The d- and f-Block Elements — NEET Chemistry video lectures, notes & PYQs

14 previous-year questions from this chapter every option and the correct answer, free · where the marks are

1. Transition elements: trends and properties

Exam focus: Zn, Cd and Hg are d-block elements but not true transition elements — their d-subshell is completely filled (d¹⁰) in both the metal and its common ions, so they miss the defining traits: variable oxidation states, coloured ions, catalytic activity. Also, second- and third-row transition metals in the same group (Zr and Hf, for instance) have almost identical atomic radii, because lanthanoid contraction cancels out the increase you'd otherwise expect going down a group.

The d- and f-Block Elements: Properties of Transition Elements

NCERT Wallah

Why transition metals are coloured, magnetic and show variable oxidation states, tied back to (n-1)d electron configuration rather than memorised as a list.

ChemistryNEET 2025show ▾

Given below are two statements: Statement I: Ferromagnetism is considered an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a

C r^{2 +}

ion (

Z = 24

) is the same as that of an

N d^{3 +}

ion (

Z = 60

). In the light of the above statements, choose the correct answer from the options given below:

A.Both Statement I and Statement II are true
B.Both Statement I and Statement II are false
C.Statement I is true but Statement II is false✓
D.Statement I is false but Statement II is true

Solution

Ferromagnetism is indeed treated as an extreme form of paramagnetism (I true). $C r^{2 +}$ is $3 d^{4}$ with 4 unpaired electrons; $N d^{3 +}$ is $4 f^{3}$ with 3 — not the same (II false).

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ChemistryNEET 2025show ▾

Match List I with List II.

List I	List II
A. Haber process	I. Fe catalyst
B. Wacker oxidation	II. $PdC l_{2}$
C. Wilkinson catalyst	III. $[(PP h_{3})_{3} RhCl]$
D. Ziegler catalyst	IV. $TiC l_{4}$ with $Al (C H_{3})_{3}$

Choose the correct answer from the options given below:

A.A-I, B-II, C-IV, D-III
B.A-II, B-III, C-I, D-IV
C.A-I, B-II, C-III, D-IV✓
D.A-I, B-IV, C-III, D-II

Solution

The Haber process uses iron (I); Wacker oxidation uses $PdC l_{2}$ (II); Wilkinson's catalyst is $[(PP h_{3})_{3} RhCl]$ (III); Ziegler's catalyst is $TiC l_{4}$ with triethyl/trimethylaluminium (IV).

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ChemistryNEET 2024show ▾

The 'spin only' magnetic moment is the same for which of the following ions? A.

T i^{3 +}

· B.

C r^{2 +}

· C.

M n^{2 +}

· D.

F e^{2 +}

· E.

S c^{3 +}

Choose the most appropriate answer from the options given below:

A.A and E only
B.B and C only
C.A and D only
D.B and D only✓

Solution

Spin-only moment depends on unpaired electrons: $T i^{3 +}$ ( $d^{1}$ ) 1, $C r^{2 +}$ ( $d^{4}$ ) 4, $M n^{2 +}$ ( $d^{5}$ ) 5, $F e^{2 +}$ ( $d^{6}$ ) 4, $S c^{3 +}$ ( $d^{0}$ ) 0. B and D both have 4 unpaired electrons, $μ = 24 \approx 4.9$ BM.

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ChemistryNEET 2024show ▾

The

E^{\circ}

value for the

M n^{3 +} /M n^{2 +}

couple is more positive than that of

C r^{3 +} /C r^{2 +}

F e^{3 +} /F e^{2 +}

due to the change of:

A. $d^{5}$ to $d^{2}$ configuration
B. $d^{4}$ to $d^{5}$ configuration✓
C. $d^{3}$ to $d^{5}$ configuration
D. $d^{5}$ to $d^{4}$ configuration

Solution

$M n^{3 +}$ is $d^{4}$ ; gaining an electron gives the stable half-filled $d^{5}$ configuration of $M n^{2 +}$ . This extra stability makes the reduction unusually favourable, i.e. $E^{\circ}$ unusually positive.

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ChemistryNEET 2023show ▾

The stability of

C u^{2 +}

salts is more than that of

C u^{+}

salts in aqueous solution due to:

A.First ionisation enthalpy
B.Enthalpy of atomisation
C.Hydration energy✓
D.Second ionisation enthalpy

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ChemistryNEET 2023show ▾

Which of the following statements are incorrect? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in

S c_{2} O_{3}

M n_{2} O_{7}

. C. Basic character increases from

V_{2} O_{3}

V_{2} O_{4}

V_{2} O_{5}

. D.

V_{2} O_{4}

dissolves in acids to give

V O_{4}^{3 -}

salts. E. CrO is basic but

C r_{2} O_{3}

is amphoteric. Choose the correct answer from the options given below:

A.A and E only
B.B and D only
C.C and D only✓
D.B and C only

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ChemistryNEET 2021show ▾

Zr (

Z = 40

) and Hf (

Z = 72

) have similar atomic and ionic radii because of:

A.Belonging to the same group
B.Diagonal relationship
C.Lanthanoid contraction✓
D.Having similar chemical properties

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ChemistryNEET 2020show ▾

Identify the incorrect statement.

A. $C r^{2 +}$ ( $d^{4}$ ) is a stronger reducing agent than $F e^{2 +}$ ( $d^{6}$ ) in water
B.Transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and form complexes
C.Interstitial compounds are those formed when small atoms like H, C or N are trapped inside the crystal lattices of metals
D.The oxidation states of chromium in $Cr O_{4}^{2 -}$ and $C r_{2} O_{7}^{2 -}$ are not the same✓

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2. Potassium dichromate and permanganate

Exam focus: MnO₄⁻'s reduction product depends entirely on the medium: Mn²⁺ (colourless, 5 electrons gained) in acid, MnO₂ (brown precipitate, 3 electrons) in neutral or faintly alkaline, MnO₄²⁻ (green, 1 electron) in strongly alkaline — equivalent-weight calculations hinge on picking the right one, and acidic is the default unless stated otherwise. Orange Cr₂O₇²⁻ and yellow CrO₄²⁻ interconvert with pH, not by any redox change.

Potassium Permanganate and Potassium Dichromate

NCERT Wallah

Single lecture covering both compounds' preparation and reactions exactly as named in the subtopic.

3. Lanthanoids and actinoids

Exam focus: Lanthanoid contraction — steady radius shrinkage across the series from poor 4f shielding — is why second- and third-row transition metals end up almost the same size element for element (Zr≈Hf, Nb≈Ta), which is exactly why those pairs are so hard to separate chemically. +3 is the dominant lanthanoid oxidation state, but watch for Ce⁴⁺ and Eu²⁺, both reaching the extra stability of an empty or half-filled f-subshell — the same logic as Cr and Cu in the d-block.

Lanthanoids and Actinoids

Competition Wallah

Dedicated lecture on lanthanoid/actinoid electronic configuration, lanthanoid contraction and properties.

ChemistryNEET 2024show ▾

The pair of lanthanoid ions which are diamagnetic is:

A. $C e^{3 +}$ and $E u^{2 +}$
B. $G d^{3 +}$ and $E u^{3 +}$
C. $P m^{3 +}$ and $S m^{3 +}$
D. $C e^{4 +}$ and $Y b^{2 +}$ ✓

Solution

Diamagnetism needs no unpaired $f$ electrons. $C e^{4 +}$ is $4 f^{0}$ and $Y b^{2 +}$ is $4 f^{14}$ — both diamagnetic. $C e^{3 +}$ ( $f^{1}$ ), $E u^{2 +}$ / $G d^{3 +}$ ( $f^{7}$ ), $E u^{3 +}$ ( $f^{6}$ ), $P m^{3 +}$ ( $f^{4}$ ), $S m^{3 +}$ ( $f^{5}$ ) are all paramagnetic.

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ChemistryNEET 2022show ▾

Gadolinium has a low value of third ionisation enthalpy because of:

A.High basic character
B.Small size
C.High exchange enthalpy✓
D.High electronegativity

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ChemistryNEET 2021show ▾

The incorrect statement among the following is:

A.Actinoid contraction is greater from element to element than lanthanoid contraction
B.Most of the trivalent lanthanoid ions are colourless in the solid state✓
C.Lanthanoids are good conductors of heat and electricity
D.Actinoids are highly reactive metals, especially when finely divided

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ChemistryNEET 2020show ▾

Identify the incorrect match.

Name	IUPAC official name
(a) Unnilunium	(i) Mendelevium
(b) Unniltrium	(ii) Lawrencium
(c) Unnilhexium	(iii) Seaborgium
(d) Unununnium	(iv) Darmstadtium

A.(a), (i)
B.(b), (ii)
C.(c), (iii)
D.(d), (iv)✓

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