JEE Main & Advanced

d- and f-Block Elements

d- and f-Block Elements for JEE Main & Advanced

1
Module 1

d-Block (Transition Elements)

General Trends and Electronic ConfigurationTopic 1

Transition Elements (d-block): Elements with partially filled $d$-orbitals in atom or in common oxidation states. Located in groups 3-12 of periodic table.

Four Series:

  • 3d series (1st transition): Sc(21) to Zn(30)
  • 4d series (2nd): Y(39) to Cd(48)
  • 5d series (3rd): La(57), Hf(72) to Hg(80)
  • 6d series (4th): Ac(89), Rf(104) onwards (mostly synthetic)

General Electronic Configuration: $(n-1)d^{1-10}\,ns^{1-2}$

3d Series Configurations:

ElementZConfig
Sc21$[Ar]\,3d^1\,4s^2$
Ti22$[Ar]\,3d^2\,4s^2$
V23$[Ar]\,3d^3\,4s^2$
Cr24$[Ar]\,3d^5\,4s^1$ (anomalous — half-filled $d$ stability)
Mn25$[Ar]\,3d^5\,4s^2$
Fe26$[Ar]\,3d^6\,4s^2$
Co27$[Ar]\,3d^7\,4s^2$
Ni28$[Ar]\,3d^8\,4s^2$
Cu29$[Ar]\,3d^{10}\,4s^1$ (anomalous — fully-filled $d$ stability)
Zn30$[Ar]\,3d^{10}\,4s^2$

Cr and Cu anomaly: Extra stability of half-filled ($d^5$) and fully-filled ($d^{10}$) configurations due to symmetric distribution and exchange energy.

Note: Zn ($d^{10}\,s^2$) and Hg are sometimes considered "non-transition" because their d-orbitals are fully filled and don't show typical transition properties. But traditionally still grouped under d-block.

Trends in Properties:

PropertyTrend
Atomic radiusDecreases from Sc → Cr; slight increase at the end (e⁻-e⁻ repulsion in larger $d^n$)
IEGenerally increases across; small differences
DensityIncreases across (heavier nuclei)
MP, BPVery high; max in middle (V, Cr)
Electrical conductivityHigh in middle

Reasons for Typical Transition Element Behavior:

  • Variable oxidation states (multiple available d-electrons)
  • Forms complex ions (vacant d-orbitals accept lone pairs)
  • Colored compounds (d-d transitions)
  • Magnetic properties (unpaired d-electrons)
  • Catalytic activity (variable oxidation states)
  • Alloy formation (similar atomic sizes)
Worked Examples
1

Why is the electronic configuration of Cr written as $3d^5\,4s^1$ instead of $3d^4\,4s^2$?

Show solution

The half-filled ($d^5$) configuration has extra stability due to:

  1. Symmetric distribution of electrons in $d$ subshell
  2. Maximum exchange energy — exchange between electrons of same spin (more exchanges possible with $d^5\,s^1$ than $d^4\,s^2$)
  3. Hund's rule maximizes multiplicity

So Cr promotes one electron from $4s$ to $3d$ to achieve $3d^5\,4s^1$.

Final Answer: Half-filled $3d^5$ has extra stability from symmetric distribution and exchange energy.

2

Why are transition metal melting points higher than s-block?

Show solution

Transition metals have $(n-1)d$ and $ns$ electrons available for metallic bonding (often $5-6$ valence electrons in middle of series). More valence electrons participate in metallic bonding → stronger metallic bond → higher MP/BP.

In contrast, s-block has only 1 or 2 valence electrons.

Final Answer: More valence electrons in d and s shells contribute to strong metallic bonding.

✎ Self-Check — 5 questions0 / 5
Q1.

d-block elements have configuration:

Q2.

Anomalous configuration of Cu:

Q3.

Atomic radius trend in 3d series:

Q4.

Zn is sometimes not considered transition metal because:

Q5.

Highest MP among 3d transition metals:

Properties — Color, Magnetism, Oxidation States, CatalysisTopic 2

Variable Oxidation States: Most transition metals show multiple ONs.

ElementCommon 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$+2$
Cu$+1, +2$
Zn$+2$ only

Maximum ON = total of $4s + 3d$ electrons (for early elements). E.g., Mn: $+7$ (in $MnO_4^-$); Cr: $+6$ (in $Cr_2O_7^{2-}$).

Stability trends:

  • $+2$ stable in middle of series (V to Cu)
  • $+3$ stable for early elements (Sc, Ti, V) and Fe, Co
  • Higher ONs less stable down the group (for early transition metals); more stable for heavier (Pt, Ru, etc.)

Color in Transition Metal Compounds:

Color due to d-d transitions: electron moves from one $d$-orbital to another absorbing visible light; observed color is complementary.

CompoundColorReason
$Cu(NO_3)_2$Blue$Cu^{2+}$ ($d^9$): d-d transition
$CuSO_4$Blue
$FeCl_3$Yellow-brown$Fe^{3+}$
$MnO_4^-$PurpleCharge transfer (Mn-O); not d-d
$CrO_4^{2-}$YellowCharge transfer (Cr-O)
$Cr_2O_7^{2-}$OrangeCharge transfer
$[Ti(H_2O)_6]^{3+}$Violet$Ti^{3+}$ ($d^1$): d-d
$ZnSO_4$Colorless$Zn^{2+}$ ($d^{10}$): no d-d possible
$Sc^{3+}$, $Cu^+$, $Cd^{2+}$Colorless$d^0$ or $d^{10}$

Magnetic Properties:

  • Diamagnetic: No unpaired electrons; weakly repelled (e.g., $Zn^{2+}$, $Cu^+$, $Cd^{2+}$)
  • Paramagnetic: Unpaired electrons; attracted

Magnetic Moment ($\mu$): "Spin-only" formula (ignoring orbital contribution): $$\mu = \sqrt{n(n+2)} \text{ B.M.}$$ where $n$ = number of unpaired electrons, B.M. = Bohr Magneton ($9.27 \times 10^{-24}$ J/T).

Examples:

  • $Fe^{2+}$ ($d^6$): 4 unpaired → $\mu = \sqrt{4 \times 6} = 4.9$ B.M.
  • $Cu^{2+}$ ($d^9$): 1 unpaired → $\mu = \sqrt{3} = 1.73$ B.M.
  • $Mn^{2+}$ ($d^5$, high spin): 5 unpaired → $\mu = \sqrt{35} = 5.92$ B.M. (maximum)

Catalytic Activity: Transition metals/compounds excellent catalysts due to:

  • Multiple oxidation states (alternate paths)
  • Surface adsorption (heterogeneous catalysis)
  • Complex formation

Examples:

  • Fe in Haber process
  • V₂O₅ in Contact process
  • Ni in hydrogenation of vegetable oils
  • Pt-Rh in Ostwald process

Complex Formation: Empty $d$-orbitals + small size + variable ONs allow coordination compound formation.

Examples: $[Fe(CN)_6]^{4-}$, $[Cu(NH_3)_4]^{2+}$, $[CoCl_4]^{2-}$.

Alloy Formation: Similar atomic sizes (within $\sim 10\%$) → form alloys easily.

  • Brass: Cu + Zn
  • Bronze: Cu + Sn
  • Steel: Fe + C (sometimes Ni, Cr)
  • Stainless steel: Fe + Cr + Ni

Interstitial Compounds: Small atoms (H, C, N, B) occupy interstitial sites in transition metal lattice. Hard, high MP, chemically inert. Examples: TiC, $Fe_3C$ (cementite), TiN, VH.

Worked Examples
1

Calculate magnetic moment of $Fe^{3+}$.

Show solution

$Fe^{3+}$: $[Ar]\,3d^5$ (lost $4s^2$ first, then $3d^1$). $d^5$: 5 unpaired electrons (Hund's rule). $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$ B.M.

Final Answer: $\mu = 5.92$ B.M.

2

Why are $ZnSO_4$ and $CuSO_4 \cdot 5H_2O$ different in color?

Show solution

$Zn^{2+}$ has $d^{10}$ configuration — no d-d transitions possible → colorless. $Cu^{2+}$ has $d^9$ configuration — d-d transition absorbing yellow light → blue color (complement). Also, water of hydration affects crystal field splitting in $[Cu(H_2O)_4]^{2+}$, enhancing color.

Final Answer: $Zn^{2+}$ has no unpaired $d$ electrons; $Cu^{2+}$ has $d^9$ allowing d-d transitions.

✎ Self-Check — 5 questions0 / 5
Q1.

$Mn^{2+}$ has unpaired electrons:

Q2.

$ZnSO_4$ is colorless because:

Q3.

Maximum oxidation state of Mn:

Q4.

Magnetic moment of $Cu^{2+}$:

Q5.

Color of $KMnO_4$ is due to:

2
Module 2

f-Block (Inner Transition)

Lanthanides — Configuration, Lanthanide ContractionTopic 1

f-Block elements: Have partially filled $4f$ (lanthanides) or $5f$ (actinides) orbitals.

Lanthanides (Ce to Lu): Elements $58-71$. Filling $4f$ orbital.

General Configuration: $[Xe]\,4f^{1-14}\,5d^{0-1}\,6s^2$

ElementZConfig
Ce58$[Xe]\,4f^1\,5d^1\,6s^2$
Pr59$[Xe]\,4f^3\,6s^2$
Nd60$[Xe]\,4f^4\,6s^2$
Pm61$[Xe]\,4f^5\,6s^2$
Sm62$[Xe]\,4f^6\,6s^2$
Eu63$[Xe]\,4f^7\,6s^2$ (half-filled $f$)
Gd64$[Xe]\,4f^7\,5d^1\,6s^2$
Tb65$[Xe]\,4f^9\,6s^2$
Dy66$[Xe]\,4f^{10}\,6s^2$
Ho67$[Xe]\,4f^{11}\,6s^2$
Er68$[Xe]\,4f^{12}\,6s^2$
Tm69$[Xe]\,4f^{13}\,6s^2$
Yb70$[Xe]\,4f^{14}\,6s^2$ (full $f$)
Lu71$[Xe]\,4f^{14}\,5d^1\,6s^2$

General Properties:

  • Silvery-white metals; soft
  • Good conductors
  • $+3$ is most common oxidation state; some show $+2$ (Eu, Yb) or $+4$ (Ce, Tb)
  • Colored ions (due to $f$-$f$ transitions, sometimes); some colorless ($La^{3+}$, $Lu^{3+}$)
  • Most $Ln^{3+}$ are paramagnetic

Lanthanide Contraction: Steady decrease in atomic and ionic radius across the lanthanide series (Ce to Lu).

Cause: Poor shielding by 4f electrons (which are diffuse and inner-located). Effective nuclear charge increases steadily, pulling electrons closer.

Consequences:

  1. Similar atomic radii of 4d (Zr, Hf etc.) and 5d (Hf, Mo etc.) elements — separation difficult
  2. Difficulty in separation of lanthanides — chemically very similar
  3. Density increase down the series
  4. Increase in basicity of $Ln(OH)_3$ from $La(OH)_3$ to $Lu(OH)_3$? Actually, decreases — smaller cation → less basic hydroxide
  5. Increase in ionization enthalpy along the series

Important point: $Zr$ and $Hf$ have almost identical radii ($160$ pm) because of lanthanide contraction; chemically very similar — costly to separate.

Use of lanthanides:

  • Misch metal ($Ce + La + Nd$): in cigarette lighters, tracer bullets
  • $CeO_2$: catalyst, polishing agent
  • $La_2O_3$: optical glasses
  • Magnets ($Nd_2Fe_{14}B$): hard disks, motors
  • Lasers ($Nd^{3+}$ in YAG laser)
Worked Examples
1

What is lanthanide contraction and its consequences?

Show solution

Lanthanide contraction: The steady decrease in atomic/ionic radii from La (or Ce) to Lu. Cause: 4f electrons (innermost, diffuse) shield poorly; effective nuclear charge increases steadily; pulls electrons closer.

Consequences:

  1. Hf has similar radius to Zr (chemically similar, hard to separate)
  2. Lanthanides are very similar to each other (chemical separation difficult)
  3. Basic strength of $Ln(OH)_3$ decreases across (smaller cation → stronger bond to OH⁻ → less ionic)
  4. Density increases across the series

Final Answer: Decrease in size across lanthanides due to poor 4f shielding.

2

Why is the separation of Zr and Hf difficult?

Show solution

Zr (4d) and Hf (5d) belong to same group. Hf comes after the lanthanide series, so it has the cumulative effect of lanthanide contraction. As a result, $Hf$ has radius ($160$ pm) very close to Zr ($160$ pm). They have nearly identical chemical properties and very similar densities — hence very difficult to separate. Modern methods like ion-exchange and solvent extraction needed.

Final Answer: Lanthanide contraction makes Zr and Hf nearly identical in size and chemistry.

✎ Self-Check — 5 questions0 / 5
Q1.

Lanthanides have configuration:

Q2.

Lanthanide contraction is due to:

Q3.

Most common oxidation state of lanthanides:

Q4.

Zr and Hf are difficult to separate because:

Q5.

Misch metal contains:

Actinides and Important Compounds (KMnO₄, K₂Cr₂O₇)Topic 2

Actinides (Th to Lr): Elements $90-103$. Filling $5f$ orbital.

General Configuration: $[Rn]\,5f^{1-14}\,6d^{0-1}\,7s^2$

Properties:

  • All radioactive
  • Th, U found in nature; rest are synthetic
  • Show many oxidation states (more than lanthanides): U shows $+3, +4, +5, +6$
  • Form complexes more readily than lanthanides

Difference from Lanthanides:

PropertyLanthanidesActinides
Stable ON$+3$ mainlyMany ($+3, +4, +5, +6$)
RadioactivityOnly PmAll
Complex formationLowMore
DensityLowerHigher
5f vs 4f4f more buried5f less penetrating; outer ones

Important Compounds:

Potassium Permanganate ($KMnO_4$):

Preparation:

  • $2MnO_2 + 4KOH + O_2 \to 2K_2MnO_4 + 2H_2O$ (Mn $+4 \to +6$, manganate)
  • $3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$ (disproportionation in acidic)
  • Or electrolytic oxidation of $MnO_4^{2-}$

Properties:

  • Dark purple crystals
  • Strong oxidizing agent in acidic, neutral, and basic media
  • Color due to charge transfer

Reactions (Acidic medium):

  • $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ ($n$-factor 5; $E = M/5 = 31.6$)
  • Oxidizes Fe²⁺: $5Fe^{2+} + MnO_4^- + 8H^+ \to 5Fe^{3+} + Mn^{2+} + 4H_2O$
  • Oxidizes $C_2O_4^{2-}$: $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \to 10CO_2 + 2Mn^{2+} + 8H_2O$
  • Oxidizes $I^-$ to $I_2$; $H_2S$ to $S$; $SO_2$ to $SO_4^{2-}$

Reactions (Basic/Neutral medium):

  • $MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$ ($n$-factor 3)

Reactions (Strongly basic, Strong reducing agent):

  • $MnO_4^- + e^- \to MnO_4^{2-}$ (ON: $+7 \to +6$; $n$-factor 1)

Uses: Oxidizing agent in lab; volumetric analysis (self-indicator); bleaching wool, silk; disinfectant.

Potassium Dichromate ($K_2Cr_2O_7$):

Preparation:

  • From chromite ore $FeCr_2O_4$:

$4FeCr_2O_4 + 16NaOH + 7O_2 \to 8Na_2CrO_4 + 2Fe_2O_3 + 8H_2O$

  • $2Na_2CrO_4 + H_2SO_4 \to Na_2Cr_2O_7 + Na_2SO_4 + H_2O$
  • $Na_2Cr_2O_7 + 2KCl \to K_2Cr_2O_7 + 2NaCl$ (less soluble)

Properties:

  • Orange-red crystals
  • Stable in air
  • Strong oxidizing agent in acidic medium

$CrO_4^{2-}$ vs $Cr_2O_7^{2-}$:

  • In acidic: $2CrO_4^{2-} + 2H^+ \to Cr_2O_7^{2-} + H_2O$ (orange)
  • In basic: $Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$ (yellow)

Reactions:

  • $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$ ($n$-factor 6; $E = M/6 = 49$)
  • Oxidizes Fe²⁺: $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O$
  • Oxidizes $I^-$ to $I_2$
  • Oxidizes $H_2S$ to $S$

Chromyl Chloride Test: $K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \to 2CrO_2Cl_2$ (orange-red vapors) $+ 4NaHSO_4 + 2KHSO_4 + 3H_2O$. Test for $Cl^-$.

Uses: Volumetric analysis (vs Fe²⁺), tanning of leather, dyes, photography.

Comparison of $KMnO_4$ vs $K_2Cr_2O_7$:

Property$KMnO_4$$K_2Cr_2O_7$
ColorPurpleOrange
StabilityLessMore
Self-indicatorYesNo
Used as primary standardNoYes
$n$-factor (acidic)$5$$6$
Worked Examples
1

Find equivalent weight of $K_2Cr_2O_7$ in acidic medium. ($M = 294$)

Show solution

In acidic: $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$. $n$-factor = 6 (electrons gained per molecule). $E = M/n = 294/6 = 49$.

Final Answer: $E = 49$.

2

Write balanced equation for: $KMnO_4 + FeSO_4$ in acidic medium.

Show solution

$Fe^{2+} \to Fe^{3+} + e^-$ (oxidation, ×5) $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ (reduction, ×1) Net: $MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$ Full molecular: $2KMnO_4 + 10FeSO_4 + 8H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O$

Final Answer: As above.

✎ Self-Check — 5 questions0 / 5
Q1.

All actinides are:

Q2.

$KMnO_4$ in acidic medium has $n$-factor:

Q3.

Color of $K_2Cr_2O_7$:

Q4.

Equivalent weight of $K_2Cr_2O_7$:

Q5.

Chromyl chloride test detects:

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