Why do actinoids show a much wider range of oxidation states compared to lanthanoids?

Why do actinoids show a much wider range of oxidation states compared to lanthanoids? +

Correct Answer

Actinoids have 5f, 6d and 7s orbitals of comparable energies, so electrons from all three subshells can participate in bonding.

Detailed Explanation

Both lanthanoids and actinoids are f-block elements, but the energy arrangement of their orbitals is different. In actinoids, the 5f, 6d and 7s orbitals have relatively close energies. Therefore, different numbers of electrons can participate in chemical bonding.

5f ≈ 6d ≈ 7s

Because these orbitals have comparable energies, actinoids can lose or share electrons from more than one subshell. Consequently, actinoids exhibit oxidation states ranging from approximately +3 to +7, with the early actinoids showing the widest variation.

Why Are Lanthanoids Different?

In lanthanoids, the 4f orbitals are more deeply buried and are relatively less available for bonding. The common oxidation state is therefore:

Ln → +3

Some lanthanoids can show +2 or +4 states, but these are generally associated with extra stability of particular electronic configurations. Thus, lanthanoids have a much narrower range of oxidation states than actinoids.

Actinoid Trend

The early actinoids show particularly large variations:

Th(+4), Pa(+5), U(+3,+4,+5,+6), Np(+3 to +7), Pu(+3 to +7)

The highest oxidation states are more accessible in the early actinoids because the 5f electrons are comparatively available for bonding. As we move toward the later actinoids, the 5f electrons become more strongly bound and the +3 state becomes increasingly important.

JEE Level Quick Trick

Actinoids → 5f, 6d & 7s energies are close → more electrons participate → many oxidation states.

Lanthanoids → 4f is buried → mainly +3 oxidation state.

30 Related MCQs with Solutions

1. Actinoids show a wider range of oxidation states mainly because:

A. 5f, 6d and 7s orbitals have comparable energies

B. 4f orbitals are very stable

C. They have no f-electrons

D. They are all gases

Answer: A

2. The most common oxidation state of lanthanoids is:

A. +1

B. +2

C. +3

D. +7

Answer: C. +3

3. Which subshell is involved in actinoid chemistry?

A. 3d

B. 4f

C. 5f

D. 6p

Answer: C. 5f

4. Which orbitals have comparable energies in actinoids?

A. 2s and 2p

B. 3d and 4s

C. 5f, 6d and 7s

D. 4f and 6p only

Answer: C

5. Which element commonly exhibits +6 oxidation state?

A. Uranium

B. Lanthanum

C. Cerium only

D. Lutetium

Answer: A. Uranium

6. Uranium can exhibit which oxidation state?

A. +2 only

B. +3 only

C. +6

D. −3 only

Answer: C. +6

7. The 5f orbitals in early actinoids are:

A. Completely inaccessible

B. Relatively available for bonding

C. Always empty

D. Identical to 4f orbitals

Answer: B

8. Which series generally shows more variable oxidation states?

A. Lanthanoids

B. Actinoids

C. Alkali metals

D. Noble gases

Answer: B. Actinoids

9. The +3 oxidation state is especially important for:

A. Lanthanoids

B. Halogens

C. Noble gases

D. Alkali metals only

Answer: A. Lanthanoids

10. Which orbitals are more deeply buried in lanthanoids?

A. 4f orbitals

B. 5f orbitals

C. 6d orbitals

D. 7s orbitals

Answer: A. 4f orbitals

11. Which actinoid can show oxidation states up to +7?

A. Neptunium

B. Lanthanum

C. Lutetium

D. Gadolinium

Answer: A. Neptunium

12. Which is a typical oxidation state of uranium?

A. +6

B. −6

C. +1 only

D. −2 only

Answer: A. +6

13. The wider oxidation-state range of actinoids is associated with:

A. Greater energy separation of 5f and 6d

B. Small energy difference among 5f, 6d and 7s

C. Absence of 7s electrons

D. Completely filled 5f subshell

Answer: B

14. Which f-block series has 5f filling?

A. Lanthanoids

B. Actinoids

C. Both

D. Neither

Answer: B. Actinoids

15. Lanthanoids mainly show +3 because:

A. 4f electrons are relatively less available for bonding

B. They have no valence electrons

C. They are noble gases

D. Their nuclear charge is zero

Answer: A

16. Which oxidation state is characteristic of many actinoids?

A. +3

B. −7

C. +8 only

D. −5 only

Answer: A. +3

17. Besides +3, actinoids commonly exhibit:

A. +4, +5 and +6

B. −1 only

C. +1 only

D. No other states

Answer: A

18. Which element is known for several oxidation states including +3 to +7?

A. Np

B. La

C. Lu

D. Ce only

Answer: A. Np

19. The 5f electrons of early actinoids are more available than:

A. Lanthanoid 4f electrons

B. Hydrogen electrons

C. Helium electrons

D. Oxygen 2p electrons

Answer: A

20. Which statement is correct?

A. Actinoids show only +3

B. Actinoids show a wide range of oxidation states

C. Lanthanoids show +7 predominantly

D. Both series show exactly the same states

Answer: B

21. The ability of actinoids to use 5f electrons in bonding leads to:

A. Variable oxidation states

B. Constant +1 state

C. Zero oxidation state only

D. Complete chemical inactivity

Answer: A

22. Which orbital is more diffuse and available for bonding in early actinoids compared with 4f?

A. 5f

B. 4f

C. 2s

D. 1s

Answer: A. 5f

23. Which actinoid commonly shows +5 oxidation state?

A. Protactinium

B. Lanthanum

C. Lutetium

D. Ytterbium

Answer: A. Protactinium

24. The higher oxidation states of actinoids are especially common in:

A. Early actinoids

B. Only the last actinoid

C. Noble gases

D. Alkali metals

Answer: A. Early actinoids

25. Towards the later actinoids, which oxidation state becomes increasingly important?

A. +3

B. +7 only

C. −3

D. +8

Answer: A. +3

26. The energy difference between 5f and 6d orbitals in actinoids is:

A. Relatively small

B. Extremely large

C. Infinite

D. Zero for every actinoid

Answer: A

27. Which pair correctly represents the f-block series and their principal differentiating subshell?

A. Lanthanoids—4f; Actinoids—5f

B. Lanthanoids—5f; Actinoids—4f

C. Both—3d

D. Both—5d

Answer: A

28. Which statement best explains the difference between lanthanoids and actinoids?

A. 5f electrons participate more readily in bonding than 4f electrons

B. 4f electrons are more reactive than 5f electrons

C. Actinoids have no f-electrons

D. Lanthanoids have no valence electrons

Answer: A

29. Which sequence correctly represents the general oxidation-state range of actinoids?

A. Mainly +3 with several higher states possible

B. Only −3

C. Only +1

D. Only +8

Answer: A

30. The most important reason for the wider oxidation-state range of actinoids is:

A. Comparable energies of 5f, 6d and 7s orbitals

B. Complete shielding of 5f electrons

C. Absence of d orbitals

D. Very low nuclear charge

Answer: A

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Why does the basic strength of lanthanoid hydroxides decrease systematically from $La(OH)_3$ to $Lu(OH)_3$?

Why does the basic strength of lanthanoid hydroxides decrease systematically from La(OH)₃ to Lu(OH)₃? La(OH)₃ से Lu(OH)₃ तक लैन्थेनाइड हाइड्रॉक्साइड्स की क्षारीयता क्रमशः क्यों घटती है? +

Correct Answer

Due to lanthanoid contraction

4f-electrons show poor shielding → effective nuclear charge increases → Ln³⁺ size decreases → hydroxide becomes less ionic → basic strength decreases.

Detailed Explanation

As we move from La to Lu, the atomic number increases and electrons are progressively added to the 4f-subshell. The 4f-electrons have a poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. This causes a gradual decrease in the size of Ln³⁺ ions. This phenomenon is called lanthanoid contraction.

La³⁺   →   Ce³⁺   →   ...   →   Lu³⁺
Size ↓

As Ln³⁺ becomes smaller, its charge density increases and its attraction for the O²⁻ part of OH⁻ becomes stronger. Therefore, the Ln–OH bond develops greater covalent character and the hydroxide becomes less ionic. Since the basic character of metal hydroxides is associated with their ionic character and ease of providing OH⁻, the basic strength decreases.

La(OH)₃ > Ce(OH)₃ > ... > Lu(OH)₃

The Complete Chain

4f-electrons added
↓
Poor shielding
↓
Z_eff increases
↓
Ln³⁺ ionic radius decreases
↓
Charge density increases
↓
Covalent character of Ln–OH increases
↓
Ionic character decreases
↓
Basic strength decreases

Important Comparison

La(OH)₃ has a relatively large La³⁺ ion and greater ionic character, so it is the most basic hydroxide in the series.

Lu(OH)₃ has the smallest Ln³⁺ ion, greater charge density and greater covalent character, so it is the least basic.

Basicity: La(OH)₃ > Lu(OH)₃

JEE Level Quick Trick

Lanthanoid contraction ↑ → Ln³⁺ size ↓ → covalent character ↑ → basicity ↓

30 Related MCQs with Solutions

1. The basic strength of lanthanoid hydroxides from La to Lu generally:

A. Increases

B. Decreases

C. Remains constant

D. First increases then decreases

Answer: B. Decreases

2. The main reason for the decrease in basicity is:

A. Lanthanoid contraction

B. Inert pair effect

C. Diagonal relationship

D. Hydrogen bonding

Answer: A. Lanthanoid contraction

3. Which hydroxide is generally the most basic?

A. La(OH)₃

B. Gd(OH)₃

C. Lu(OH)₃

D. Yb(OH)₃

Answer: A. La(OH)₃

4. Which hydroxide is least basic in the lanthanoid series?

A. La(OH)₃

B. Ce(OH)₃

C. Lu(OH)₃

D. Pr(OH)₃

Answer: C. Lu(OH)₃

5. Lanthanoid contraction is mainly due to poor shielding by:

A. 3d-electrons

B. 4f-electrons

C. 5d-electrons

D. 6p-electrons

Answer: B. 4f-electrons

6. From La³⁺ to Lu³⁺, ionic radius:

A. Increases

B. Decreases

C. Remains constant

D. Becomes zero

Answer: B. Decreases

7. Smaller Ln³⁺ ions have:

A. Lower charge density

B. Higher charge density

C. No charge density

D. Constant charge density

Answer: B. Higher charge density

8. Smaller Ln³⁺ ions tend to form bonds with greater:

A. Ionic character

B. Covalent character

C. Metallic character

D. Hydrogen character

Answer: B. Covalent character

9. Which has greater ionic character?

A. La(OH)₃

B. Lu(OH)₃

C. Both equal

D. Neither

Answer: A. La(OH)₃

10. Basicity of a metal hydroxide generally increases with:

A. Increasing covalent character

B. Increasing ionic character

C. Increasing polarization

D. Decreasing electropositivity

Answer: B. Increasing ionic character

11. Correct order of basicity is:

A. La(OH)₃ < Lu(OH)₃

B. La(OH)₃ > Lu(OH)₃

C. La(OH)₃ = Lu(OH)₃

D. Cannot be predicted

Answer: B

12. The effective nuclear charge across lanthanoids generally:

A. Decreases

B. Increases

C. Remains constant

D. Becomes negative

Answer: B. Increases

13. Which electron type has poor shielding ability?

A. s

B. p

C. d

D. f

Answer: D. f

14. The basicity trend is directly related to:

A. Lanthanoid contraction

B. Atomic mass only

C. Melting point only

D. Neutron number only

Answer: A. Lanthanoid contraction

15. Which ion has the greatest charge density?

A. La³⁺

B. Lu³⁺

C. Both equal

D. La²⁺

Answer: B. Lu³⁺

16. As ionic radius decreases, polarization of OH⁻:

A. Decreases

B. Increases

C. Remains unchanged

D. Becomes zero

Answer: B. Increases

17. Increased polarization generally leads to:

A. Greater covalent character

B. Greater ionic character

C. No bonding

D. Metallic bonding only

Answer: A. Greater covalent character

18. Which factor does NOT primarily control the basicity trend?

A. Ionic size

B. Charge density

C. Covalent character

D. Colour of the hydroxide

Answer: D

19. The smallest Ln³⁺ ion among the following is:

A. La³⁺

B. Ce³⁺

C. Gd³⁺

D. Lu³⁺

Answer: D. Lu³⁺

20. Which statement explains the basicity trend most accurately?

A. Ionic size increases from La to Lu

B. Ionic size decreases and covalent character increases

C. Nuclear charge decreases

D. 4f shielding becomes stronger

Answer: B

21. Lanthanoid contraction results in an increase in:

A. Ionic radius

B. Charge density

C. Atomic volume

D. Ionic character

Answer: B. Charge density

22. Which hydroxide has the highest covalent character?

A. La(OH)₃

B. Ce(OH)₃

C. Lu(OH)₃

D. Pr(OH)₃

Answer: C. Lu(OH)₃

23. Which hydroxide has the lowest covalent character?

A. La(OH)₃

B. Lu(OH)₃

C. Yb(OH)₃

D. Er(OH)₃

Answer: A. La(OH)₃

24. The basicity trend from La to Lu is:

A. La(OH)₃ > Ce(OH)₃ > ... > Lu(OH)₃

B. Lu(OH)₃ > ... > La(OH)₃

C. Constant

D. Random

Answer: A

25. Which phenomenon causes the gradual decrease in Ln³⁺ radius?

A. Lanthanoid contraction

B. Inert pair effect

C. Resonance

D. Hydrogen bonding

Answer: A

26. Poor shielding by 4f-electrons causes:

A. Decrease in Z_eff

B. Increase in Z_eff

C. No change in Z_eff

D. Negative Z_eff

Answer: B. Increase in Z_eff

27. Greater charge density of Lu³⁺ causes:

A. Lower polarization

B. Higher polarization of OH⁻

C. No interaction

D. Complete dissociation always

Answer: B

28. Which relation is correct?

A. Ionic radius ↑ → covalent character ↑

B. Ionic radius ↓ → covalent character ↑

C. Ionic radius ↓ → charge density ↓

D. Ionic radius ↑ → polarization ↑

Answer: B

29. Which factor makes La(OH)₃ more basic than Lu(OH)₃?

A. La³⁺ is larger and the hydroxide has greater ionic character

B. La³⁺ is smaller

C. La has greater 4f shielding

D. Lu has lower nuclear charge

Answer: A

30. The best one-line explanation for decreasing basicity of lanthanoid hydroxides is:

A. Lanthanoid contraction decreases ionic size and increases covalent character

B. Atomic mass decreases from La to Lu

C. Nuclear charge decreases

D. 4f-electrons shield increasingly well

Answer: A

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Chemistry Chapter 3: Full 100 MCQs (101-200)

Chemistry Chapter 3: Periodicity - Full Set (Questions 101-200)

[Full 100-200 MCQs Set Complete]

Chemistry Chapter 3: Full MCQs 51-100 (JEE/NEET)

Chemistry Chapter 3: Periodicity - Full Set (Questions 51-100)

[Full 50 Questions (51-100) Set Complete for JEE/NEET Level]

Chemistry Chapter 3: Full 50 JEE/NEET MCQ

Chemistry Chapter 3: Classification of Elements (50 Full MCQs)

1. The IUPAC name of element with atomic number 119 is:

A. Ununennium

B. Unbinilium

C. Ununnilium

D. Ununseptium

Answer: A

Solution: For 119: 1=un, 1=un, 9=enn. Name: Ununennium. Symbol: Uue.

2. Maximum number of elements in the 6th period is:

A. 18

B. 32

C. 8

D. 50

Answer: B

Solution: 6th period includes 6s, 4f, 5d, and 6p orbitals (2+14+10+6 = 32 elements).

3. Which block contains metals, non-metals, and metalloids?

A. s-block

B. p-block

C. d-block

D. f-block

Answer: B

Solution: The p-block is the only block containing all three types of elements.

4. The element with Z=114 belongs to:

A. Group 14, Period 7

B. Group 16, Period 7

C. Group 14, Period 6

D. Group 13, Period 7

Answer: A

Solution: It follows Lead (Z=82) in Group 14 (Carbon family).

5. General configuration of f-block elements is:

A. (n-2)f^1-14(n-1)d^0-1ns²

B. (n-1)f^1-14nd^0-1ns²

C. (n-2)f^0-14(n-1)d¹⁰ns²

D. nf^1-14(n-1)d¹ns²

Answer: A

Solution: Electrons fill the (n-2)f subshell.

6. Total number of elements in s-block is:

A. 12

B. 14

C. 10

D. 13

Answer: B

Solution: 2 from Group 1, 2 from Group 2 across 7 periods, but only 14 total (including H and He).

7. Screening effect order for the same shell is:

A. s > p > d > f

B. f > d > p > s

C. s = p = d = f

D. p > s > d > f

Answer: A

Solution: s-orbitals are closer to the nucleus and provide better shielding.

8. Element with Z=33 belongs to:

A. Period 3

B. Period 4

C. Group 13

D. Group 17

Answer: B

Solution: Configuration: [Ar] 3d¹⁰ 4s² 4p³. Highest n=4, so Period 4.

9. Which group contains only metals?

A. Group 1

B. Group 2

C. Group 17

D. Group 18

Answer: B

Solution: Group 1 contains Hydrogen (non-metal), while Group 2 contains only metals.

10. First noble gas is:

A. Neon

B. Helium

C. Argon

D. Radon

Answer: B

Solution: Helium (Z=2) is the first noble gas in Period 1.

11. Largest ionic radius among isoelectronic species O^2-, F^-, Na⁺, Mg²⁺ is:

A. Mg²⁺

B. Na⁺

C. F^-

D. O^2-

Answer: D

Solution: Size increases as nuclear charge (Z) decreases in isoelectronic species. O^2- has Z=8.

12. Atomic radius of Gallium (Ga) is less than Aluminum (Al) due to:

A. Lanthanide contraction

B. Poor shielding by 3d electrons

C. Inert pair effect

D. Large size

Answer: B

Solution: 3d electrons shield poorly, increasing Z_eff and shrinking the atom.

13. Across a period, atomic size decreases due to:

A. Decrease in Z_eff

B. Increase in Z_eff

C. Increase in shielding

D. Addition of shells

Answer: B

Solution: Effective nuclear charge increases while electrons are added to the same shell.

14. Which is the correct size order?

A. I^- > I > I⁺

B. I⁺ > I > I^-

C. I > I^- > I⁺

D. I^- > I⁺ > I

Answer: A

Solution: Anion > Parent Atom > Cation.

15. Lanthanide contraction is caused by:

A. 4f electrons

B. Poor shielding of 4f electrons

C. 5d electrons

D. 6s electrons

Answer: B

Solution: Poor shielding of 4f electrons causes the nucleus to pull 6s electrons more strongly.

16. van der Waals radius is ______ than covalent radius.

A. Smaller

B. Larger

C. Equal

D. Half

Answer: B

Solution: van der Waals radius accounts for non-bonding distance between atoms.

17. Which gas has the largest atomic radius in Period 2?

A. Lithium

B. Fluorine

C. Neon

D. Oxygen

Answer: C

Solution: Noble gases are measured by van der Waals radius, which is much larger than covalent radii.

18. Radius of Fe²⁺ is ______ than Fe³⁺.

A. Larger

B. Smaller

C. Same

D. Triple

Answer: A

Solution: Higher positive charge leads to greater contraction of the electron cloud.

19. Atomic radius increases down a group due to:

A. Increase in Z_eff

B. Addition of new shell

C. Decrease in Z

D. Shielding effect

Answer: B

Solution: Adding a principal energy level outweighs the increase in nuclear charge.

20. Zr and Hf have similar radii due to:

A. Being in same group

B. Lanthanide contraction

C. Same Z

D. Shielding

Answer: B

Solution: Contraction in the 5d series (Hf) cancels out the expected size increase from Zr.

21. 1st Ionization Enthalpy is highest for:

A. B

B. C

C. N

D. O

Answer: C

Solution: Nitrogen has a stable half-filled 2p³ configuration.

22. Element with lowest 1st I.E. is:

A. Li

B. Cs

C. Fr

D. He

Answer: B

Solution: Cesium has the largest size and lowest effective nuclear pull on valence electrons.

23. Successive I.E. values are always:

A. I.E.1 > I.E.2

B. I.E.2 > I.E.1

C. I.E.1 = I.E.2

D. Random

Answer: B

Solution: Removing an electron from a positive ion is always harder due to increased Z/e ratio.

24. Be has higher I.E. than B due to:

A. Penetration effect of 2s

B. Size

C. Nuclear charge

D. Shielding

Answer: A

Solution: 2s electrons of Be are more penetrated towards the nucleus than 2p of B.

25. Noble gases have very high I.E. because:

A. Stable octet

B. Small size

C. High Z_eff

D. All of these

Answer: D

Solution: Full shells and high Z_eff make electron removal extremely difficult.

26. Order of 1st I.E. in Group 13:

A. B > Al > Ga > In > Tl

B. B > Tl > Ga > Al > In

C. B > Ga > Al > Tl > In

D. Al > B > Ga

Answer: B

Solution: Irregular trend due to d-block and f-block contractions.

27. Which ion is isoelectronic with Neon?

A. O^-

B. Mg²⁺

C. Cl^-

D. K⁺

Answer: B

Solution: Mg²⁺ has 10 electrons, same as Ne.

28. I.E. decreases down the group due to:

A. Size increase

B. Shielding increase

C. Both A and B

D. Z increase

Answer: C

Solution: Larger distance and more inner shells reduce the nuclear pull.

29. Highest second I.E. is of:

A. Na

B. Mg

C. Al

D. Si

Answer: A

Solution: Removing 2nd electron from Na⁺ involves breaking a stable noble gas core (2p⁶).

30. Metallic character is highest for:

A. s-block

B. p-block

C. d-block

D. f-block

Answer: A

Solution: Low I.E. allows s-block elements to lose electrons easily.

31. Most negative electron gain enthalpy is of:

A. F

B. Cl

C. Br

D. I

Answer: B

Solution: F is too small, causing interelectronic repulsion for the incoming electron.

32. Most electronegative element is:

A. O

B. F

C. Cl

D. N

Answer: B

Solution: Fluorine has the highest tendency to attract shared electrons (Pauling scale = 4.0).

33. Noble gases have ______ electron gain enthalpy.

A. Highly negative

B. Positive

C. Zero

D. Negative

Answer: B

Solution: Energy must be supplied to force an electron into a higher shell of a stable atom.

34. Electronegativity increases across a period because:

A. Size increases

B. Z_eff increases

C. Shielding increases

D. I.E. decreases

Answer: B

Solution: Stronger nuclear pull attracts bonding electrons more effectively.

35. Order of electronegativity:

A. F > O > N

B. N > O > F

C. O > F > N

D. F > N > O

Answer: A

Solution: Values are roughly F(4.0) > O(3.5) > N(3.0).

36. Pauling scale is based on:

A. Bond energies

B. I.E. and E.A.

C. Atomic size

D. Metallic character

Answer: A

Solution: It relates electronegativity to the extra stability of polar bonds.

37. Element with highest electron affinity:

A. Cl

B. F

C. O

D. S

Answer: A

Solution: Chlorine releases the most energy when accepting an electron.

38. Across a period, non-metallic character:

A. Increases

B. Decreases

C. Constant

D. First increases

Answer: A

Solution: Higher I.E. and EN favor non-metallic behavior.

39. Which is most electropositive?

A. Li

B. Na

C. K

D. Cs

Answer: D

Solution: Cesium loses its valence electron most easily.

40. Nature of oxides changes from left to right as:

A. Acidic to Basic

B. Basic to Acidic

C. Basic to Neutral

D. Neutral to Basic

Answer: B

Solution: Metals form basic oxides; non-metals form acidic oxides.

41. Diagonal relationship is between:

A. Li-Mg

B. Be-Al

C. B-Si

D. All of these

Answer: D

Solution: Diagonal elements share similar charge/radius ratios.

42. Amphoteric oxide is:

A. Na₂O

B. Al₂O₃

C. CO₂

D. SO₂

Answer: B

Solution: Aluminum oxide reacts with both acids and bases.

43. Strongest reducing agent in aqueous solution is:

A. Li

B. Na

C. K

D. Cs

Answer: A

Solution: Lithium has the highest hydration enthalpy due to small size.

44. Strongest oxidizing agent is:

A. F₂

B. Cl₂

C. I₂

D. O₂

Answer: A

Solution: Fluorine has the highest reduction potential.

45. Liquid non-metal is:

A. Hg

B. Br₂

C. Ga

D. S

Answer: B

Solution: Bromine is the only non-metal that is liquid at STP.

46. Most basic hydroxide:

A. NaOH

B. KOH

C. CsOH

D. LiOH

Answer: C

Solution: Basic strength of Group 1 hydroxides increases down the group.

47. Inert pair effect is most prominent in:

A. C

B. Si

C. Ge

D. Pb

Answer: D

Solution: s-electrons of the valence shell become reluctant to bond in heavier p-block elements.

48. Typical elements are:

A. Period 2

B. Period 3

C. Period 4

D. Period 1

Answer: B

Solution: Period 3 elements represent the properties of their respective groups.

49. Nature of Cl₂O₇:

A. Strongly acidic

B. Strongly basic

C. Amphoteric

D. Neutral

Answer: A

Solution: It reacts with water to form Perchloric acid (HClO₄), the strongest mineral acid.

50. Triad system was given by:

A. Mendeleev

B. Dobereiner

C. Newlands

D. Moseley

Answer: B

Solution: Johann Dobereiner arranged elements in groups of three with similar properties.

Screening constant (Sigma) calculation using Slater's rule for 4s electron.

1. Screening constant (σ) calculation using Slater's rule for 4s electron +

Correct Answer — General Method

For a 4s electron, calculate σ by Slater's rules according to the electron configuration of the atom.

Slater's Rules for a 4s Electron

For an electron in the ns or np orbital, such as 4s, Slater's rules are:

σ = 0.35 × (other electrons in the same 4s,4p group) + 0.85 × (electrons in the n−1 shell) + 1.00 × (electrons in n−2 or lower shells)

The electron being considered is not counted in calculating σ.

For a 4s electron:

• Other electrons in 4s, 4p → 0.35 each
• Electrons in 3s, 3p, 3d → 0.85 each
• Electrons in 1s, 2s, 2p → 1.00 each

Example: Calculate σ for a 4s Electron of Potassium (K, Z = 19)

Electronic configuration of K:

K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

We want the shielding constant for the 4s¹ electron.

Step 1: Same group

There are no other electrons in the 4s/4p group.

0 × 0.35 = 0

Step 2: n−1 shell

The n−1 shell is the third shell: 3s² + 3p⁶ = 8 electrons

8 × 0.85 = 6.80

Step 3: n−2 and lower shells

1s² + 2s² + 2p⁶ = 10 electrons

10 × 1.00 = 10

Step 4: Total screening constant

σ = 0 + 6.80 + 10
σ = 16.80

Therefore:

For K, σ(4s) = 16.80

Effective Nuclear Charge

After calculating σ, effective nuclear charge is:

Z_eff = Z − σ

For K:

Z_eff = 19 − 16.80
Z_eff = 2.20

Thus, the 4s electron of potassium experiences an effective nuclear charge of approximately +2.20.

Important Exam Point

Do not use one fixed σ value for every 4s electron. The screening constant depends on the atomic number and electronic configuration of the atom/ion being considered. For example, for Ca:

Ca: [Ar]4s²

For one 4s electron, the other 4s electron contributes:

1 × 0.35

while the eight 3s/3p electrons contribute:

8 × 0.85

and the ten electrons in n−2 and lower shells contribute:

10 × 1.00

Hence:

σ = 0.35 + 6.80 + 10
σ = 17.15

JEE/NEET Shortcut

For a 4s/4p electron, remember:

Same shell/group → 0.35
One shell below → 0.85
Two or more shells below → 1.00

Then use:

Z_eff = Z − σ

30 Related MCQs with Solutions

1. According to Slater's rule, the contribution of another electron in the same ns/np group is:

A. 0.10

B. 0.35

C. 0.85

D. 1.00

Answer: B. 0.35

2. For a 4s electron, electrons in the 3s and 3p orbitals contribute:

A. 0.35 each

B. 0.50 each

C. 0.85 each

D. 1.00 each

Answer: C. 0.85 each

3. For a 4s electron, electrons in 1s, 2s and 2p contribute:

A. 0.35 each

B. 0.50 each

C. 0.85 each

D. 1.00 each

Answer: D. 1.00 each

4. Which electron is excluded while calculating its own screening?

A. All inner electrons

B. The electron under consideration

C. All 1s electrons

D. All d electrons

Answer: B

5. The electronic configuration of K is:

A. [Ar]4s¹

B. [Ne]3s²

C. [Ar]3d¹

D. [Kr]5s¹

Answer: A. [Ar]4s¹

6. The screening constant for the 4s electron of K is:

A. 10.00

B. 16.80

C. 18.15

D. 19.00

Answer: B. 16.80

7. For K, the 3s²3p⁶ electrons contribute:

A. 6.80

B. 8.00

C. 10.00

D. 2.80

Answer: A. 6.80

8. For K, the electrons in 1s²2s²2p⁶ contribute:

A. 3.50

B. 6.80

C. 8.50

D. 10.00

Answer: D. 10.00

9. Effective nuclear charge is calculated using:

A. Z + σ

B. Z − σ

C. σ − Z

D. Z × σ

Answer: B. Z − σ

10. The Z_eff of the 4s electron in K is:

A. 1.20

B. 2.20

C. 3.20

D. 4.20

Answer: B. 2.20

11. Slater's rules are used to estimate:

A. Atomic mass

B. Effective nuclear charge

C. Neutron number

D. Melting point

Answer: B

12. The shielding effect causes the outer electron to experience:

A. Full nuclear charge

B. Reduced effective nuclear attraction

C. No nuclear attraction

D. Negative nuclear charge

Answer: B

13. For a 4s electron, which orbital belongs to the same principal shell?

A. 3p

B. 3d

C. 4p

D. 2p

Answer: C. 4p

14. For an ns/np electron, the n−1 shell contributes:

A. 0.35 per electron

B. 0.50 per electron

C. 0.85 per electron

D. 1.50 per electron

Answer: C

15. For a 4s electron, the n−2 shell corresponds to:

A. n=3

B. n=2

C. n=1

D. n=5

Answer: B. n=2

16. For K, the number of electrons contributing 1.00 each is:

A. 2

B. 8

C. 10

D. 18

Answer: C. 10

17. For Ca ([Ar]4s²), another 4s electron contributes:

A. 0.35

B. 0.85

C. 1.00

D. 2.00

Answer: A. 0.35

18. The screening constant for one 4s electron of Ca is:

A. 16.80

B. 17.15

C. 18.00

D. 20.00

Answer: B. 17.15

19. For Ca, the approximate Z_eff of a 4s electron is:

A. 1.85

B. 2.85

C. 3.85

D. 4.85

Answer: B. 2.85

20. Which quantity increases when σ decreases for a fixed Z?

A. Z_eff

B. Atomic mass

C. Neutron number

D. Principal quantum number

Answer: A. Z_eff

21. Greater shielding generally causes:

A. Greater effective nuclear attraction

B. Lower effective nuclear attraction

C. No change in attraction

D. Nuclear decay

Answer: B

22. Which electrons shield a 4s electron most effectively according to Slater's rules?

A. Same 4s/4p group: 0.35 each

B. 3s/3p: 0.85 each

C. 2s/2p: 1.00 each

D. All have equal contribution

Answer: C. 2s/2p electrons contribute 1.00 each

23. Which factor is NOT directly used in calculating σ by Slater's rule?

A. Number of electrons

B. Electron grouping

C. Principal shell position

D. Atomic mass

Answer: D. Atomic mass

24. If σ increases while Z remains constant, Z_eff:

A. Increases

B. Decreases

C. Becomes infinite

D. Becomes negative necessarily

Answer: B. Decreases

25. The 4s electron of K is shielded by how many inner-shell electrons?

A. 8

B. 10

C. 18

D. 19

Answer: C. 18 inner-shell electrons

26. In Slater's method, the shielding contribution of a 3d electron to a 4s electron is:

A. 0.35

B. 0.50

C. 0.85

D. 1.00

Answer: C. 0.85

27. For a 4s electron in a transition-metal atom, which group is especially important to check?

A. 1s only

B. 3d electrons

C. 5s electrons only

D. 6p electrons

Answer: B. 3d electrons

28. The main purpose of the shielding constant σ is to represent:

A. Nuclear charge reduction due to other electrons

B. Number of protons

C. Number of neutrons

D. Atomic volume

Answer: A

29. For a 4s electron, which contribution is strongest per electron?

A. Same-group electron

B. n−1 electron

C. n−2 or lower electron

D. All are equal

Answer: C. 1.00 per electron

30. The correct sequence for calculating Z_eff is:

A. Find Z → find σ → calculate Z−σ

B. Find σ → calculate Z+σ

C. Find atomic mass → calculate σ

D. Find neutrons → calculate σ

Answer: A

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