Vardaan Watermark
Vardaan Learning Institute
Created by Team Vardaan with Powered by VARDAAN COMET

The d- and f-Block Elements

1. Introduction & Position in the Periodic Table

The d-block of the periodic table contains elements of Groups 3 to 12 in which the inner penultimate $(n-1)d$ subshell orbitals are progressively filled across four long periods ($3d, 4d, 5d, 6d$). The f-block consists of elements in which $4f$ and $5f$ orbitals are progressively filled, placed in a separate two-row panel at the bottom of the periodic table.

Position of d-block and f-block elements in the Periodic Table
Position of d-Block (Groups 3-12) and f-Block Elements in the Periodic Table
IUPAC Definition of Transition Elements Transition Metal: IUPAC strictly defines a transition metal as an element that has an incompletely (partially) filled d-subshell in its neutral ground state atom or in any one of its common oxidation states.

Distinction: "d-Block Elements" is a structural group name (Groups 3 to 12), whereas "Transition Metals" is a functional definition based on incomplete d-orbitals ($d^{1-9}$).
EXCEPTION 1: IUPAC DEFINITION & GROUP 12 NON-TYPICAL ELEMENTS

The Four Transition Series

NCERT Example 4.1 / Solved Problem Question 1: On what ground can you say that Scandium ($Z = 21$) is a transition element but Zinc ($Z = 30$) is not?
Solution:
Scandium atom in its ground state has an incompletely filled $3d$ orbital ($3d^1 4s^2$). According to IUPAC definition, it is a transition element.
On the other hand, a Zinc atom has completely filled $d$-orbitals ($3d^{10} 4s^2$) in its ground state, as well as in its oxidized $+2$ state ($Zn^{2+}: 3d^{10}$). Hence, Zinc is not regarded as a transition element.
NCERT Intext Question 4.1 Question 2: Silver atom ($Ag, Z = 47$) has completely filled d-orbitals ($4d^{10} 5s^1$) in its ground state. How can you say that it is a transition element?
Solution:
Silver exhibits a $+2$ oxidation state in compounds such as Silver(II) fluoride ($AgF_2$) and Silver(II) oxide ($AgO$). In the $+2$ state, its electronic configuration becomes $4d^9$, which has an incompletely filled $d$-orbital. Therefore, silver is classified as a transition element.

2. Electronic Configurations of d-Block Elements

The general outer electronic configuration of d-block elements is $(n-1)d^{1-10} ns^{1-2}$, where $(n-1)$ stands for the inner (penultimate) d-orbitals and $n$ is the outermost principal shell.

Why the Configuration Exceptions? (Cr, Cu, Pd, Pt, Au) The unusual configurations of Chromium ($Cr$), Copper ($Cu$), Palladium ($Pd$), Platinum ($Pt$), and Gold ($Au$) arise due to two primary factors:

1. Very Small Energy Gap: The energy difference between the $(n-1)d$ and $ns$ orbitals is minimal, making electron transfer energetically favorable.
2. Extra Stability of Half-filled ($d^5, f^7$) and Fully-filled ($d^{10}, f^{14}$) Subshells:
Exchange Energy Pairs for 3d5 vs 3d4 Configurations
Exchange Energy Pairs for 3d5 (10 Pairs) vs 3d4 (6 Pairs) Configurations
Key Exceptions to Memorize for Boards & Entrance Exams:
Element Symbol & Z Expected Config Actual Config Key Exception Reason
Chromium $Cr (Z=24)$ $[Ar] 3d^4 4s^2$ $\mathbf{[Ar] 3d^5 4s^1}$ Half-filled $3d^5$ subshell stability
Copper $Cu (Z=29)$ $[Ar] 3d^9 4s^2$ $\mathbf{[Ar] 3d^{10} 4s^1}$ Fully-filled $3d^{10}$ subshell stability
Niobium $Nb (Z=41)$ $[Kr] 4d^3 5s^2$ $[Kr] 4d^4 5s^1$ Small $4d-5s$ energy gap
Molybdenum $Mo (Z=42)$ $[Kr] 4d^4 5s^2$ $[Kr] 4d^5 5s^1$ Half-filled $4d^5$ stability
Palladium $Pd (Z=46)$ $[Kr] 4d^8 5s^2$ $\mathbf{[Kr] 4d^{10} 5s^0}$ CRUCIAL: Only element with $5s^0$!
Platinum $Pt (Z=78)$ $[Xe] 4f^{14} 5d^8 6s^2$ $[Xe] 4f^{14} 5d^9 6s^1$ $5d^9 6s^1$ stability
Gold $Au (Z=79)$ $[Xe] 4f^{14} 5d^9 6s^2$ $[Xe] 4f^{14} 5d^{10} 6s^1$ Fully-filled $5d^{10}$ stability
The Crucial 4s Orbital Rule (Filling vs Ionization) - Filling (Aufbau Principle): The $4s$ orbital is filled BEFORE the $3d$ orbital because in neutral isolated atoms, $4s$ has lower energy than $3d$.
- Ionization (Loss of Electrons): When transition metals form cations, $ns$ electrons are lost BEFORE $(n-1)d$ electrons! Once $3d$ orbitals are filled, nuclear pull increases, making $3d$ lower in energy than $4s$.
NCERT Table 4.1 & 4.2: Electronic Configurations of 3d Transition Series
Element Symbol Z Neutral Atom (M) $M^+$ Ion $M^{2+}$ Ion $M^{3+}$ Ion
Scandium Sc 21 $[Ar] 3d^1 4s^2$ $[Ar] 3d^1 4s^1$ $[Ar] 3d^1$ $[Ar] 3d^0$
Titanium Ti 22 $[Ar] 3d^2 4s^2$ $[Ar] 3d^2 4s^1$ $[Ar] 3d^2$ $[Ar] 3d^1$
Vanadium V 23 $[Ar] 3d^3 4s^2$ $[Ar] 3d^3 4s^1$ $[Ar] 3d^3$ $[Ar] 3d^2$
Chromium Cr 24 $[Ar] 3d^5 4s^1$ $[Ar] 3d^5$ $[Ar] 3d^4$ $[Ar] 3d^3$
Manganese Mn 25 $[Ar] 3d^5 4s^2$ $[Ar] 3d^5 4s^1$ $[Ar] 3d^5$ $[Ar] 3d^4$
Iron Fe 26 $[Ar] 3d^6 4s^2$ $[Ar] 3d^6 4s^1$ $[Ar] 3d^6$ $[Ar] 3d^5$
Cobalt Co 27 $[Ar] 3d^7 4s^2$ $[Ar] 3d^7 4s^1$ $[Ar] 3d^7$ $[Ar] 3d^6$
Nickel Ni 28 $[Ar] 3d^8 4s^2$ $[Ar] 3d^8 4s^1$ $[Ar] 3d^8$ $[Ar] 3d^7$
Copper Cu 29 $[Ar] 3d^{10} 4s^1$ $[Ar] 3d^{10}$ $[Ar] 3d^9$ --
Zinc Zn 30 $[Ar] 3d^{10} 4s^2$ $[Ar] 3d^{10} 4s^1$ $[Ar] 3d^{10}$ --
Practice Problem 3 Question: Write the outer electronic configurations of $Fe^{3+}$, $Cu^+$, $Cr^{3+}$, and $Mn^{2+}$.
Solution:
1. $Fe (Z=26)$: $[Ar] 3d^6 4s^2 \xrightarrow{-2e^-(4s) -1e^-(3d)} Fe^{3+}: \mathbf{[Ar] 3d^5}$ (Stable half-filled).
2. $Cu (Z=29)$: $[Ar] 3d^{10} 4s^1 \xrightarrow{-1e^-(4s)} Cu^+: \mathbf{[Ar] 3d^{10}}$ (Stable fully-filled).
3. $Cr (Z=24)$: $[Ar] 3d^5 4s^1 \xrightarrow{-1e^-(4s) -2e^-(3d)} Cr^{3+}: \mathbf{[Ar] 3d^3}$ (Half-filled $t_{2g}^3$ in aq medium).
4. $Mn (Z=25)$: $[Ar] 3d^5 4s^2 \xrightarrow{-2e^-(4s)} Mn^{2+}: \mathbf{[Ar] 3d^5}$ (Stable half-filled).

3. General Physical Properties & Periodic Trends (d-Block)

3.1 Metallic Characteristics & Lattice Structures

Nearly all transition metals exhibit typical metallic properties: high tensile strength, ductility, malleability, high thermal and electrical conductivity, and metallic lustre. With the exceptions of $Zn, Cd, Hg,$ and $Mn$, they adopt typical metallic crystal lattices at normal temperatures:

NCERT Lattice Structures Summary

3.2 Melting Points & Enthalpies of Atomisation (NCERT Fig 4.1 & 4.2)

NCERT Fig. 4.1 & 4.2: Trends in Melting Points and Enthalpies of Atomisation
NCERT Fig. 4.1 & 4.2: Trends in Melting Points and Enthalpies of Atomisation (3d, 4d, 5d Series)
EXCEPTION 3: MELTING POINT & ATOMISATION DIPS (Mn, Tc, Zn, Hg)
NCERT Intext Question 4.2 Question 4: In the series $Sc (Z = 21)$ to $Zn (Z = 30)$, the enthalpy of atomisation of zinc is the lowest, i.e., $126 \text{ kJ mol}^{-1}$. Why?
Solution:
In zinc, all $3d$ orbitals are completely filled ($3d^{10} 4s^2$) and there are no unpaired d-electrons available for interatomic bonding. Consequently, the metallic bonds in zinc are very weak, giving it the lowest enthalpy of atomisation in the 3d series.

3.4 Variation in Atomic and Ionic Radii (NCERT Fig 4.3)

NCERT Fig. 4.3: Trends in Atomic Radii of Transition Elements (3d, 4d, 5d)
NCERT Fig. 4.3: Trends in Atomic Radii of Transition Elements (3d, 4d, 5d Series)
The Tug-of-War: Atomic Radius along a 3d Period The atomic radius across a transition series is governed by two opposing factors:

1. Increasing Nuclear Charge ($Z_{eff}$): As atomic number increases, nuclear charge increases, pulling outer electrons inward (Size $\downarrow$).
2. Shielding (Screening) Effect: Electrons added to the inner $(n-1)d$ subshell shield outer $ns$ electrons from nuclear pull (Size $\uparrow$).

Shielding Effect and Lanthanoid Contraction Infographic
Shielding Effect & Lanthanoid Contraction Orbital Mechanism
Three Distinct Regions in the 3d Period:
EXCEPTION 4: LANTHANOID CONTRACTION & IDENTICAL 4d/5d RADII Normally, moving down a group increases atomic size due to addition of a shell ($3d < 4d$). Thus, $4d$ elements are larger than $3d$ elements.

However, $5d$ series elements have almost IDENTICAL radii to corresponding $4d$ series elements!
Example: Zirconium ($Zr, 4d$: $160 \text{ pm}$) and Hafnium ($Hf, 5d$: $159 \text{ pm}$); Niobium ($Nb$: $146 \text{ pm}$) and Tantalum ($Ta$: $146 \text{ pm}$); Molybdenum ($Mo$: $139 \text{ pm}$) and Tungsten ($W$: $139 \text{ pm}$).

Reason (Lanthanoid Contraction): Before the $5d$ series begins, $14$ lanthanoid elements are filled, placing electrons into $4f$ orbitals. The $4f$ electrons have highly diffused shapes and exert extremely poor shielding effect. Consequently, the nuclear charge increases by $+14$ units, exerting a powerful inward pull on outer electrons. This contraction exactly cancels out the expected shell-addition size increase from $4d$ to $5d$.
Practice Problem 5 Question: Why do Zirconium ($Zr, Z=40$) and Hafnium ($Hf, Z=72$) exhibit almost identical atomic radii ($160 \text{ pm}$ vs $159 \text{ pm}$) and similar chemical properties?
Solution:
This phenomenon is a direct consequence of Lanthanoid Contraction. Hafnium ($5d$) is preceded by 14 lanthanoid elements ($4f^{14}$). The poor shielding offered by $4f$ electrons causes the effective nuclear charge on Hafnium's outer electrons to increase significantly. The resulting contraction in size counteracts the normal size increase expected when moving down a group from $Zr$ to $Hf$, rendering their atomic and ionic radii nearly identical.

3.5 Density Trends

From Titanium ($Z=22, d=4.1 \text{ g cm}^{-3}$) to Copper ($Z=29, d=8.9 \text{ g cm}^{-3}$), there is a steady increase in density. This is due to the decrease in atomic radius coupled with a simultaneous increase in atomic mass across the period. Density drops slightly at Zinc ($7.1 \text{ g cm}^{-3}$) due to its larger atomic radius.

3.6 Ionisation Enthalpies ($\Delta_i H^\circ$)

4. Oxidation States & Standard Electrode Potentials ($E^\circ$)

4.1 Variety of Oxidation States

One of the most notable features of transition metals is their display of variable oxidation states. This variability arises because the energy gap between $(n-1)d$ and $ns$ orbitals is extremely small, allowing electrons from both subshells to participate in bond formation.

NCERT Table 4.3: Oxidation States of 3d Metals
Element Sc Ti V Cr Mn Fe Co Ni Cu Zn
Oxidation States +3 +2,+3,
+4
+2,+3,
+4,+5
+2,+3,
+4,+5,+6
+2,+3,
+4,+5,
+6,+7
+2,+3,
+4,+6
+2,+3,
+4
+2 +1,
+2
+2

*Bold numbers represent the most common / stable oxidation states in aqueous solution or solids.

EXCEPTION 5: OXIDATION STATE ANOMALIES & GROUP STABILITY TRENDS 1. Minimum & Maximum States: Minimum state is usually $+2$ (loss of two $ns$ electrons). Maximum state increases up to $Mn$ in the middle ($+7$ in $MnO_4^-$), equal to the sum of $4s$ and $3d$ electrons, followed by a sharp drop ($Fe: +6$, $Co, Ni: +4$, $Cu: +2$, $Zn: +2$).
2. NO VARIABLE STATEScandium & Zinc: Scandium exhibits ONLY $+3$ (Forms $Sc^{3+}: 3d^0$). Zinc exhibits ONLY $+2$ ($Zn^{2+}: 3d^{10}$).
3. d-Block vs p-Block Variability: 4. PARADOX TRENDGroup Stability Trends (d-Block vs p-Block Exception): 5. ZERO STATELow Oxidation States ($0, +1$): Stabilized when complexes contain ligands with $\pi$-acceptor character (e.g., $Ni(CO)_4$ and $Fe(CO)_5$ where oxidation state of metal is ZERO).

4.2 Standard Electrode Potentials ($E^\circ$) (NCERT Fig 4.4)

The standard electrode potential ($E^\circ$) for the $M^{2+}/M$ couple measures the tendency of solid metal to form hydrated divalent cations in aqueous solution. It is NOT determined by ionization energy alone, but by a thermochemical balance of three energy terms:

The E° Energy Balance Equation $$\mathbf{\Delta H_{\text{total}} = \Delta_a H^\circ + IE_1 + IE_2 + \Delta_{\text{hyd}}H^\circ}$$

($\Delta_a H^\circ$: Sublimation  |  $IE_1 + IE_2$: Ionization  |  $\Delta_{\text{hyd}}H^\circ$: Hydration)

For a metal to have a negative $E^\circ$ (strong reducing agent), the energy released during Hydration ($\Delta_{\text{hyd}}H^\circ$, negative) must overcome the energy required for Sublimation ($\Delta_a H^\circ$, positive) and Ionization ($IE_1+IE_2$, positive).
NCERT Fig. 4.4: Standard Electrode Potential E°(M2+/M) for 3d Elements
NCERT Fig. 4.4: Standard Electrode Potential E°(M2+/M) for 3d Transition Metals
NCERT Thermochemical Parameters & E° Values (Table 4.4)
Metal $\Delta_a H^\circ$ $IE_1$ $IE_2$ $\Delta_{hyd}H^\circ$ $E^\circ(M^{2+}/M)$ $E^\circ(M^{3+}/M^{2+})$
Ti 469 656 1309 -1866 -1.63 V -0.37 V
V 515 650 1414 -1895 -1.18 V -0.26 V
Cr 398 653 1592 -1925 -0.90 V -0.41 V
Mn 279 717 1509 -1862 -1.18 V +1.57 V
Fe 418 762 1561 -1998 -0.44 V +0.77 V
Co 427 758 1644 -2079 -0.28 V +1.97 V
Ni 431 736 1752 -2121 -0.25 V --
Cu 339 745 1958 -2121 +0.34 V --
Zn 130 906 1734 -2059 -0.76 V --
EXCEPTION 6: THE COPPER ANOMALY & Cr2+ / Mn3+ REDOX PARADOX 1. COPPER ANOMALYThe Copper Anomaly ($E^\circ = +0.34 \text{ V}$): Copper is the ONLY metal in the 3d series with a positive $E^\circ(M^{2+}/M)$ value. Thus, it cannot liberate $H_2$ gas from dilute mineral acids (reacts only with oxidising acids like $\text{HNO}_3$ or hot conc. $\text{H}_2\text{SO}_4$).
Reason: High enthalpy of sublimation ($\Delta_a H^\circ$) combined with very high second ionization enthalpy ($IE_1+IE_2$) is NOT balanced by its hydration enthalpy ($\Delta_{hyd}H^\circ$).

2. More Negative $E^\circ$ for Mn, Ni, and Zn: 3. REDOX PARADOX$E^\circ(M^{3+}/M^{2+})$ Redox Couple Trends (Cr2+ vs Mn3+):
NCERT Example 4.4 / Solved Problem Question 6: Why is $Cr^{2+}$ reducing and $Mn^{3+}$ oxidising when both have a $d^4$ configuration?
Solution:
- $Cr^{2+}$ acts as a reducing agent because its configuration changes from $d^4$ to $d^3$ upon oxidation ($Cr^{3+}$). In aqueous solution, $d^3$ corresponds to an exceptionally stable half-filled $t_{2g}^3$ level.
- $Mn^{3+}$ acts as an oxidising agent because gaining an electron converts it from $d^4$ to $d^5$ ($Mn^{2+}$), attaining the extra stability of a half-filled $d^5$ subshell.

4.3 Trends in Stability of Halides and Oxides

5. Important Characteristics and Phenomena

5.1 Magnetic Properties

Transition metal ions generally exhibit magnetic behavior due to unpaired electrons. When a magnetic field is applied, two main behaviors are observed: Diamagnetism (repelled by magnetic field, all paired electrons) and Paramagnetism (attracted by magnetic field, unpaired electrons). Extreme paramagnetism is called Ferromagnetism ($Fe, Co, Ni$).

The "Spin-Only" Formula For 3d transition series elements, contribution of orbital angular momentum is effectively quenched. The magnetic moment ($\mu$) depends solely on the spin angular momentum: $$\mathbf{\mu = \sqrt{n(n+2)} \text{ BM}}$$ Where $n$ = number of unpaired electrons, and BM = Bohr Magneton (unit of magnetic moment).
Quick Calculation Guide: $n=1 \rightarrow 1.73 \text{ BM}$; $n=2 \rightarrow 2.84 \text{ BM}$; $n=3 \rightarrow 3.87 \text{ BM}$; $n=4 \rightarrow 4.90 \text{ BM}$; $n=5 \rightarrow 5.92 \text{ BM}$.
NCERT Table 4.7: Calculated and Observed Magnetic Moments (BM)
Ion Config Unpaired $e^-$ ($n$) Calculated $\mu$ Observed $\mu$
$Sc^{3+}, Ti^{4+}$ $3d^0$ 0 0 BM 0 BM
$Ti^{3+}, V^{4+}$ $3d^1$ 1 1.73 BM 1.75 - 1.76 BM
$V^{3+}, Ti^{2+}$ $3d^2$ 2 2.84 BM 2.76 - 2.86 BM
$V^{2+}, Cr^{3+}$ $3d^3$ 3 3.87 BM 3.86 - 3.87 BM
$Cr^{2+}, Mn^{3+}$ $3d^4$ 4 4.90 BM 4.80 - 4.90 BM
$Mn^{2+}, Fe^{3+}$ $3d^5$ 5 5.92 BM 5.96 BM
$Fe^{2+}$ $3d^6$ 4 4.90 BM 5.3 - 5.5 BM
$Co^{2+}$ $3d^7$ 3 3.87 BM 4.4 - 5.2 BM
$Ni^{2+}$ $3d^8$ 2 2.84 BM 2.9 - 3.4 BM
$Cu^{2+}$ $3d^9$ 1 1.73 BM 1.8 - 2.2 BM
$Zn^{2+}, Cu^+$ $3d^{10}$ 0 0 BM 0 BM
NCERT Intext Question 4.8 Question 7: Calculate the 'spin-only' magnetic moment of $M^{2+}(aq)$ ion where atomic number $Z = 27$.
Solution:
1. Atomic number $Z=27$ is Cobalt ($Co$).
2. Neutral $Co$: $[Ar] 3d^7 4s^2$. Cation $Co^{2+}$: $[Ar] 3d^7$.
3. Filling $3d^7$ orbital using Hund's Rule: $\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow \implies n = 3$ unpaired electrons.
4. $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} = \mathbf{3.87 \text{ BM}}$.

5.2 Formation of Coloured Ions

Most transition metal compounds display vibrant colors in solid or aquated states.

Mechanism of d-d Transition When ligands or water molecules approach a transition metal ion, the five degenerate d-orbitals split into sets of different energy levels ($t_{2g}$ and $e_g$). When visible light strikes the ion, an unpaired d-electron absorbs light of a specific wavelength (frequency) and jumps from a lower energy d-orbital to a higher energy d-orbital. This is called a d-d transition. The transmitted (unabsorbed) light gives the complementary color.

Crucial Rule: Color requires partially filled d-orbitals ($d^{1-9}$). Ions with $d^0$ ($Sc^{3+}, Ti^{4+}$) or $d^{10}$ ($Cu^+, Zn^{2+}$) cannot undergo d-d transitions and are colorless / white.
Crystal Field Splitting & d-d Transition Mechanism
Crystal Field Splitting of d-Orbitals & d-d Transition Mechanism
NCERT Table 4.8: Colors of Hydrated 3d Transition Metal Ions
Ion Configuration Observed Aquated Color
$Sc^{3+}, Ti^{4+}$ $3d^0$ Colourless
$Ti^{3+}$ $3d^1$ Purple
$V^{4+}$ $3d^1$ Blue
$V^{3+}$ $3d^2$ Green
$V^{2+}, Cr^{3+}, Mn^{3+}$ $3d^3, 3d^4$ Violet / Purple
$Cr^{2+}$ $3d^4$ Blue
$Mn^{2+}$ $3d^5$ Pink
$Fe^{3+}$ $3d^5$ Yellow
$Fe^{2+}$ $3d^6$ Green
$Co^{2+}$ $3d^7$ Pink / Blue-pink
$Ni^{2+}$ $3d^8$ Green
$Cu^{2+}$ $3d^9$ Blue
$Zn^{2+}$ $3d^{10}$ Colourless

5.3 Formation of Complex Compounds

Transition metals form a vast array of coordination complexes (such as $[Fe(CN)_6]^{3-}, [Fe(CN)_6]^{4-}, [Cu(NH_3)_4]^{2+}, [PtCl_4]^{2-}$). This tendency is due to:
1. Comparatively small ionic size and high nuclear charge density.
2. Availability of vacant d-orbitals of suitable energy to accept lone pairs donated by ligands.

5.4 Catalytic Properties

Transition metals and their compounds are widely used as industrial catalysts. Reasons include:
1. Ability to adopt variable oxidation states and form unstable intermediate compounds, lowering activation energy.
2. Providing a large surface area with free valencies for reactant molecules to adsorb onto.

NCERT Key Catalysts & Catalytic Mechanism Mechanism Example ($Fe^{3+}$ in Iodide-Persulphate Reaction):
Overall Reaction: $$\mathbf{2I^- + S_2O_8^{2-} \xrightarrow{Fe^{3+}} I_2 + 2SO_4^{2-}}$$ Step 1: $\mathbf{2Fe^{3+} + 2I^- \longrightarrow 2Fe^{2+} + I_2}$
Step 2: $\mathbf{2Fe^{2+} + S_2O_8^{2-} \longrightarrow 2Fe^{3+} + 2SO_4^{2-}}$

5.5 Formation of Interstitial Compounds

Interstitial compounds are formed when small non-metal atoms like Hydrogen ($H$), Carbon ($C$), or Nitrogen ($N$) get trapped inside the vacant interstitial spaces of metal crystal lattices. They are usually non-stoichiometric (e.g., $TiC, Mn_4N, Fe_3H, VH_{0.56}, TiH_{1.7}$).

5.6 Alloy Formation

An alloy is a blend of metals (solid solution). Transition metals readily form alloys with one another (e.g., Ferrous alloys, stainless steel, Manganese steel) and with non-transition metals (Brass: $Cu+Zn$, Bronze: $Cu+Sn$).
Reason: Their atomic radii are within 15% of each other, allowing atoms of one metal to substitute easily into the crystal lattice of another without major distortion.

6. Important Compounds of Transition Metals ($K_2Cr_2O_7$ & $KMnO_4$)

6.1 Potassium Dichromate ($K_2Cr_2O_7$)

Potassium dichromate is an orange crystalline solid, widely used as an oxidizing agent and primary standard in volumetric analysis.

Preparation from Chromite Ore (FeCr2O4) Step 1 (Roasting with Alkali): Fusion of finely powdered chromite ore with Sodium Carbonate ($Na_2CO_3$) in excess air yields yellow Sodium Chromate.
$$\mathbf{4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \longrightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2}$$ Step 2 (Acidification): The yellow solution is filtered and acidified with dilute sulfuric acid to convert chromate to orange Sodium Dichromate.
$$\mathbf{2Na_2CrO_4 + 2H^+ \longrightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O}$$ Step 3 (Potassium Salt Conversion): Sodium dichromate is highly soluble. Treatment with Potassium Chloride ($KCl$) precipitates less soluble orange Potassium Dichromate crystals.
$$\mathbf{Na_2Cr_2O_7 + 2KCl \longrightarrow K_2Cr_2O_7(s) + 2NaCl}$$
3-Step Preparation Flowchart of Potassium Dichromate (K2Cr2O7)
3-Step Preparation Flowchart of Potassium Dichromate (K2Cr2O7) from Chromite Ore
Chromate-Dichromate pH Equilibrium & Structures Chromate ($CrO_4^{2-}$, Yellow) and Dichromate ($Cr_2O_7^{2-}$, Orange) are interconvertible depending on pH. Oxidation state of Chromium is +6 in BOTH ions! No redox occurs.

- Acidic Medium ($\text{pH} < 7$): $$\mathbf{2CrO_4^{2-} \text{ (Yellow)} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} \text{ (Orange)} + H_2O}$$ - Basic Medium ($\text{pH} > 7$): $$\mathbf{Cr_2O_7^{2-} \text{ (Orange)} + 2OH^- \rightleftharpoons 2CrO_4^{2-} \text{ (Yellow)} + H_2O}$$
NCERT Fig. 4.5: Structures of Chromate Ion (CrO4 2-) and Dichromate Ion (Cr2O7 2-)
NCERT Fig. 4.5: Structures of Chromate Ion (CrO4 2-) and Dichromate Ion (Cr2O7 2-)
Structural Parameters (NCERT Details):
- Chromate ($CrO_4^{2-}$): Tetrahedral geometry with four identical $Cr-O$ bond lengths due to resonance.
- Dichromate ($Cr_2O_7^{2-}$): Consists of two $CrO_4$ tetrahedra sharing one corner oxygen atom. The $Cr-O-Cr$ bond angle is $126^\circ$. Bridging $Cr-O$ bond length is $179 \text{ pm}$, whereas terminal $Cr-O$ bond length is $163 \text{ pm}$.
Oxidising Reactions of Acidified K2Cr2O7 (E° = 1.33 V) Half-Reaction: $$\mathbf{Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} \text{ (Green)} + 7H_2O}$$ 1. Oxidises Iodide to Iodine: $$\mathbf{Cr_2O_7^{2-} + 14H^+ + 6I^- \longrightarrow 2Cr^{3+} + 7H_2O + 3I_2}$$ 2. Oxidises Iron(II) to Iron(III): $$\mathbf{Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}}$$ 3. Oxidises Hydrogen Sulfide to Sulfur: $$\mathbf{Cr_2O_7^{2-} + 8H^+ + 3H_2S \longrightarrow 2Cr^{3+} + 7H_2O + 3S}$$ 4. Oxidises Tin(II) to Tin(IV): $$\mathbf{Cr_2O_7^{2-} + 14H^+ + 3Sn^{2+} \longrightarrow 2Cr^{3+} + 7H_2O + 3Sn^{4+}}$$

6.2 Potassium Permanganate ($KMnO_4$)

Potassium permanganate is a dark purple (almost black) crystalline solid. It is an exceptionally powerful oxidizing agent in acidic, neutral, and alkaline media.

Preparation from Pyrolusite Ore (MnO2) Step 1 (Alkaline Fusion): Pyrolusite ore ($MnO_2$) is fused with $KOH$ in the presence of $O_2$ or an oxidising agent ($KNO_3$) to form dark green Potassium Manganate ($K_2MnO_4$). $$\mathbf{2MnO_2 + 4KOH + O_2 \longrightarrow 2K_2MnO_4 + 2H_2O}$$ Step 2 (Disproportionation / Electrolytic Oxidation): Laboratory Preparation from $Mn^{2+}$ Salt (with Peroxodisulphate): $$\mathbf{2Mn^{2+} + 5S_2O_8^{2-} + 8H_2O \longrightarrow 2MnO_4^- + 10SO_4^{2-} + 16H^+}$$
Preparation & Reactions of Potassium Permanganate (KMnO4)
Industrial Preparation Flowchart of Potassium Permanganate (KMnO4)
Physical & Structural Properties of KMnO4 - Isostructural: $KMnO_4$ crystals are isostructural with $KClO_4$.
- Thermal Decomposition (513 K): $$\mathbf{2KMnO_4 \xrightarrow{513 \text{ K}} K_2MnO_4 + MnO_2 + O_2}$$ - Color & Magnetism Mechanism: Permanganate ion ($MnO_4^-$) has $Mn(VII)$ with $3d^0$ configuration. It is diamagnetic. Its intense dark purple color is NOT due to d-d transitions, but due to Ligand-to-Metal Charge Transfer (LMCT) ($p$-orbital of $O^{2-} \rightarrow d$-orbital of $Mn^{7+}$).
- Manganate ion ($MnO_4^{2-}$) has $Mn(VI)$ with $3d^1$ configuration. It is paramagnetic (green).
- Both $MnO_4^-$ and $MnO_4^{2-}$ are tetrahedral.
Oxidising Reactions of KMnO4 (Acidic vs Alkaline) 1. In Acidic Medium ($E^\circ = +1.52 \text{ V}$): $$\mathbf{MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} \text{ (Colourless)} + 4H_2O}$$ 2. In Neutral or Faintly Alkaline Medium ($E^\circ = +1.69 \text{ V}$): $$\mathbf{MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 \text{ (Brown Solid)} + 4OH^-}$$ Warning (Hydrochloric Acid): Permanganate titrations in the presence of $HCl$ are unsatisfactory because $KMnO_4$ oxidises $HCl$ to chlorine gas ($Cl_2$).

7. The f-Block Elements: Lanthanoids

The f-block consists of two inner transition series: Lanthanoids (filling of $4f$ subshell, 14 elements from Cerium $Ce, Z=58$ to Lutetium $Lu, Z=71$) and Actinoids (filling of $5f$ subshell, 14 elements from Thorium $Th, Z=90$ to Lawrencium $Lr, Z=103$).

NCERT Table 4.9: Lanthanoids Configurations & Radii
Z Name Sym Atom (Ln) $Ln^{2+}$ $Ln^{3+}$ $Ln^{4+}$ Radii (pm)
57 Lanthanum La $[Xe]5d^1 6s^2$ $5d^1$ $4f^0$ -- 106
58 Cerium Ce $[Xe]4f^1 5d^1 6s^2$ $4f^2$ $4f^1$ $4f^0$ 103
59 Praseodymium Pr $[Xe]4f^3 6s^2$ $4f^3$ $4f^2$ $4f^1$ 101
60 Neodymium Nd $[Xe]4f^4 6s^2$ $4f^4$ $4f^3$ $4f^2$ 99
61 Promethium Pm $[Xe]4f^5 6s^2$ $4f^5$ $4f^4$ -- 98
62 Samarium Sm $[Xe]4f^6 6s^2$ $4f^6$ $4f^5$ -- 96
63 Europium Eu $[Xe]4f^7 6s^2$ $4f^7$ $4f^6$ -- 95
64 Gadolinium Gd $[Xe]4f^7 5d^1 6s^2$ $4f^7 5d^1$ $4f^7$ -- 94
65 Terbium Tb $[Xe]4f^9 6s^2$ $4f^9$ $4f^8$ $4f^7$ 92
66 Dysprosium Dy $[Xe]4f^{10} 6s^2$ $4f^{10}$ $4f^9$ $4f^8$ 91
67 Holmium Ho $[Xe]4f^{11} 6s^2$ $4f^{11}$ $4f^{10}$ -- 89
68 Erbium Er $[Xe]4f^{12} 6s^2$ $4f^{12}$ $4f^{11}$ -- 88
69 Thulium Tm $[Xe]4f^{13} 6s^2$ $4f^{13}$ $4f^{12}$ -- 87
70 Ytterbium Yb $[Xe]4f^{14} 6s^2$ $4f^{14}$ $4f^{13}$ -- 86
71 Lutetium Lu $[Xe]4f^{14} 5d^1 6s^2$ $4f^{14} 5d^1$ $4f^{14}$ -- 86

7.1 Lanthanoid Contraction (NCERT Fig 4.6)

NCERT Fig. 4.6: Trends in Ionic Radii of Ln3+ Ions (Lanthanoid Contraction)
NCERT Fig. 4.6: Trends in Ionic Radii of Ln3+ Ions (Lanthanoid Contraction)
Definition, Cause & Consequences Lanthanoid Contraction: The steady, cumulative decrease in atomic and ionic radii ($Ln^{3+}$) across the lanthanoid series from Lanthanum ($106 \text{ pm}$) to Lutetium ($86 \text{ pm}$).

Cause: With increasing atomic number, nuclear charge increases by $+1$ unit at each step, while the new electron enters the inner $4f$ subshell. The mutual shielding of $4f$ electrons is extremely poor (even worse than d-electrons). Consequently, effective nuclear charge increases, pulling the electron cloud strongly inward.

Three Major Consequences:
  1. Identical Radii of 4d and 5d Transition Series Pairs: Radii of $4d$ and $5d$ elements in the same group are nearly identical ($Zr=160\text{ pm}$ vs $Hf=159\text{ pm}$; $Nb=146\text{ pm}$ vs $Ta=146\text{ pm}$).
  2. Difficulty in Separation: Due to nearly identical ionic radii and chemical properties, pure separation of lanthanoids is extremely difficult.
  3. Decrease in Basic Strength of Hydroxides: As ionic radius of $Ln^{3+}$ decreases from $La^{3+}$ to $Lu^{3+}$, the covalent character of the $Ln-OH$ bond increases (Fajans' Rule). Hence, $La(OH)_3$ is the most basic, while $Lu(OH)_3$ is the least basic.
EXCEPTION 10: LANTHANOID & ACTINOID ALL SPECIAL EXCEPTIONS

7.3 Chemical Reactivity & Mischmetall

Burns in $\text{O}_2$: $\mathbf{Ln + O_2 \longrightarrow Ln_2O_3}$ Reacts with $\text{H}_2\text{O}$: $\mathbf{Ln + H_2O \longrightarrow Ln(OH)_3 + H_2(g)}$
With Acids: $\mathbf{Ln + H^+ \longrightarrow Ln^{3+} + H_2(g)}$ With Halogens: $\mathbf{Ln + X_2 \longrightarrow LnX_3}$
With $\text{N}_2$: $\mathbf{Ln + N_2 \xrightarrow{\Delta} LnN}$ With Carbon: $\mathbf{Ln + C \xrightarrow{2773\text{ K}} LnC_2}$
Mischmetall Composition & Uses Mischmetall: A well-known lanthanoid alloy consisting of: Uses: Used in Magnesium-based alloys to produce bullets, shell casings, and sparkling flints for cigarette lighters.

8. The f-Block Elements: Actinoids

The Actinoid series consists of 14 elements from Thorium ($Th, Z=90$) to Lawrencium ($Lr, Z=103$), involving the progressive filling of $5f$ orbitals.

Comparative Summary: Lanthanoids vs. Actinoids
Property Lanthanoids ($4f$ Series) Actinoids ($5f$ Series)
Orbital Filling $4f$ orbitals filled $5f$ orbitals filled
Oxidation States Mainly $+3$; occasionally $+2, +4$. Wide variety ($+3, +4, +5, +6, +7$).
Contraction Magnitude Smaller element-to-element contraction. Greater element-to-element contraction (poorer $5f$ shielding).
Radioactivity Non-radioactive (except Promethium, $Pm$). ALL actinoids are radioactive.
Complex Formation Lesser tendency to form complexes. Greater tendency to form complexes (smaller, higher charge).
Chemical Reactivity Less reactive than actinoids. Highly reactive (especially when finely divided).
Basic Strength Hydroxides less basic than actinoid hydroxides. Hydroxides more basic.

9. Applications of d- and f-Block Elements

10. High-Yield NCERT Solved Examples & Practice Problems

Practice Problem 8 (NCERT Solved Example 4.5) Question: How would you account for the increasing oxidising power in the series $VO_2^+ < Cr_2O_7^{2-} < MnO_4^-$?
Solution:
This trend is due to the increasing stability of the lower oxidation species to which these oxoanions are reduced. $VO_2^+$ is reduced to $VO^{2+}$ ($V^{IV}$), $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$ ($Cr^{III}$), and $MnO_4^-$ is reduced to $Mn^{2+}$ ($Mn^{II}$). As the stability of the resulting reduced state increases across the series, the oxidising power of the parent oxoanion increases.
Practice Problem 9 (NCERT Intext 4.7) Question: Which is a stronger reducing agent, $Cr^{2+}$ or $Fe^{2+}$, and why?
Solution:
$Cr^{2+}$ is a stronger reducing agent than $Fe^{2+}$.
Reason: Oxidation of $Cr^{2+}$ ($3d^4$) to $Cr^{3+}$ yields a $3d^3$ configuration, which corresponds to a stable half-filled $t_{2g}^3$ level in aqueous medium. On the other hand, oxidation of $Fe^{2+}$ ($3d^6$) to $Fe^{3+}$ yields a $3d^5$ configuration. The $d^4 \rightarrow d^3$ change involves a larger gain in stabilization energy in water than $d^6 \rightarrow d^5$.
Practice Problem 10 (NCERT Intext 4.9) Question: Explain why $Cu^+$ ion is not stable in aqueous solutions.
Solution:
In aqueous solution, $Cu^+$ undergoes disproportionation:
$$\mathbf{2Cu^+(aq) \longrightarrow Cu^{2+}(aq) + Cu(s)}$$
Although removing a second electron from copper requires high energy ($IE_2$), the hydration enthalpy of $Cu^{2+}(aq)$ is much more negative than that of $Cu^+(aq)$ due to its smaller size and higher charge. This huge hydration energy release more than compensates for $IE_2$, making $Cu^{2+}(aq)$ far more stable.
Practice Problem 11 (NCERT Solved Example 4.10) Question: Name a member of the lanthanoid series which is well known to exhibit a $+4$ oxidation state. Why does it do so?
Solution:
Cerium ($Ce, Z=58$) is well known to exhibit a $+4$ oxidation state. Its ground state electronic configuration is $[Xe] 4f^1 5d^1 6s^2$. By losing 4 electrons to form $Ce^{4+}$, it attains the highly stable noble gas configuration of Xenon ($[Xe] 4f^0$).
Practice Problem 12 (NCERT Intext 4.10) Question: Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Solution:
In actinoids, electrons are added to $5f$ orbitals, whereas in lanthanoids they are added to $4f$ orbitals. $5f$ orbitals extend further in space and are less buried than $4f$ orbitals, making the mutual shielding effect of $5f$ electrons even poorer than that of $4f$ electrons. Consequently, the effective nuclear charge increases more sharply across actinoids, causing a greater contraction from element to element.