorest shielding ability?

A. s

B. p

C. d

D. f

Answer: D. f

8. Correct order of shielding effectiveness is:

A. f > d > p > s

B. s > p > d > f

C. p > s > f > d

D. d > f > s > p

Answer: B. s > p > d > f

9. Due to lanthanoid contraction, the size of Ln³⁺ ions:

A. Increases

B. Decreases

C. Remains unchanged

D. First increases then decreases

Answer: B. Decreases

10. Lanthanoid contraction results mainly from increasing:

A. Atomic volume

B. Effective nuclear charge

C. Shielding effect

D. Number of shells

Answer: B. Effective nuclear charge

11. Which pair has unusually similar atomic radii because of lanthanoid contraction?

A. Na and K

B. Zr and Hf

C. Li and Na

D. Mg and Ca

Answer: B. Zr and Hf

12. The 4f-electrons are added progressively in:

A. Alkali metals

B. Lanthanoids

C. Halogens

D. Noble gases

Answer: B. Lanthanoids

13. Lanthanoid contraction causes the basicity of Ln(OH)₃ to generally:

A. Increase across the series

B. Decrease across the series

C. Remain constant

D. Become zero

Answer: B. Decrease

14. Which statement is correct?

A. 4f-electrons shield very effectively

B. 4f-electrons shield poorly

C. 4f-electrons are absent in lanthanoids

D. 4f-electrons increase atomic radius

Answer: B

15. The effective nuclear charge increases because:

A. Nuclear charge increases faster than shielding

B. Nuclear charge decreases

C. 4f shielding is perfect

D. Electrons disappear

Answer: A

16. Which series shows lanthanoid contraction?

A. Li to Ne

B. Sc to Zn

C. La to Lu

D. Na to Ar

Answer: C. La to Lu

17. Which of the following is NOT a consequence of lanthanoid contraction?

A. Similarity of Zr and Hf

B. Decrease in basicity of hydroxides

C. Gradual decrease in Ln³⁺ radii

D. Large increase in atomic radius from La to Lu

Answer: D

18. Poor shielding is mainly due to the nature of:

A. s-orbitals

B. p-orbitals

C. d-orbitals

D. f-orbitals

Answer: D. f-orbitals

19. As atomic number increases across lanthanoids, nuclear charge:

A. Decreases

B. Increases

C. Remains constant

D. Becomes negative

Answer: B. Increases

20. Lanthanoid contraction is most directly associated with:

A. 4f filling

B. 3d filling

C. 5p filling

D. 6s filling

Answer: A. 4f filling

21. Which statement best explains lanthanoid contraction?

A. Added 4f electrons completely cancel added nuclear charge

B. Added 4f electrons poorly shield the increasing nuclear charge

C. Nuclear charge decreases

D. Number of shells decreases

Answer: B

22. The radius of Hf is unexpectedly similar to that of:

A. Ti

B. Zr

C. Sc

D. Y

Answer: B. Zr

23. Which electrons have the least penetrating power among s, p, d and f?

A. s

B. p

C. d

D. f

Answer: D. f

24. Lanthanoid contraction is a result of competition between:

A. Nuclear charge and shielding

B. Mass and volume

C. Ionization and electron affinity only

D. Melting point and boiling point

Answer: A

25. Which ion has a smaller radius: La³⁺ or Lu³⁺?

A. La³⁺

B. Lu³⁺

C. Both are equal

D. Cannot be predicted

Answer: B. Lu³⁺

26. Which property generally decreases from La(OH)₃ to Lu(OH)₃?

A. Basicity

B. Atomic number

C. Nuclear charge

D. Number of electrons

Answer: A. Basicity

27. Which effect explains the similarity of the second and third transition series?

A. Inert pair effect

B. Lanthanoid contraction

C. Diagonal effect

D. Photoelectric effect

Answer: B. Lanthanoid contraction

28. Which is the correct sequence of shielding effectiveness?

A. f > d > p > s

B. s > p > d > f

C. s > d > p > f

D. p > s > d > f

Answer: B

29. The net effect of lanthanoid contraction is:

A. Increasing size

B. Decreasing size

C. No change in size

D. Random change in size

Answer: B. Decreasing size

30. The best one-line explanation of lanthanoid contraction is:

A. Strong shielding by 4f-electrons

B. Poor shielding by 4f-electrons causing increasing effective nuclear charge

C. Decreasing nuclear charge

D. Addition of new shells

Answer: B

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Which of the following statements regarding interstitial compounds is incorrect?

Which of the following statements regarding interstitial compounds is incorrect? +

Correct Answer

Incorrect statement: Interstitial compounds are always non-stoichiometric.

Detailed Explanation

Interstitial compounds are formed when small atoms such as H, B, C or N occupy the interstitial spaces (holes) in the crystal lattice of transition metals. Common examples include:

Fe₃C   |   TiC   |   VN   |   Fe₄N

Their important characteristics are:

• They are generally hard and rigid.
• They often have high melting points.
• They generally retain the metallic conductivity of the parent metal.
• They are usually formed by small atoms occupying interstitial sites in a metal lattice.
• They may be non-stoichiometric, but they are not necessarily always non-stoichiometric.

Therefore, saying that all interstitial compounds are necessarily non-stoichiometric is incorrect.

Why are they called Interstitial Compounds?

The small atoms do not normally replace metal atoms in the crystal lattice. Instead, they occupy the empty spaces between the metal atoms.

Metal lattice + small atom in interstitial site → Interstitial compound

For example, carbon atoms can occupy the spaces in the iron lattice to form iron carbide.

JEE / NEET Quick Revision

Interstitial compounds → small atoms in lattice holes → hard + high m.p. + metallic conductivity → often non-stoichiometric, but not always.

30 Related MCQs with Solutions

1. Interstitial compounds are formed when small atoms occupy:

A. Substitutional sites only

B. Interstitial sites

C. Nuclei of metal atoms

D. Valence shells only

Answer: B. Interstitial sites

2. Which of the following atoms commonly forms interstitial compounds?

A. Na

B. K

C. C

D. Cs

Answer: C. Carbon

3. Interstitial compounds are mainly associated with:

A. Alkali metals

B. Transition metals

C. Noble gases

D. Halogens

Answer: B. Transition metals

4. Which is an example of an interstitial compound?

A. NaCl

B. TiC

C. H₂O

D. CO₂

Answer: B. TiC

5. Interstitial compounds are generally:

A. Soft

B. Hard

C. Gaseous

D. Volatile liquids

Answer: B. Hard

6. Interstitial compounds generally have:

A. Very low melting points

B. High melting points

C. No melting point

D. Melting points below room temperature

Answer: B. High melting points

7. Which property is usually retained by interstitial compounds?

A. Metallic conductivity

B. Molecular volatility

C. Ionic conductivity only

D. Gas-like behaviour

Answer: A. Metallic conductivity

8. Which statement is incorrect?

A. Small atoms occupy lattice holes

B. They are generally hard

C. They always have fixed stoichiometry

D. They can retain metallic character

Answer: C.
Interstitial compounds can often show variable composition.

9. Which element can occupy interstitial positions in transition-metal lattices?

A. Hydrogen

B. Carbon

C. Nitrogen

D. All of these

Answer: D. All of these

10. The small atoms in interstitial compounds are generally:

A. Larger than metal atoms

B. Smaller than metal atoms

C. Equal in size to metal atoms

D. Always noble gases

Answer: B. Smaller than metal atoms

11. Which compound is associated with iron and carbon?

A. FeCl₃

B. Fe₃C

C. FeSO₄

D. Fe(NO₃)₃

Answer: B. Fe₃C

12. Which of the following is a carbide-type interstitial compound?

A. TiC

B. NaCl

C. MgO

D. CaCO₃

Answer: A. TiC

13. Interstitial compounds are generally formed by:

A. Large non-metals entering lattice sites

B. Small atoms entering lattice holes

C. Complete replacement of metal atoms

D. Noble gases entering the nucleus

Answer: B

14. Which is NOT a typical property of interstitial compounds?

A. High hardness

B. High melting point

C. Metallic conductivity

D. High volatility

Answer: D. High volatility

15. Interstitial compounds may be:

A. Only stoichiometric

B. Only non-stoichiometric

C. Stoichiometric or non-stoichiometric

D. Always molecular

Answer: C

16. The term "interstitial" refers to:

A. Surface of a crystal

B. Space between atoms in a lattice

C. Nucleus of an atom

D. Electron cloud

Answer: B

17. Which element is particularly small and can enter metal lattice holes?

A. H

B. Cs

C. Rb

D. Ba

Answer: A. Hydrogen

18. Interstitial compounds are usually:

A. Non-metallic insulators

B. Metallic or metal-like in conductivity

C. Molecular gases

D. Ionic liquids

Answer: B

19. Which statement about stoichiometry is correct?

A. All interstitial compounds must be non-stoichiometric

B. Some interstitial compounds have definite stoichiometry

C. None have definite composition

D. Stoichiometry is impossible

Answer: B

20. Which pair contains interstitial compounds?

A. TiC and VN

B. NaCl and KBr

C. H₂O and NH₃

D. CO₂ and SO₂

Answer: A. TiC and VN

21. Interstitial atoms mainly cause:

A. Weakening of the metallic lattice

B. Distortion and strengthening of the lattice

C. Complete destruction of lattice

D. Formation of a gas

Answer: B

22. Which is an important industrial feature of many interstitial compounds?

A. High hardness

B. Extreme volatility

C. Low thermal stability

D. Low melting point

Answer: A. High hardness

23. Carbon in iron lattice can form:

A. Fe₃C

B. FeCl₃

C. FeO₄

D. Fe(NO₃)₂

Answer: A. Fe₃C

24. Which small atoms are commonly involved in interstitial compounds?

A. H, B, C, N

B. Na, K, Rb, Cs

C. Cl, Br, I, At

D. Xe, Kr, Ar, Ne

Answer: A

25. Which statement is true?

A. Interstitial compounds are always ionic

B. Interstitial compounds are generally associated with metallic lattices

C. Interstitial compounds are always molecular

D. They contain no metals

Answer: B

26. Which property results from strong interaction between the interstitial atom and metal lattice?

A. Increased hardness

B. Increased volatility

C. Gas formation

D. Loss of all conductivity

Answer: A

27. Which statement is incorrect regarding interstitial compounds?

A. Small atoms occupy lattice holes

B. They are generally hard

C. They are always non-stoichiometric

D. They may have high melting points

Answer: C. They are always non-stoichiometric

28. VN is an example of an interstitial compound involving:

A. Vanadium and nitrogen

B. Vanadium and sodium

C. Vanadium and neon

D. Vanadium and chlorine

Answer: A. Vanadium and nitrogen

29. The electrical conductivity of many interstitial compounds is:

A. Metallic

B. Always zero

C. Only ionic

D. Molecular

Answer: A. Metallic

30. The best description of interstitial compounds is:

A. Small atoms occupy interstitial sites of a metal lattice

B. Large atoms replace all metal atoms

C. Only non-metals form them

D. They are always gaseous

Answer: A

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The intense deep purple color of Potassium Permanganate ($KMnO_4$) is attributed to which of the following?

The intense deep purple color of Potassium Permanganate (KMnO₄) is attributed to which of the following? +

Correct Answer

Ligand-to-Metal Charge Transfer (LMCT) from O²⁻ to Mn⁷⁺.

Detailed Explanation

In potassium permanganate:

KMnO₄ → K⁺ + MnO₄⁻

The oxidation state of manganese is:

+1 + x + 4(−2) = 0
x = +7

Thus, manganese is present in the very high +7 oxidation state. Mn⁷⁺ has a d⁰ configuration. Therefore, the intense purple colour is not due to ordinary d–d transitions. Instead, when light falls on MnO₄⁻, an electron can be transferred from an oxygen-based orbital to a manganese-based orbital:

O → Mn charge-transfer transition

This is called a Ligand-to-Metal Charge Transfer (LMCT) transition. Because charge-transfer transitions are highly allowed, they absorb strongly in the visible region, producing the characteristic intense deep-purple colour.

Why Not d–d Transition?

A common competitive-exam trap is to say that the purple colour is caused by a d–d transition. This is incorrect. Mn in MnO₄⁻ is:

Mn⁷⁺ = d⁰

Since there are no d-electrons, a conventional d–d transition cannot occur. Therefore:

Deep purple colour → LMCT, not d–d transition

JEE / NEET Quick Trick

d⁰ or d¹⁰ species can still be coloured due to charge-transfer transitions.

KMnO₄: Mn⁷⁺ → d⁰ → LMCT → Purple

30 Related MCQs with Solutions

1. The intense purple colour of KMnO₄ is due to:

A. d–d transition

B. Ligand-to-metal charge transfer

C. Metal-to-ligand charge transfer

D. f–f transition

Answer: B. Ligand-to-metal charge transfer (LMCT)

2. Oxidation state of Mn in MnO₄⁻ is:

A. +2

B. +4

C. +6

D. +7

Answer: D. +7

3. The d-electron configuration of Mn⁷⁺ is:

A. d⁰

B. d²

C. d⁵

D. d¹⁰

Answer: A. d⁰

4. Can MnO₄⁻ show a normal d–d transition?

A. Yes

B. No

C. Only in solid state

D. Only at high pressure

Answer: B. No
Mn⁷⁺ is d⁰.

5. LMCT means:

A. Ligand-to-Metal Charge Transfer

B. Ligand-Metal Crystal Transition

C. Light-Metal Charge Transfer

D. Lone-pair Metal Transfer

Answer: A

6. In LMCT of permanganate, electron density moves mainly:

A. Mn → O

B. O → Mn

C. K → O

D. K → Mn

Answer: B. O → Mn

7. Which ion is responsible for the purple colour of KMnO₄?

A. K⁺

B. Mn²⁺

C. MnO₄⁻

D. O²⁻

Answer: C. MnO₄⁻

8. Which transition is generally intense?

A. Spin-forbidden d–d

B. Charge-transfer transition

C. Nuclear transition

D. Rotational transition

Answer: B. Charge-transfer transition

9. The colour of permanganate ion is:

A. Deep purple

B. Green

C. Yellow

D. Colourless

Answer: A. Deep purple

10. Which statement is correct?

A. MnO₄⁻ is coloured because Mn⁷⁺ has many d-electrons

B. MnO₄⁻ is coloured due to charge transfer

C. KMnO₄ is coloured due to K⁺

D. MnO₄⁻ has an f–f transition

Answer: B

11. MnO₄⁻ contains manganese in:

A. +2 oxidation state

B. +3 oxidation state

C. +6 oxidation state

D. +7 oxidation state

Answer: D

12. Mn⁷⁺ has how many d-electrons?

A. 0

B. 2

C. 5

D. 7

Answer: A. 0

13. A d⁰ complex can be coloured mainly because of:

A. d–d transitions

B. Charge-transfer transitions

C. f–f transitions

D. Nuclear decay

Answer: B

14. Which orbital is mainly the electron donor in LMCT of permanganate?

A. Oxygen-based orbital

B. Potassium orbital

C. Hydrogen orbital

D. Chlorine orbital

Answer: A. Oxygen-based orbital

15. Which is NOT responsible for the purple colour of MnO₄⁻?

A. Charge transfer

B. O → Mn electron transfer

C. Conventional d–d transition

D. Absorption of visible light

Answer: C

16. The charge on permanganate ion is:

A. +1

B. −1

C. −2

D. +2

Answer: B. −1

17. KMnO₄ contains:

A. K²⁺ and MnO₄⁻

B. K⁺ and MnO₄⁻

C. K⁺ and MnO₄²⁻

D. K²⁺ and MnO₄²⁻

Answer: B

18. Charge-transfer bands are usually:

A. Very weak

B. Relatively intense

C. Always invisible

D. Nuclear in origin

Answer: B. Relatively intense

19. Which ion is commonly cited as a d⁰ coloured species?

A. MnO₄⁻

B. Zn²⁺

C. Sc³⁺

D. Ca²⁺

Answer: A. MnO₄⁻

20. Why is KMnO₄ a strong oxidising agent?

A. Mn is in a high +7 oxidation state

B. K is a noble gas

C. It has d¹⁰ Mn

D. It contains no oxygen

Answer: A

21. Which statement about MnO₄⁻ is correct?

A. It is d⁵

B. It is d⁰

C. It is d¹⁰

D. It is f⁷

Answer: B. d⁰

22. The purple colour results from absorption of:

A. Visible light

B. Only radio waves

C. Only microwaves

D. Only X-rays

Answer: A. Visible light

23. Which is the best explanation of the colour?

A. Mn⁷⁺ undergoes d–d transition

B. Electron transfer from O to Mn occurs upon light absorption

C. K⁺ absorbs visible light strongly

D. Mn has 10 d-electrons

Answer: B

24. Which oxidation state corresponds to d⁰ manganese?

A. +2

B. +4

C. +6

D. +7

Answer: D. +7

25. The ligand in permanganate responsible for LMCT is:

A. O

B. K

C. Mn

D. None

Answer: A. Oxygen

26. Which type of transition is generally much more intense than a d–d transition?

A. Charge-transfer transition

B. Spin transition only

C. Nuclear transition

D. Vibrational transition

Answer: A

27. In LMCT, the electron moves:

A. From ligand to metal

B. From metal to ligand

C. From K to O

D. From nucleus to ligand

Answer: A

28. Which species is responsible for the characteristic colour of permanganate solutions?

A. K⁺

B. MnO₄⁻

C. H₂O

D. Mn metal

Answer: B

29. The oxidation state of Mn in KMnO₄ is:

A. +2

B. +4

C. +6

D. +7

Answer: D. +7

30. The correct match is:

A. KMnO₄ — d–d transition

B. KMnO₄ — LMCT — deep purple

C. KMnO₄ — f–f transition

D. KMnO₄ — K⁺ transition

Answer: B. LMCT — deep purple

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What is the chemical formula of the reddish-brown vapors evolved during a positive Chromyl Chloride test, and what is the oxidation state of chromium in it?

What is the chemical formula of the reddish-brown vapors evolved during a positive Chromyl Chloride test, and what is the oxidation state of chromium in it? +

Correct Answer

Reddish-brown vapors are Chromyl chloride (CrO₂Cl₂).
The oxidation state of chromium is +6.

Why?

In the Chromyl Chloride Test, a chloride-containing salt is heated with potassium dichromate (K₂Cr₂O₇) and concentrated H₂SO₄. If chloride ions are present, reddish-brown vapours of chromyl chloride are produced.

CrO₂Cl₂

To calculate the oxidation state of Cr: Let oxidation state of Cr = x. Oxygen has −2 oxidation state and chlorine has −1.

x + 2(−2) + 2(−1) = 0
x − 4 − 2 = 0
x = +6

Therefore, chromium is in the +6 oxidation state.

Chromyl Chloride Test — Key Reaction

A simplified representation is:

Cr₂O₇²⁻ + 4Cl⁻ + 6H⁺ → 2CrO₂Cl₂ + 3H₂O

The characteristic reddish-brown vapours indicate a positive Chromyl Chloride Test for chloride.

Important Exam Point

Reddish-brown vapour → CrO₂Cl₂ → Cr = +6

30 Related MCQs with Solutions

1. The reddish-brown vapours in the Chromyl Chloride Test are:

A. CrCl₃

B. CrO₂Cl₂

C. CrO₃

D. CrCl₂

Answer: B. CrO₂Cl₂

2. The common name of CrO₂Cl₂ is:

A. Chromium chloride

B. Chromyl chloride

C. Chromous chloride

D. Chromic chloride

Answer: B. Chromyl chloride

3. Oxidation state of chromium in CrO₂Cl₂ is:

A. +2

B. +3

C. +5

D. +6

Answer: D. +6

4. Which colour is characteristic of chromyl chloride vapours?

A. Green

B. Blue

C. Reddish-brown

D. Yellow

Answer: C. Reddish-brown

5. Chromyl Chloride Test is primarily used to detect:

A. Sulphate

B. Nitrate

C. Chloride

D. Carbonate

Answer: C. Chloride

6. The oxidising agent commonly used in the test is:

A. KMnO₄

B. K₂Cr₂O₇

C. NaOH

D. HCl

Answer: B. K₂Cr₂O₇

7. Which acid is used along with dichromate in the test?

A. Dilute HCl

B. Concentrated H₂SO₄

C. HNO₃

D. CH₃COOH

Answer: B. Concentrated H₂SO₄

8. Oxidation state of oxygen in CrO₂Cl₂ is:

A. −1

B. −2

C. +1

D. +2

Answer: B. −2

9. Oxidation state of chlorine in CrO₂Cl₂ is:

A. +1

B. 0

C. −1

D. +7

Answer: C. −1

10. In CrO₂Cl₂, the total contribution of oxygen to oxidation number is:

A. −2

B. −4

C. −6

D. +4

Answer: B. −4

11. Chromyl chloride has how many oxygen atoms?

A. 1

B. 2

C. 3

D. 4

Answer: B. 2

12. Chromyl chloride has how many chlorine atoms?

A. 1

B. 2

C. 3

D. 4

Answer: B. 2

13. The formula CrO₂Cl₂ is electrically:

A. +2 charged

B. −2 charged

C. Neutral

D. +6 charged

Answer: C. Neutral

14. The chromium species responsible for the reddish-brown vapour contains Cr in:

A. +2 state

B. +3 state

C. +6 state

D. 0 state

Answer: C. +6 state

15. Which ion is converted into chromyl chloride during the test?

A. Br⁻

B. I⁻

C. Cl⁻

D. SO₄²⁻

Answer: C. Cl⁻

16. A positive Chromyl Chloride Test gives:

A. Reddish-brown fumes

B. White precipitate only

C. Blue solution

D. Black precipitate

Answer: A. Reddish-brown fumes

17. Which chromium compound is used as the chromium source in the test?

A. K₂Cr₂O₇

B. CrCl₃

C. CrO

D. Cr₂O₃

Answer: A. K₂Cr₂O₇

18. The oxidation state of Cr in dichromate ion is:

A. +3

B. +4

C. +6

D. +7

Answer: C. +6

19. Which observation confirms chloride in the Chromyl Chloride Test?

A. Reddish-brown vapours

B. Green precipitate

C. Blue precipitate

D. No colour change

Answer: A

20. The correct formula is:

A. CrOCl₂

B. CrO₂Cl

C. CrO₂Cl₂

D. Cr₂OCl₂

Answer: C. CrO₂Cl₂

21. The oxidation state calculation for CrO₂Cl₂ is:

A. x − 4 − 2 = 0

B. x + 4 + 2 = 0

C. x − 2 + 2 = 0

D. x + 2 − 4 = 0

Answer: A

22. Chromium in +6 oxidation state belongs to:

A. Chromous state

B. Chromic state

C. High oxidation state

D. Zero state

Answer: C. High oxidation state

23. Chromyl chloride is best described as:

A. CrO₂Cl₂

B. CrCl₃

C. CrO₃

D. K₂Cr₂O₇

Answer: A

24. The colour of chromyl chloride vapour is mainly described as:

A. Reddish-brown

B. Violet

C. Pale green

D. Colourless

Answer: A

25. The Chromyl Chloride Test belongs to qualitative analysis of:

A. Anions

B. Cations only

C. Organic compounds only

D. Noble gases

Answer: A. Anions

26. Which oxidation state is common to Cr in both dichromate and chromyl chloride?

A. +2

B. +3

C. +6

D. +7

Answer: C. +6

27. If chloride is absent, the characteristic chromyl chloride vapour:

A. Should not be formed

B. Must always form

C. Becomes blue

D. Becomes green

Answer: A

28. Which element is present in chromyl chloride besides Cr, O and Cl?

A. Na

B. K

C. No additional element

D. S

Answer: C

29. Chromyl chloride is a useful test because its vapour has a:

A. Characteristic reddish-brown colour

B. Characteristic black colour

C. Characteristic blue precipitate

D. Characteristic white colour

Answer: A

30. The correct match is:

A. CrO₂Cl₂ — Cr(+3)

B. CrO₂Cl₂ — Cr(+6)

C. CrCl₃ — Cr(+6)

D. CrO₃ — Cr(+2)

Answer: B. CrO₂Cl₂ — Cr(+6)

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The standard reduction potential ($E^\circ$) for the $Mn^{3+}/Mn^{2+}$ couple is highly positive ($+1.57\text{ V}$), whereas it is negative for the $Cr^{3+}/Cr^{2+}$ couple. Why?

The standard reduction potential (E°) for the Mn³⁺/Mn²⁺ couple is highly positive (+1.57 V), whereas it is negative for the Cr³⁺/Cr²⁺ couple. Why? +

Correct Answer

Mn³⁺ → Mn²⁺ is highly favourable because Mn²⁺ has a particularly stable 3d⁵ half-filled configuration, whereas Cr³⁺ → Cr²⁺ requires destroying the stable 3d³ arrangement of Cr³⁺.

Electronic Configurations

For manganese:

Mn: [Ar] 3d⁵4s²
Mn²⁺: [Ar] 3d⁵
Mn³⁺: [Ar] 3d⁴

Thus:

Mn³⁺(3d⁴) + e⁻ → Mn²⁺(3d⁵)

The product Mn²⁺ has a half-filled 3d⁵ configuration, which has extra stability due to exchange energy and symmetrical distribution of electrons. Therefore, Mn³⁺ has a strong tendency to accept an electron. Hence:

E°(Mn³⁺/Mn²⁺) = +1.57 V

What happens in Chromium?

Chromium has:

Cr: [Ar] 3d⁵4s¹
Cr³⁺: [Ar] 3d³
Cr²⁺: [Ar] 3d⁴

The reduction reaction is:

Cr³⁺(3d³) + e⁻ → Cr²⁺(3d⁴)

Cr³⁺ has a relatively stable 3d³ configuration. Reduction to Cr²⁺ gives 3d⁴, which does not gain the special half-filled d⁵ stability. Therefore, Cr³⁺ is less willing to accept an electron and the reduction is less favourable. Consequently:

E°(Cr³⁺/Cr²⁺) < 0

Why is the Difference So Large?

The standard reduction potential is determined by the overall free-energy change of the redox process. Electronic configuration is a major factor, but it is not the only factor. For Mn:

3d⁴ → 3d⁵

This produces a particularly stable half-filled d-subshell. For Cr:

3d³ → 3d⁴

This does not produce an especially stable half-filled or fully filled configuration. Thus the thermodynamic driving force for reduction is much greater for Mn³⁺ than for Cr³⁺.

JEE-Level Shortcut

Remember:

Mn³⁺ → Mn²⁺: 3d⁴ → 3d⁵ → half-filled stability → high +E°

Cr³⁺ → Cr²⁺: 3d³ → 3d⁴ → no special extra stability → low/negative E°

30 Related MCQs with Solutions

1. The electronic configuration of Mn²⁺ is:

A. 3d³

B. 3d⁴

C. 3d⁵

D. 3d⁶

Answer: C. 3d⁵

2. The configuration of Mn³⁺ is:

A. 3d³

B. 3d⁴

C. 3d⁵

D. 3d⁶

Answer: B. 3d⁴

3. Mn³⁺ reduction gives:

A. 3d³

B. 3d⁴

C. 3d⁵

D. 3d⁶

Answer: C. 3d⁵

4. Mn²⁺ is particularly stable because of:

A. 3d²

B. 3d³

C. 3d⁵

D. 3d¹⁰

Answer: C. Half-filled 3d⁵

5. The standard reduction potential of Mn³⁺/Mn²⁺ is approximately:

A. −1.57 V

B. −0.26 V

C. +1.57 V

D. 0 V

Answer: C. +1.57 V

6. A positive E° for Mn³⁺/Mn²⁺ indicates:

A. Reduction is favourable

B. Oxidation is impossible

C. No redox reaction occurs

D. Mn²⁺ is unstable

Answer: A

7. Cr³⁺ has which d-electron configuration?

A. 3d²

B. 3d³

C. 3d⁴

D. 3d⁵

Answer: B. 3d³

8. Cr²⁺ has:

A. 3d²

B. 3d³

C. 3d⁴

D. 3d⁵

Answer: C. 3d⁴

9. Which ion has a half-filled d-subshell?

A. Cr³⁺

B. Mn²⁺

C. Cr²⁺

D. Mn³⁺

Answer: B. Mn²⁺

10. The half-filled d⁵ configuration is stabilised by:

A. Exchange energy

B. Zero electron density

C. Low nuclear charge

D. Large atomic radius

Answer: A. Exchange energy

11. Reduction of Mn³⁺ involves:

A. Loss of electron

B. Gain of electron

C. Loss of proton

D. Gain of neutron

Answer: B. Gain of electron

12. Reduction of Cr³⁺ gives:

A. Cr⁺

B. Cr²⁺

C. Cr⁴⁺

D. Cr⁵⁺

Answer: B. Cr²⁺

13. Which reduction is more favourable?

A. Mn³⁺ → Mn²⁺

B. Cr³⁺ → Cr²⁺

C. Both equally

D. Neither

Answer: A

14. The unusually high E° of Mn³⁺/Mn²⁺ is mainly associated with formation of:

A. 3d³

B. 3d⁴

C. 3d⁵

D. 3d¹⁰

Answer: C. 3d⁵

15. Which configuration is especially stable among the following?

A. d⁴

B. d⁵

C. d⁶

D. d⁷

Answer: B. d⁵

16. Cr³⁺ is comparatively stable because of its:

A. d³ configuration

B. d⁵ configuration

C. d¹⁰ configuration

D. d¹ configuration

Answer: A. d³

17. Cr³⁺ → Cr²⁺ changes:

A. d³ → d²

B. d³ → d⁴

C. d⁴ → d⁵

D. d⁵ → d⁴

Answer: B

18. Which statement is correct about E°?

A. More positive E° generally means greater tendency for reduction

B. More positive E° means oxidation is always spontaneous

C. E° has no relation to redox tendency

D. E° is always zero

Answer: A

19. Mn²⁺ has how many unpaired d-electrons in the free-ion picture?

A. 1

B. 2

C. 3

D. 5

Answer: D. 5

20. Which configuration has maximum exchange energy among these?

A. d⁴

B. d⁵

C. d⁶

D. d⁸

Answer: B. d⁵

21. The standard reduction potential of Cr³⁺/Cr²⁺ is:

A. Highly positive

B. Negative

C. Exactly +1.57 V

D. Always zero

Answer: B. Negative

22. The reduction potential depends on:

A. Only electronic configuration

B. Overall thermodynamics of the redox process

C. Only atomic mass

D. Only colour

Answer: B

23. Mn³⁺ is a strong:

A. Reducing agent

B. Oxidising agent

C. Neutral molecule

D. Ligand

Answer: B. Oxidising agent
Its high positive reduction potential means it readily accepts an electron.

24. Cr²⁺ tends to act as a:

A. Strong reducing agent

B. Strong oxidising agent only

C. Noble gas

D. Non-redox species

Answer: A. Strong reducing agent

25. The stable configuration of Mn²⁺ is:

A. d²

B. d³

C. d⁵

D. d⁸

Answer: C. d⁵

26. Which transition-metal ion is strongly associated with a half-filled d-subshell?

A. Mn²⁺

B. Cr³⁺

C. Cu²⁺

D. Ni²⁺

Answer: A. Mn²⁺

27. Which change gives a half-filled d-subshell?

A. Mn³⁺ + e⁻ → Mn²⁺

B. Cr³⁺ + e⁻ → Cr²⁺

C. Fe³⁺ + e⁻ → Fe²⁺

D. Cu²⁺ + e⁻ → Cu⁺

Answer: A

28. A highly positive standard reduction potential indicates:

A. Strong tendency to gain electrons

B. Strong tendency to lose electrons

C. No electron transfer

D. Metallic bonding only

Answer: A

29. The unusual E° values of Cr and Mn are related to:

A. d-electron configurations and their relative stability

B. Only atomic size

C. Only density

D. Only melting point

Answer: A

30. The best explanation for the difference between Mn³⁺/Mn²⁺ and Cr³⁺/Cr²⁺ is:

A. Mn³⁺ reduction produces stable d⁵, whereas Cr³⁺ reduction produces d⁴

B. Both produce d⁵

C. Cr³⁺ has no d-electrons

D. Mn²⁺ has d¹⁰

Answer: A

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Why is the $Cu^{2+}$ ion more stable in an aqueous solution than the $Cu^+$ ion, despite $Cu^+$ having a stable, fully-filled $3d^{10}$ configuration?

Why is the Cu²⁺ ion more stable in an aqueous solution than the Cu⁺ ion, despite Cu⁺ having a stable, fully-filled 3d¹⁰ configuration? +

Correct Answer

Cu²⁺ is more stable in aqueous solution mainly because it has a much larger hydration enthalpy than Cu⁺.

Detailed Explanation

The electronic configurations are:

Cu: [Ar] 3d¹⁰4s¹

Cu⁺: [Ar] 3d¹⁰
Cu²⁺: [Ar] 3d⁹

At first sight, Cu⁺ appears more stable because it possesses the completely filled 3d¹⁰ subshell. However, in an aqueous solution, electronic configuration alone does not determine the overall stability of an ion. The important factor is the balance between:

Ionisation enthalpy + Hydration enthalpy

Cu²⁺ has a charge of +2 and a relatively small ionic radius. Therefore, it attracts water molecules much more strongly than Cu⁺. Hence:

|Hydration enthalpy of Cu²⁺| >> |Hydration enthalpy of Cu⁺|

The large negative hydration enthalpy releases a considerable amount of energy and stabilises Cu²⁺ in water. Although the formation of Cu²⁺ requires removal of another electron from Cu⁺, the energy released during hydration of Cu²⁺ more than compensates for this unfavourable ionisation step.

Why Cu⁺ is unstable in Water

Cu⁺ tends to undergo disproportionation in aqueous solution:

2Cu⁺ → Cu²⁺ + Cu

Thus, Cu⁺ simultaneously undergoes oxidation and reduction:

• One Cu⁺ is oxidised to Cu²⁺.
• Another Cu⁺ is reduced to metallic Cu.

This reaction is thermodynamically favourable in aqueous medium. Therefore, Cu⁺ is comparatively unstable in water, while Cu²⁺ is much better stabilised by hydration.

JEE-Level Concept

The key point is:

Stable electronic configuration ≠ maximum stability in every medium

For aqueous ions, the total stability depends on several energetic factors, especially:

• Ionisation enthalpy
• Hydration enthalpy
• Crystal/lattice effects where applicable
• Overall thermodynamic feasibility

For Cu:

Cu⁺ → Cu²⁺

The second ionisation is energetically costly, but Cu²⁺ receives a much greater hydration stabilisation because of its higher charge density.

Cu²⁺ + H₂O → strongly hydrated Cu²⁺

This makes Cu²⁺ thermodynamically favoured in aqueous solution.

Quick Revision

Cu⁺: 3d¹⁰ but relatively poor hydration
Cu²⁺: 3d⁹ but much stronger hydration

Therefore, in aqueous medium: Cu²⁺ > Cu⁺ in stability

30 Related MCQs with Solutions

1. Electronic configuration of Cu⁺ is:

A. 3d⁹

B. 3d¹⁰

C. 3d⁸4s¹

D. 3d¹⁰4s¹

Answer: B. 3d¹⁰

2. Electronic configuration of Cu²⁺ is:

A. 3d¹⁰

B. 3d⁹

C. 3d⁸

D. 3d¹⁰4s¹

Answer: B. 3d⁹

3. Which ion is more stable in aqueous solution?

A. Cu⁺

B. Cu²⁺

C. Both equally

D. Neither

Answer: B. Cu²⁺

4. The major factor stabilising Cu²⁺ in aqueous solution is:

A. 3d¹⁰ configuration

B. Large hydration enthalpy

C. Low electronegativity

D. Large size

Answer: B. Large hydration enthalpy

5. Which ion has greater charge density?

A. Cu⁺

B. Cu²⁺

C. Both same

D. Cu atom

Answer: B. Cu²⁺

6. Hydration enthalpy of Cu²⁺ is generally:

A. Less negative than Cu⁺

B. More negative than Cu⁺

C. Zero

D. Positive

Answer: B. More negative than Cu⁺

7. Cu⁺ in aqueous solution tends to undergo:

A. Hydrolysis only

B. Disproportionation

C. Polymerisation

D. Sublimation

Answer: B. Disproportionation

8. The disproportionation reaction of Cu⁺ is:

A. Cu⁺ → Cu

B. 2Cu⁺ → Cu²⁺ + Cu

C. Cu²⁺ → Cu⁺

D. Cu → Cu²⁺

Answer: B

9. Cu⁺ has a:

A. Half-filled d-subshell

B. Completely filled d-subshell

C. Empty d-subshell

D. Half-filled p-subshell

Answer: B. Completely filled 3d¹⁰ subshell

10. Cu²⁺ has how many d-electrons?

A. 7

B. 8

C. 9

D. 10

Answer: C. 9

11. The hydration process involves:

A. Removal of electrons by water

B. Solvation of an ion by water molecules

C. Formation of a metal

D. Removal of water

Answer: B

12. Higher ionic charge generally results in:

A. Weaker hydration

B. Stronger hydration

C. No hydration

D. Lower charge density

Answer: B. Stronger hydration

13. Cu²⁺ is better hydrated than Cu⁺ because:

A. It has lower charge

B. It has higher charge density

C. It has no d-electrons

D. It is neutral

Answer: B

14. Which factor cannot be ignored when comparing ionic stability in water?

A. Hydration enthalpy

B. Only electronic configuration

C. Atomic number only

D. Colour only

Answer: A

15. Cu⁺ disproportionation produces:

A. Cu and Cu²⁺

B. Only Cu²⁺

C. Only Cu

D. Cu³⁺

Answer: A

16. The filled 3d¹⁰ configuration of Cu⁺ makes it:

A. Electronically favourable

B. Always the most stable ion in water

C. Impossible to oxidise

D. A noble gas

Answer: A
It provides electronic stability, but total aqueous stability also depends strongly on hydration and redox energetics.

17. Which ion experiences stronger electrostatic attraction toward water molecules?

A. Cu⁺

B. Cu²⁺

C. Cu

D. Both Cu⁺ and Cu²⁺ equally

Answer: B. Cu²⁺

18. Which statement is correct?

A. Electronic configuration alone determines ionic stability

B. Hydration can alter the relative stability of ions in water

C. Cu⁺ cannot exist

D. Cu²⁺ has 3d¹⁰

Answer: B

19. The oxidation state of Cu in CuSO₄ is:

A. 0

B. +1

C. +2

D. +3

Answer: C. +2

20. CuCl contains copper mainly in:

A. +1 state

B. +2 state

C. +3 state

D. 0 state

Answer: A. +1

21. Which process releases substantial energy for a highly charged small ion?

A. Hydration

B. Sublimation

C. Ionisation only

D. Reduction only

Answer: A. Hydration

22. The stability of an ion in solution is best understood using:

A. Only Hund's rule

B. Only Aufbau principle

C. Overall thermodynamics

D. Only Pauli principle

Answer: C. Overall thermodynamics

23. Cu⁺ is particularly susceptible to disproportionation in:

A. Aqueous solution

B. Vacuum

C. Solid copper metal only

D. Dry nitrogen only

Answer: A. Aqueous solution

24. Which configuration belongs to neutral copper?

A. [Ar]3d⁹4s²

B. [Ar]3d¹⁰4s¹

C. [Ar]3d⁸4s²

D. [Ar]3d¹⁰

Answer: B

25. Which statement about Cu⁺ is false?

A. It has 3d¹⁰ configuration

B. It can undergo disproportionation

C. It is strongly stabilised by hydration compared with Cu²⁺

D. It can form CuCl

Answer: C
Cu²⁺ is much more strongly hydrated.

26. Which ion has the greater hydration enthalpy magnitude?

A. Cu⁺

B. Cu²⁺

C. Both equal

D. Cu atom

Answer: B. Cu²⁺

27. The main reason Cu²⁺ is favoured in water is:

A. It has a completely filled d-subshell

B. Strong hydration compensates for ionisation energy

C. It is neutral

D. It has no unpaired electrons

Answer: B

28. In the reaction 2Cu⁺ → Cu²⁺ + Cu, Cu⁺ acts as:

A. Only oxidising agent

B. Only reducing agent

C. Both oxidising and reducing species

D. Catalyst

Answer: C

29. Which concept explains why Cu²⁺ can be more stable than the electronically attractive Cu⁺ in water?

A. Hydration enthalpy

B. Boyle's law

C. Raoult's law

D. Le Chatelier's principle only

Answer: A. Hydration enthalpy

30. Which is the best conclusion?

A. Cu⁺ is always more stable because of 3d¹⁰

B. Cu²⁺ is favoured in aqueous medium because of strong hydration and overall thermodynamics

C. Cu²⁺ cannot exist in water

D. Cu⁺ has a 3d⁹ configuration

Answer: B

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