d- and f-Block Elements : Class 12 Chemistry Notes (Board Exams, NEET, JEE, CUET)

d- and f-Block Elements

The d- and f-Block Elements chapter is mainly about the elements in which the differentiating electron enters a d-orbital or f-orbital. For Class 12, this chapter becomes much easier when you understand the reason behind their variable oxidation states, coloured ions, magnetic behaviour, catalytic activity, complex formation, alloy formation and lanthanoid/actinoid properties.

For board examinations, focus especially on electronic configuration, oxidation states, trends, important compounds of transition elements, lanthanoid contraction and KMnO₄/K₂Cr₂O₇ reactions. For NEET/JEE, the reasons behind these properties and exceptions are particularly important.

1. What are d-Block Elements?

The elements in which the differentiating electron enters the d-subshell of the penultimate shell are called d-block elements.

They are placed mainly in Groups 3–12 of the periodic table.

The general electronic configuration is:

(n − 1)d¹–¹⁰ ns⁰–²

Here:

n = outermost shell
(n − 1)d = penultimate shell d-subshell
The d-subshell can accommodate a maximum of 10 electrons.

Example

Scandium:

Sc = [Ar] 3d¹ 4s²

Titanium:

Ti = [Ar] 3d² 4s²

Iron:

Fe = [Ar] 3d⁶ 4s²

Copper:

Cu = [Ar] 3d¹⁰ 4s¹

The unusual configuration of Cu is important. Instead of the expected 3d⁹ 4s², it has:

Cu = [Ar] 3d¹⁰ 4s¹

This happens because a completely filled 3d¹⁰ subshell has additional stability.

2. What are Transition Elements?

A transition element is an element whose atom or at least one of its ions has an incompletely filled d-subshell.

This definition creates an important distinction:

Every transition element is a d-block element, but every d-block element is not necessarily a transition element.

Why?

Consider zinc:

Zn = [Ar] 3d¹⁰ 4s²

Zn²⁺:

Zn²⁺ = [Ar] 3d¹⁰

The d-subshell remains completely filled in both Zn and Zn²⁺. Therefore, Zn is a d-block element but not a transition element.

Similarly:

Cd → not a transition element
Hg → not a transition element

Important examples

Element	Configuration	Transition element?
Sc	3d¹ 4s²	Yes
Ti	3d² 4s²	Yes
Fe	3d⁶ 4s²	Yes
Cu	3d¹⁰ 4s¹	Yes
Zn	3d¹⁰ 4s²	No

3. Position of d-Block Elements

The four commonly discussed transition series are:

Series	Elements
3d series	Sc to Zn
4d series	Y to Cd
5d series	Hf to Hg
6d series	Rf onwards

For Class 12, the 3d series is especially important.

3d series

Sc → Ti → V → Cr → Mn → Fe → Co → Ni → Cu → Zn

Atomic numbers:

21 → 30

4. Electronic Configuration of 3d-Series

The general pattern is:

Sc: [Ar] 3d¹ 4s²

Ti: [Ar] 3d² 4s²

V: [Ar] 3d³ 4s²

Cr: [Ar] 3d⁵ 4s¹

Mn: [Ar] 3d⁵ 4s²

Fe: [Ar] 3d⁶ 4s²

Co: [Ar] 3d⁷ 4s²

Ni: [Ar] 3d⁸ 4s²

Cu: [Ar] 3d¹⁰ 4s¹

Zn: [Ar] 3d¹⁰ 4s²

Important exceptions: Cr and Cu

Expected configuration of Cr:

[Ar] 3d⁴ 4s²

Actual:

[Ar] 3d⁵ 4s¹

Expected configuration of Cu:

[Ar] 3d⁹ 4s²

Actual:

[Ar] 3d¹⁰ 4s¹

The reason is the extra stability associated with half-filled d⁵ and completely filled d¹⁰ subshells.

5. Why Do Transition Elements Show Variable Oxidation States?

This is one of the most important concepts in this chapter.

The energies of the ns and (n−1)d electrons are relatively close. Therefore, electrons from both subshells can participate in bonding.

As a result, transition elements can show more than one oxidation state.

Example: Iron

Fe:

[Ar] 3d⁶ 4s²

It can lose:

2 electrons → Fe²⁺
3 electrons → Fe³⁺

Therefore iron commonly shows +2 and +3 oxidation states.

Important oxidation states of 3d elements

Element	Important oxidation states
Sc	+3
Ti	+2, +3, +4
V	+2, +3, +4, +5
Cr	+2, +3, +6
Mn	+2, +3, +4, +6, +7
Fe	+2, +3
Co	+2, +3
Ni	+2, +3
Cu	+1, +2
Zn	+2

Important observation

The maximum oxidation state increases from Sc to Mn and then generally decreases towards Zn.

Manganese can reach +7 because it has seven valence electrons available:

Mn = 3d⁵ 4s²

6. Why is +2 Oxidation State Common?

In many first-row transition elements, the 4s electrons are lost first.

For example:

Fe → Fe²⁺ + 2e⁻

Fe:

[Ar] 3d⁶ 4s²

Fe²⁺:

[Ar] 3d⁶

Therefore, +2 is a common oxidation state.

Important exception: Sc

Sc³⁺ has:

Sc³⁺ = [Ar]

It loses both 4s electrons and one 3d electron.

7. Why Do Transition Elements Show Colour?

Many transition-metal ions are coloured because of d–d electronic transitions.

When light falls on a transition-metal ion, an electron can absorb a particular wavelength of visible light and move from one d-energy level to another.

The remaining transmitted/reflected light gives the substance its observed colour.

Example

Cu²⁺ compounds are often blue or blue-green.

MnO₄⁻ is purple, although its colour is mainly due to charge-transfer transition, not a simple d–d transition.

Important exception

Ions with:

d⁰ or d¹⁰ configuration

generally do not show d–d transitions.

Examples:

Sc³⁺ → d⁰ → colourless
Ti⁴⁺ → d⁰ → colourless
Zn²⁺ → d¹⁰ → colourless
Cu⁺ → d¹⁰ → generally colourless

Common confusion: Not every coloured transition-metal compound gets its colour from a d–d transition. Charge-transfer transitions can also produce intense colours.

8. Why are Transition Elements Paramagnetic?

Paramagnetism occurs when a substance contains unpaired electrons.

The magnetic moment can be estimated using:

μ = √[n(n + 2)] BM

where:

μ = magnetic moment
n = number of unpaired electrons
BM = Bohr Magneton

Example

For Fe³⁺:

Fe:

[Ar] 3d⁶ 4s²

Fe³⁺:

[Ar] 3d⁵

There are 5 unpaired electrons.

Therefore:

μ = √[5(5 + 2)]

μ = √35 BM

Important point

More unpaired electrons generally means stronger paramagnetism.

9. Why Do Transition Elements Form Complex Compounds?

Transition-metal ions have:

small size
relatively high charge
vacant orbitals
ability to accept electron pairs from ligands

Therefore, they readily form coordination compounds.

Example

[Cu(NH₃)₄]²⁺

Here:

Cu²⁺ = central metal ion
NH₃ = ligand
four NH₃ molecules coordinate with Cu²⁺

Another example:

[Fe(CN)₆]⁴⁻

The CN⁻ ions act as ligands.

10. Why are Transition Elements Good Catalysts?

Transition metals and their compounds often act as catalysts because they can:

show variable oxidation states
form intermediate compounds
provide a suitable surface for adsorption
facilitate electron transfer

Examples

Fe is used as a catalyst in the Haber process:

N₂ + 3H₂ ⇌ 2NH₃

V₂O₅ is used in the Contact process:

2SO₂ + O₂ ⇌ 2SO₃

Ni is used in hydrogenation reactions:

RCH=CHR + H₂ → RCH₂–CH₂R

Why variable oxidation states help

A catalyst can temporarily change its oxidation state during a reaction and then return to its original state.

11. Alloy Formation

Transition metals readily form alloys with one another because their atomic sizes are relatively similar.

An alloy is a homogeneous or heterogeneous mixture of metals, or a metal with another element, having useful properties.

Examples:

Steel → Fe + C
Stainless steel → Fe + Cr + Ni + C
Brass → Cu + Zn
Bronze → Cu + Sn

Transition metals are important in alloy production because alloys can have improved:

strength
hardness
corrosion resistance
thermal properties

12. Interstitial Compounds

Transition metals can form compounds by incorporating small atoms such as:

into spaces or interstices in their crystal lattice.

These are called interstitial compounds.

Examples include carbides and nitrides of transition metals.

Properties

They are generally:

hard
high-melting
relatively stable
often retain metallic conductivity

13. Trends in the First Transition Series

Atomic Radii

Across the 3d series, atomic radii generally decrease initially and then become nearly constant.

The decrease is not very large because the added d-electrons provide some shielding.

Why?

As nuclear charge increases:

attraction between nucleus and electrons increases
d-electrons partly shield one another

Therefore, the effective change in atomic size is relatively small.

Ionisation Enthalpy

Ionisation enthalpy generally increases across the transition series, but the trend is not perfectly regular.

This happens because both:

increasing nuclear charge
increasing electron–electron repulsion

affect the energy required to remove an electron.

14. Oxidation States Across the 3d Series

The maximum oxidation state increases up to Mn.

Element	Maximum oxidation state
Sc	+3
Ti	+4
V	+5
Cr	+6
Mn	+7

After Mn, the highest oxidation state generally decreases.

Important idea

Higher oxidation states are often stabilised by highly electronegative elements such as O and F.

Examples:

MnO₄⁻ → Mn(+7)
Cr₂O₇²⁻ → Cr(+6)
CrO₄²⁻ → Cr(+6)

15. Why is Mn²⁺ Particularly Stable?

Mn: [Ar] 3d⁵ 4s²

Mn²⁺: [Ar] 3d⁵

The d⁵ configuration is half-filled, which has extra stability.

Therefore Mn²⁺ is relatively stable.

This is a favourite conceptual question:

Why is Mn²⁺ more stable than Mn³⁺?

Mn²⁺ has 3d⁵, a half-filled stable configuration, whereas Mn³⁺ has 3d⁴.

16. Why is Fe³⁺ Relatively Stable?

Fe:

[Ar] 3d⁶ 4s²

Fe³⁺:

[Ar] 3d⁵

Fe³⁺ therefore has a half-filled d-subshell and gains additional stability.

Fe²⁺:

[Ar] 3d⁶

Hence Fe³⁺ has a particularly stable d⁵ configuration.

17. Why is Cu⁺ Stable in Some Compounds?

Cu: [Ar] 3d¹⁰ 4s¹

Cu⁺: [Ar] 3d¹⁰

Cu⁺ therefore has a completely filled d-subshell.

However, in aqueous solution Cu²⁺ is generally more stable than Cu⁺ because hydration energy strongly favours Cu²⁺.

This distinction is important in exam questions.

18. Important Compounds of Chromium

The most important chromium compounds for Class 12 are:

Potassium dichromate, K₂Cr₂O₇
Potassium chromate, K₂CrO₄

Chromium is present in:

K₂Cr₂O₇ → Cr = +6

K₂CrO₄ → Cr = +6

18.1 Chromate–Dichromate Equilibrium

Chromate ion:

CrO₄²⁻

Dichromate ion:

Cr₂O₇²⁻

In basic medium:

Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O

In acidic medium:

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O

Colour

Chromate ion → yellow
Dichromate ion → orange

Memory trick

Acid → Dichromate → Orange

Base → Chromate → Yellow

19. Potassium Dichromate as an Oxidising Agent

K₂Cr₂O₇ is a strong oxidising agent, especially in acidic medium.

In acidic solution:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

The oxidation state of Cr changes:

+6 → +3

Therefore dichromate accepts electrons and acts as an oxidising agent.

Oxidation of Fe²⁺

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

Here:

Fe²⁺ → Fe³⁺ = oxidation
Cr⁶⁺ → Cr³⁺ = reduction

20. Potassium Permanganate, KMnO₄

KMnO₄ contains manganese in:

+7 oxidation state

It is a powerful oxidising agent.

Its behaviour depends strongly on the medium.

20.1 KMnO₄ in Acidic Medium

In acidic medium:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Mn changes:

+7 → +2

This is the strongest reducing conversion commonly used in acidic redox reactions.

Oxidation of Fe²⁺

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

20.2 KMnO₄ in Neutral or Weakly Basic Medium

Permanganate can be reduced to MnO₂:

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Mn changes:

+7 → +4

MnO₂ is a brown/black solid.

20.3 KMnO₄ in Strongly Basic Medium

Permanganate can form manganate:

MnO₄⁻ + e⁻ → MnO₄²⁻

Mn changes:

+7 → +6

Manganate ion is green.

Therefore remember:

Medium	Main Mn product	Oxidation state	Typical colour
Acidic	Mn²⁺	+2	Very pale pink/near colourless
Neutral/weakly basic	MnO₂	+4	Brown
Strongly basic	MnO₄²⁻	+6	Green

21. Lanthanoids

The lanthanoids are the elements in which the differentiating electron enters the 4f subshell.

They are associated with the period-6 f-block.

General electronic configuration:

[Xe] 4f⁰–¹⁴ 5d⁰–¹ 6s²

The commonly considered lanthanoid series runs from:

La to Lu

The 4f orbitals are progressively filled across the series.

22. Why are Lanthanoids Called Inner Transition Elements?

The differentiating electron enters an inner f-subshell, rather than the outermost shell.

Therefore, lanthanoids and actinoids are called inner transition elements.

23. Common Oxidation State of Lanthanoids

The most common oxidation state is: +3

This occurs because lanthanoids commonly lose:

two 6s electrons
one additional electron from 5d or 4f

Examples

La³⁺

Ce³⁺

Nd³⁺

Gd³⁺

Lu³⁺

23.1 Exceptions to +3 Oxidation State

Some lanthanoids also show +2 or +4 states.

Ce⁴⁺

Ce⁴⁺ is relatively stable because it gives a favourable electronic arrangement related to the empty 4f subshell.

Eu²⁺

Eu²⁺ has:

4f⁷

This is a half-filled f-subshell and is relatively stable.

Yb²⁺

Yb²⁺ has:

4f¹⁴

This is a completely filled f-subshell.

These are important exceptions.

24. Lanthanoid Contraction

One of the most important topics in the chapter is lanthanoid contraction.

Definition

The gradual decrease in the atomic and ionic radii of lanthanoids with increasing atomic number is called lanthanoid contraction.

Why does it happen?

As atomic number increases:

nuclear charge increases
electrons are added to the 4f subshell
4f electrons shield nuclear charge poorly

Therefore, the effective nuclear attraction increases.

As a result:

atomic/ionic size gradually decreases.

25. Consequences of Lanthanoid Contraction

25.1 Similarity of Lanthanoids

Because the ionic radii change only gradually, lanthanoids have very similar chemical properties.

This makes their separation difficult.

25.2 Similarity Between 4d and 5d Elements

Lanthanoid contraction explains why some 4d and 5d elements have very similar sizes.

For example:

Zr and Hf

have very similar atomic radii.

Therefore, their chemical properties are also quite similar.

25.3 Basicity of Hydroxides

The basicity of lanthanoid hydroxides generally decreases across the series.

As ionic size decreases, the charge density of Ln³⁺ increases.

Therefore, the Ln–O interaction becomes stronger and hydroxide becomes less basic.

General trend:

La(OH)₃ > … > Lu(OH)₃

in basic character.

26. Colour of Lanthanoid Ions

Many lanthanoid ions are coloured because of f–f transitions.

The colour is generally weaker than that of many transition-metal ions because f–f transitions are relatively less intense.

Important point

Lanthanoid ions with:

f⁰
f¹⁴

configurations are generally colourless.

Examples:

La³⁺ → 4f⁰ → colourless

Lu³⁺ → 4f¹⁴ → colourless

27. Magnetic Properties of Lanthanoids

Lanthanoid ions can show paramagnetism due to unpaired electrons in the 4f orbitals.

The magnetic behaviour is particularly influenced by the f-electrons.

For Class 12, remember:

More unpaired f-electrons generally means stronger paramagnetic behaviour.

28. Actinoids

The elements in which the differentiating electron enters the 5f subshell are called actinoids.

They are associated with period 7.

General electronic configuration:

[Rn] 5f⁰–¹⁴ 6d⁰–¹ 7s²

The series is generally considered from:

Ac to Lr

29. Oxidation States of Actinoids

Actinoids show greater variation in oxidation states than lanthanoids.

Common oxidation states include:

+3, +4, +5, +6

Some actinoids can show even higher states.

Why is variation greater?

The energies of:

orbitals are relatively close.

Therefore, electrons from these orbitals can participate in bonding.

30. Why are Actinoids More Reactive?

The 5f electrons are less deeply buried than 4f electrons.

Therefore, actinoids can participate more readily in bonding.

This contributes to:

greater variability in oxidation states
greater chemical reactivity
greater tendency to form complexes

31. Actinoid Contraction

Similar to lanthanoids, actinoids show a gradual decrease in atomic and ionic radii across the series.

This is called:

actinoid contraction

It occurs mainly because 5f electrons do not shield the increasing nuclear charge very effectively.

32. Lanthanoids vs Actinoids

Property	Lanthanoids	Actinoids
Subshell filled	4f	5f
Period	6	7
Common oxidation state	+3	+3
Variable oxidation states	Less common	More common
Radioactivity	Mostly non-radioactive except certain isotopes	All are radioactive
Contraction	Lanthanoid contraction	Actinoid contraction
Complex formation	Comparatively less	Generally greater
4f/5f involvement	4f electrons are more deeply buried	5f electrons are more available

33. Why are Actinoids More Complex Than Lanthanoids?

The 5f orbitals are spatially more extended than 4f orbitals.

Therefore, 5f electrons can participate more in:

bonding
complex formation
variable oxidation states

Hence actinoids generally show more complex chemistry.

34. Important Comparison: Transition Elements vs Inner Transition Elements

Feature	Transition elements	Inner transition elements
Differentiating electron	d-orbital	f-orbital
Main series	d-block	f-block
Examples	Fe, Co, Ni	Ce, Nd, U
Variable oxidation states	Common	Especially common in actinoids
Coloured ions	Common	Common for many ions
Complex formation	Strong tendency	Also possible

35. Important NCERT-Based Facts to Remember

Zn, Cd and Hg are d-block elements but not transition elements.
Transition elements generally have partially filled d-orbitals in their atoms or common ions.
Cr and Cu have exceptional ground-state configurations.
Transition elements commonly show variable oxidation states.
Many transition-metal ions are coloured.
d⁰ and d¹⁰ ions generally do not show d–d transitions.
Transition metals often act as catalysts.
Transition metals form alloys and interstitial compounds.
Mn²⁺ = 3d⁵ is especially stable.
Fe³⁺ = 3d⁵ is especially stable.
Cu⁺ = 3d¹⁰ has a completely filled d-subshell.
Chromate is yellow.
Dichromate is orange.
Permanganate is purple.
Manganate is green.
MnO₂ is brown/black.
Lanthanoids predominantly show +3 oxidation state.
Lanthanoid contraction results from poor shielding by 4f electrons.
Actinoids show more variable oxidation states than lanthanoids.
All actinoids are radioactive.

36. Common Student Mistakes

Mistake 1: Saying Zn is a transition element

Incorrect.

Zn is a d-block element, but Zn²⁺ has a completely filled 3d¹⁰ configuration.

Mistake 2: Thinking all coloured transition-metal ions have d–d transitions

Not always.

For example, the intense colour of MnO₄⁻ is mainly due to charge-transfer transition.

Mistake 3: Removing 3d electrons before 4s electrons

For transition-metal cations, electrons are generally removed from 4s before 3d.

Example:

Fe:

[Ar] 3d⁶ 4s²

Fe²⁺:

[Ar] 3d⁶

not 3d⁴ 4s².

Mistake 4: Confusing oxidation state with number of d-electrons

For example:

Fe²⁺ = 3d⁶

Fe³⁺ = 3d⁵

The oxidation state tells you how many electrons were removed relative to the neutral atom; it does not directly tell you the d-electron count without considering the original configuration.

Mistake 5: Confusing chromate and dichromate

CrO₄²⁻ → yellow
Cr₂O₇²⁻ → orange

Mistake 6: Forgetting the medium in KMnO₄ reactions

The product of reduction depends on the medium:

Acidic → Mn²⁺
Neutral/weakly basic → MnO₂
Strongly basic → MnO₄²⁻

37. Board Exam Important Questions

1. Define transition elements.

A transition element is an element whose atom or at least one of its ions has an incompletely filled d-subshell.

2. Why is Zn not a transition element?

Zn and Zn²⁺ have completely filled 3d¹⁰ configurations. Hence Zn does not satisfy the definition of a transition element.

3. Why do transition elements show variable oxidation states?

Because the energies of ns and (n−1)d electrons are relatively close, electrons from both can participate in bonding.

4. Why are transition-metal ions coloured?

Generally because of electronic transitions between split d-orbitals. Charge-transfer transitions can also cause intense colours.

5. Why are transition elements good catalysts?

They can show variable oxidation states, form intermediate compounds and provide suitable surfaces for adsorption.

6. What is lanthanoid contraction?

It is the gradual decrease in atomic and ionic radii across the lanthanoid series due to poor shielding by 4f electrons.

7. Give two consequences of lanthanoid contraction.

Similarity in properties of lanthanoids makes their separation difficult.
Zr and Hf have very similar sizes and chemical properties.

8. Why do actinoids show more oxidation states than lanthanoids?

Because the energies of 5f, 6d and 7s orbitals are relatively close.

38. NEET/JEE-Oriented Concept Points

For objective examinations, pay special attention to these areas:

Cr: 3d⁵ 4s¹
Cu: 3d¹⁰ 4s¹
Zn is not a transition element.
Mn²⁺ → d⁵
Fe³⁺ → d⁵
Cu⁺ → d¹⁰
Sc³⁺ → d⁰
Zn²⁺ → d¹⁰
Maximum oxidation state in the first transition series reaches +7 for Mn.
d⁰ and d¹⁰ ions generally do not show d–d transitions.
Chromate ↔ dichromate equilibrium depends on pH.
KMnO₄ reduction product depends on the medium.
Lanthanoid contraction is caused by poor shielding by 4f electrons.
Actinoids show greater oxidation-state variability.
Zr and Hf similarity is associated with lanthanoid contraction.

39. Practice Questions

Try solving these without looking back at the notes.

Conceptual Questions

Why is Zn classified as a d-block element but not a transition element?
Why do transition elements show variable oxidation states?
Why are many transition-metal compounds coloured?
Why is Mn²⁺ particularly stable?
Why is Fe³⁺ relatively stable?
Why do transition metals form coordination compounds?
Why do transition elements form alloys?
What causes lanthanoid contraction?
Why are actinoids more variable in oxidation state than lanthanoids?
Explain why Zr and Hf have similar properties.

Reaction-Based Questions

Complete and balance the acidic reduction half-reaction of permanganate.
Write the reaction between Fe²⁺ and MnO₄⁻ in acidic medium.
Write the chromate–dichromate equilibrium in acidic medium.
Write the chromate–dichromate equilibrium in basic medium.
Explain the reduction products of KMnO₄ in acidic, neutral and strongly basic media.

Numerical/Applied Question

Calculate the spin-only magnetic moment of an ion containing three unpaired electrons.

Use:

μ = √[n(n + 2)] BM

Answer

For n = 3:

μ = √[3(3 + 2)]

μ = √15 BM

40. 20 Important MCQs

1. Which of the following is a d-block element but not a transition element?

(A) Fe
(B) Cu
(C) Zn
(D) Mn

Correct Answer: (C) Zn

Zn has a completely filled 3d¹⁰ configuration in both Zn and Zn²⁺.

2. The electronic configuration of chromium is:

(A) [Ar] 3d⁴ 4s²
(B) [Ar] 3d⁵ 4s¹
(C) [Ar] 3d³ 4s³
(D) [Ar] 3d⁶ 4s⁰

Correct Answer: (B) [Ar] 3d⁵ 4s¹

Cr gains extra stability from its half-filled 3d⁵ configuration.

3. Which ion has a d⁵ configuration?

(A) Fe²⁺
(B) Mn²⁺
(C) Cu²⁺
(D) Zn²⁺

Correct Answer: (B) Mn²⁺

Mn²⁺ = [Ar] 3d⁵.

4. Which ion has a d¹⁰ configuration?

(A) Fe³⁺
(B) Mn²⁺
(C) Cu⁺
(D) Cr³⁺

Correct Answer: (C) Cu⁺

Cu⁺ = [Ar] 3d¹⁰.

5. The colour of dichromate ion is:

(A) Green
(B) Orange
(C) Yellow
(D) Purple

Correct Answer: (B) Orange

Cr₂O₇²⁻ is orange.

6. The colour of chromate ion is:

(A) Yellow
(B) Orange
(C) Purple
(D) Blue

Correct Answer: (A) Yellow

CrO₄²⁻ is yellow.

7. In acidic medium, MnO₄⁻ is reduced to:

(A) MnO₄²⁻
(B) MnO₂
(C) Mn²⁺
(D) Mn³⁺

Correct Answer: (C) Mn²⁺

In acidic medium:

MnO₄⁻ → Mn²⁺

8. The maximum oxidation state shown by Mn in its common compounds is:

(A) +2
(B) +4
(C) +6
(D) +7

Correct Answer: (D) +7

Mn reaches +7 in species such as MnO₄⁻.

9. Which ion is generally colourless because it has d⁰ configuration?

(A) Ti⁴⁺
(B) Fe²⁺
(C) Cu²⁺
(D) Ni²⁺

Correct Answer: (A) Ti⁴⁺

Ti⁴⁺ has a d⁰ configuration and cannot undergo d–d transition.

10. Lanthanoid contraction is mainly caused by:

(A) Strong shielding by 4f electrons
(B) Poor shielding by 4f electrons
(C) Increasing atomic mass
(D) Decreasing nuclear charge

Correct Answer: (B) Poor shielding by 4f electrons

The 4f electrons shield the increasing nuclear charge poorly.

11. Which pair has very similar atomic sizes due to lanthanoid contraction?

(A) Na and K
(B) Mg and Ca
(C) Zr and Hf
(D) Li and Cs

Correct Answer: (C) Zr and Hf

Lanthanoid contraction makes Hf unusually similar in size to Zr.

12. The most common oxidation state of lanthanoids is:

(A) +1
(B) +2
(C) +3
(D) +6

Correct Answer: (C) +3

The +3 state is dominant for most lanthanoids.

13. Which lanthanoid ion has a particularly stable half-filled 4f⁷ configuration?

(A) Eu²⁺
(B) La³⁺
(C) Lu³⁺
(D) Ce⁴⁺

Correct Answer: (A) Eu²⁺

Eu²⁺ has 4f⁷, a half-filled f-subshell.

14. Which ion has a completely filled f-subshell?

(A) Eu²⁺
(B) Yb²⁺
(C) Ce³⁺
(D) Nd³⁺

Correct Answer: (B) Yb²⁺

Yb²⁺ has 4f¹⁴.

15. Actinoids show greater variability in oxidation states mainly because:

(A) 5f, 6d and 7s orbitals have comparable energies
(B) They have no f-electrons
(C) They have completely filled 5f orbitals
(D) Their atomic size never changes

Correct Answer: (A) 5f, 6d and 7s orbitals have comparable energies

Electrons from these orbitals can participate in bonding.

16. The green species formed from permanganate in strongly basic medium is:

(A) Mn²⁺
(B) MnO₂
(C) MnO₄²⁻
(D) Mn₂O₃

Correct Answer: (C) MnO₄²⁻

Manganate ion, MnO₄²⁻, is green.

17. The number of unpaired electrons in Fe³⁺ is:

(A) 1
(B) 3
(C) 5
(D) 6

Correct Answer: (C) 5

Fe³⁺ = [Ar] 3d⁵.

18. The spin-only magnetic moment of an ion having 2 unpaired electrons is:

(A) √3 BM
(B) √8 BM
(C) √15 BM
(D) √24 BM

Correct Answer: (B) √8 BM

μ = √[2(2 + 2)] = √8 BM.

19. Which element is used as a catalyst in the Haber process?

(A) Ni
(B) Fe
(C) Cu
(D) Zn

Correct Answer: (B) Fe

Iron-based catalysts are used for ammonia synthesis.

20. Which statement about transition elements is correct?

(A) All d-block elements are transition elements.
(B) All transition elements have d⁰ configurations.
(C) Transition elements commonly show variable oxidation states.
(D) Transition elements never form complexes.

Correct Answer: (C) Transition elements commonly show variable oxidation states.

The close energies of ns and (n−1)d orbitals allow different numbers of electrons to participate in bonding.

41. Frequently Asked Questions

1. Are all d-block elements transition elements?

No. Zn, Cd and Hg are d-block elements but are not transition elements because their atoms and common ions have completely filled d-subshells.

2. Why are transition-metal ions often coloured?

Usually because of d–d transitions between split d-orbitals. However, charge-transfer transitions can also produce colour.

3. Why is Zn²⁺ colourless?

Zn²⁺ has a 3d¹⁰ configuration, so d–d transition is not possible.

4. Why is Sc³⁺ colourless?

Sc³⁺ has a 3d⁰ configuration, so there are no d-electrons available for a d–d transition.

5. Why does Mn show +7 oxidation state?

Mn has seven valence electrons in its 3d and 4s orbitals, allowing a maximum common oxidation state of +7, as in permanganate.

6. Why is KMnO₄ a strong oxidising agent?

Mn is in the high +7 oxidation state and can readily accept electrons, undergoing reduction to lower oxidation states.

7. What is the difference between chromate and dichromate?

Chromate is CrO₄²⁻ and yellow, whereas dichromate is Cr₂O₇²⁻ and orange. Their relative amounts depend on the acidity/basicity of the medium.

8. What causes lanthanoid contraction?

Poor shielding of increasing nuclear charge by the 4f electrons causes the gradual decrease in lanthanoid ionic size.

9. Why are lanthanoids difficult to separate?

Their ions have very similar sizes and predominantly show the +3 oxidation state, resulting in very similar chemical properties.

10. Why do actinoids show more oxidation states than lanthanoids?

The energies of 5f, 6d and 7s orbitals are relatively close, allowing different numbers of electrons to participate in bonding.

42. Quick Revision: One-Page Memory Notes

d-Block

General configuration:

(n−1)d¹–¹⁰ ns⁰–²

Transition elements

Atom or ion must have an incomplete d-subshell.

Important exceptions

Cr = 3d⁵ 4s¹

Cu = 3d¹⁰ 4s¹

Major properties

Transition elements show:

variable oxidation states
coloured ions
paramagnetism
complex formation
catalytic activity
alloy formation
interstitial compounds

Important stable configurations

Mn²⁺ → d⁵

Fe³⁺ → d⁵

Cu⁺ → d¹⁰

Sc³⁺ → d⁰

Zn²⁺ → d¹⁰

Chromium

CrO₄²⁻ → yellow

Cr₂O₇²⁻ → orange

Acid favours dichromate; base favours chromate.

Permanganate

Acidic → Mn²⁺

Neutral/weakly basic → MnO₂

Strongly basic → MnO₄²⁻

Lanthanoids

4f filling

Common oxidation state:

Major concept:

Lanthanoid contraction

Cause:

Poor shielding by 4f electrons

Actinoids

5f filling

Common oxidation state:

Show more variable oxidation states.

All actinoids are radioactive.

Final Exam Checklist

Before your Class 12 Chemistry exam, make sure you can confidently answer:

What is a transition element?
Why is Zn not a transition element?
Why do transition elements show variable oxidation states?
Why are transition-metal compounds coloured?
Calculate magnetic moment using μ = √[n(n+2)] BM.
Explain the catalytic behaviour of transition elements.
Write configurations of Cr, Cu, Mn²⁺ and Fe³⁺.
Explain chromate–dichromate equilibrium.
Balance redox reactions involving KMnO₄ and K₂Cr₂O₇.
Explain the effect of acidic, neutral and basic medium on KMnO₄.
Define lanthanoid contraction and explain its causes.
Explain the consequences of lanthanoid contraction.
Compare lanthanoids and actinoids.
Explain why actinoids show more variable oxidation states.

If these concepts are clear, the chapter becomes much more manageable because most of the apparently separate facts—colour, magnetic behaviour, oxidation states, catalytic activity and complex formation—are connected to electronic configuration and orbital energies.