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d- and f-Block Elements. Master transition metal chemistry, oxidation states, complexes, and Lanthanoids.
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General Properties & Atomic Trends
A transition element is defined as an element that has a partially filled d-orbital in its ground state or in any of its common oxidation states.
Zinc ($Zn$), Cadmium ($Cd$), and Mercury ($Hg$) of Group 12 have completely filled d-orbitals ($d^{10}$) in both their ground state and their common +2 oxidation state. Hence, they are technically not considered transition elements.
The general valence shell electronic configuration of d-block elements is:
$(n-1)d^{1-10} \, ns^{1-2}$
Where $(n-1)$ stands for the inner penultimate shell.
Transition elements have a large number of unpaired electrons in their $(n-1)d$ orbitals. This allows for exceptionally strong interatomic interactions (strong metallic bonding) between atoms in the solid crystal lattice. Breaking these strong metallic bonds requires very high energy, hence the high enthalpy of atomization.
Zinc has the electronic configuration $3d^{10} \, 4s^2$. It has no unpaired electrons in its d-orbitals. Consequently, metallic bonding in zinc is comparatively very weak, resulting in the lowest enthalpy of atomization in the 3d series.
As we move from left to right, the atomic radii first decrease (Sc to V) because the increase in effective nuclear charge overpowers the weak shielding of d-electrons.
Then, it becomes nearly constant (Cr to Cu) as increased shielding by added d-electrons balances the increased nuclear charge.
Finally, it slightly increases at the end (Zn) due to high inter-electronic repulsions in completely filled $3d^{10}$ orbitals.
Transition elements have their valence electrons in two different sets of orbitals: the $(n-1)d$ and the $ns$ orbitals. Because the energy difference between these two orbitals is very small, electrons from both can easily participate in bond formation under different conditions, leading to variable oxidation states.
Manganese ($Mn$) exhibits the maximum number of oxidation states (from +2 to +7).
Reason: It has the maximum number of unpaired electrons available for bonding in its valence shell (electronic configuration $3d^5 \, 4s^2$, giving $5+2 = 7$ valence electrons).
Although nuclear charge increases across the series, the added electrons enter the inner $(n-1)d$ subshell. These inner d-electrons provide substantial shielding (screening effect) to the outer $ns$ electrons. The opposing forces of increased nuclear charge and increased shielding result in irregular and very small variations in ionization enthalpies.
The standard electrode potential depends on three thermodynamic factors: Enthalpy of Atomization ($\Delta H_{at}$), Ionization Enthalpy ($IE_1 + IE_2$), and Hydration Enthalpy ($\Delta H_{hyd}$). Because ionization enthalpies and atomization enthalpies vary irregularly across the 3d series, their sum creates an irregular trend in $E^\circ$ values.
Copper is the only metal in the 3d series with a positive reduction potential. This is because the sum of its high Enthalpy of Atomization and high Ionization Enthalpies ($IE_1 + IE_2$) is not balanced by its Hydration Enthalpy. Thus, transforming solid $Cu$ to aqueous $Cu^{2+}$ is thermodynamically unfavorable, making it a poor reducing agent.
Aqueous Chemistry & Stability
In aqueous solution, $Cu^+$ undergoes spontaneous disproportionation to form $Cu^{2+}$ and solid $Cu$:
$2Cu^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + Cu_{(s)}$
This happens because the much higher hydration enthalpy of the highly charged $Cu^{2+}$ ion completely compensates for the second ionization enthalpy required to form it, making $Cu^{2+}$ thermodynamically much more stable in water.
$Cr^{2+}$ acts as a reducing agent (loses an electron) because it changes its configuration from $d^4$ to $d^3$. In an aqueous medium, $d^3$ is exceptionally stable due to a half-filled $t_{2g}$ crystal field level.
Conversely, $Mn^{3+}$ acts as an oxidizing agent (gains an electron) because it changes from $d^4$ to $d^5$, achieving an extremely stable exactly half-filled d-subshell.
Oxygen and Fluorine are highly electronegative atoms with very small sizes. They can easily draw away electron density and force the transition metal to share all its valence electrons, thus stabilizing it in its highest possible oxidation state (e.g., $Mn_2O_7$, $OsF_6$).
While fluorine is more electronegative, oxygen is capable of forming multiple bonds (pi bonds) with transition metals (e.g., in $MnO_4^-$). This ability to form double bonds allows oxygen to stabilize extremely high oxidation states better than fluorine, which can only form single bonds.
A disproportionation reaction is a specific type of redox reaction where the exact same element is simultaneously oxidized and reduced.
Example: In an acidic medium, the green manganate ion ($Mn^{+6}$) disproportionates into purple permanganate ($Mn^{+7}$) and solid manganese dioxide ($Mn^{+4}$).
$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$.
The thermodynamic stability depends on ionization enthalpies.
For the +2 state, the sum of first two ionization enthalpies $(IE_1 + IE_2)$ is less for $Ni$ than for $Pt$, making $Ni^{2+}$ more stable.
For the +4 state, the sum of the first four ionization enthalpies $(IE_1 + IE_2 + IE_3 + IE_4)$ is less for $Pt$ than for $Ni$, making $Pt^{4+}$ more stable.
A highly positive value indicates a strong tendency to get reduced. $Mn^{3+}$ has a $d^4$ configuration. Gaining an electron reduces it to $Mn^{2+}$, which has a exactly half-filled, highly stable $d^5$ configuration. This massive gain in stability is the driving force behind the highly positive $E^\circ$ value.
Titanium has the configuration $3d^2 \, 4s^2$.
$Ti^{3+}$ is $3d^1$.
$Ti^{4+}$ is $3d^0$.
$Ti^{4+}$ is much more stable because losing all 4 valence electrons allows it to attain the completely filled, highly stable noble gas electronic configuration of Argon.
In the presence of strong field ligands (like $NH_3$ or $CN^-$), $Co^{3+}$ (which is $d^6$) undergoes pairing of all its electrons in the lower energy $t_{2g}$ level, resulting in an exceptionally large Crystal Field Stabilization Energy (CFSE). This massive energy release easily compensates for the third ionization enthalpy, causing $Co^{2+}$ to oxidize to $Co^{3+}$.
In lower oxidation states (e.g., $MnO$), the metal retains some valence electrons and low charge density, making the M-O bond highly ionic and the oxide basic.
In higher oxidation states (e.g., $Mn_2O_7$), the metal has a very high charge density, polarizing the oxygen heavily. This makes the M-O bond largely covalent. Covalent oxides react with water to yield $H^+$, making them highly acidic.
Physical & Chemical Properties
Transition metal ions typically have partially filled d-orbitals. In a complex, ligands cause these d-orbitals to split into two different energy levels. Electrons absorb specific wavelengths of visible light to jump from the lower d-level to the higher d-level (d-d transition). The transmitted light (complementary color) is what our eyes perceive.
$Sc^{3+}$ has a $3d^0$ configuration (empty d-orbitals).
$Zn^{2+}$ has a $3d^{10}$ configuration (completely filled d-orbitals).
Since neither ion possesses partially filled d-orbitals, no d-d electron transition is physically possible. Therefore, they cannot absorb visible light and appear colorless.
Paramagnetism arises due to the presence of unpaired electrons in the d-orbitals. Each unpaired electron acts like a tiny magnet due to its spin. The "spin-only" magnetic moment ($\mu$) is calculated by:
$\mu = \sqrt{n(n+2)} \text{ B.M.}$
Where $n$ = number of unpaired electrons, and B.M. = Bohr Magneton.
Atomic Number 25 is Manganese ($Mn$).
$Mn$ ground state: $[Ar] 3d^5 \, 4s^2$.
$Mn^{2+}$ configuration: $[Ar] 3d^5$. Number of unpaired electrons ($n$) = 5.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx \mathbf{5.92 \text{ B.M.}}$
- Variable Oxidation States: They can easily easily adopt multiple oxidation states to form unstable intermediate compounds with reactants, providing a lower-energy pathway.
- Large Surface Area: Solid metals (like finely divided $Fe$ or $Ni$) provide extensive surface area with free valencies to adsorb reactant molecules, weakening their bonds.
- They have comparatively smaller ionic sizes and high effective nuclear charge (ionic charge), allowing them to attract ligands strongly.
- They possess vacant d-orbitals of appropriate energy to accept lone pairs of electrons donated by ligands.
Interstitial compounds are formed when small atoms like Hydrogen, Carbon, or Nitrogen are trapped inside the tiny spaces (interstices) of the transition metal crystal lattice.
Transition metals form them because they have large crystal structures with appropriately sized voids that can accommodate these small non-metal atoms without disrupting the lattice.
- They are extremely hard (some approach diamond in hardness).
- They have much higher melting points than pure metals.
- They retain metallic conductivity.
- They are chemically highly inert.
Transition metals have very similar atomic radii (within 15% of each other). Because of this size similarity, atoms of one transition metal can easily and seamlessly replace atoms of another transition metal in a solid crystal lattice to form solid solutions (alloys) like Brass or Bronze.
Because $Mn^{+7}$ has no d-electrons ($d^0$), d-d transitions are impossible. The intense purple color is entirely due to Ligand-to-Metal Charge Transfer (LMCT). An electron is temporarily transferred from the oxygen ligand to the vacant d-orbital of the highly charged Manganese atom, absorbing green light and appearing purple.
Dichromate & Permanganate
Step 1: Fusion of chromite ore with $Na_2CO_3$ in excess air to form yellow sodium chromate.
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
Step 2: Acidification of sodium chromate extract to form orange sodium dichromate.
$2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$
Step 3: Conversion of sodium dichromate to potassium dichromate by adding KCl. Because $K_2Cr_2O_7$ is much less soluble in cold water than the sodium salt, it crystallizes out as orange crystals.
$Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 \downarrow + 2NaCl$
Chromate (yellow) and dichromate (orange) ions are interconvertible depending on the pH of the solution. They exist in equilibrium:
$2CrO_4^{2-} \text{ (yellow)} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} \text{ (orange)} + H_2O$
In acidic medium (pH < 7), the equilibrium shifts to the right (orange).
In basic medium (pH > 7), $OH^-$ removes $H^+$, shifting equilibrium to the left (yellow).
Dichromate acts as a strong oxidizing agent in acidic medium, getting reduced to $Cr^{3+}$. It oxidizes $Fe^{2+}$ to $Fe^{3+}$.
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
The orange dichromate solution turns green due to the formation of $Cr^{3+}$ ions, and it oxidizes hydrogen sulfide ($H_2S$) to pale yellow solid sulphur.
$Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S\downarrow + 7H_2O$
Step 1: Pyrolusite ore ($MnO_2$) is fused with KOH in the presence of an oxidizing agent ($O_2$ or $KNO_3$) to yield green Potassium Manganate.
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
Step 2: The green manganate is then electrolytically oxidized (or disproportionated in acidic medium) to purple Potassium Permanganate.
$MnO_4^{2-} \xrightarrow{\text{electrolysis}} MnO_4^- + e^-$
Permanganate oxidizes oxalate ($C_2O_4^{2-}$) to carbon dioxide ($CO_2$). This specific reaction requires heating to about $60^\circ\text{C}$ to initiate.
$2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 10CO_2\uparrow + 8H_2O$
In a neutral or alkaline medium, permanganate ($Mn^{+7}$) is reduced to manganese dioxide ($MnO_2$, $+4$), and it remarkably oxidizes iodide ($I^-$) all the way up to Iodate ($IO_3^-$).
$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2\downarrow + 2OH^- + IO_3^-$
When heated to 513 K, $KMnO_4$ decomposes to yield Potassium Manganate, Manganese dioxide, and Oxygen gas.
$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2\uparrow$
The Chromate ion ($CrO_4^{2-}$) has a perfect tetrahedral geometry with Cr at the center.
The Dichromate ion ($Cr_2O_7^{2-}$) consists of two tetrahedra sharing one corner. It features a $Cr-O-Cr$ bridge with a bond angle of approximately $126^\circ$.
The f-Block Elements
Lanthanoids (inner transition elements) are a series of 14 elements from Cerium (Ce, Z=58) to Lutetium (Lu, Z=71) where the 4f subshell is progressively filled.
General Configuration: $[Xe] 4f^{1-14} \, 5d^{0-1} \, 6s^2$.
The steady and regular decrease in atomic and ionic radii of lanthanoid elements with increasing atomic number is called the lanthanoid contraction.
Cause: The 4f electrons provide incredibly poor shielding. As nuclear charge increases across the series, the 4f electrons fail to screen each other effectively, causing the nucleus to pull the valence shell inward with increasing force.
Because of this contraction, the elements of the second and third transition series (e.g., Zirconium $Zr$ and Hafnium $Hf$) have almost identical atomic radii. Due to identical sizes and similar configurations, they have nearly identical physical and chemical properties, making them extremely difficult to separate.
The principal and most stable oxidation state of all lanthanoids is +3.
Cerium ($Ce$) prominently exhibits a +4 oxidation state ($Ce^{4+}$) because doing so leaves it with a highly stable noble gas (Xenon) $f^0$ configuration.
Although $Ce^{4+}$ is stable due to its $f^0$ configuration, the ultimate stable state for all lanthanoids is +3. Therefore, $Ce^{4+}$ strongly desires to gain an electron to revert back to the $Ce^{3+}$ state. This makes it an excellent and strong oxidizing agent.
Europium forms a $Eu^{2+}$ ion because doing so grants it an exactly half-filled, stable $4f^7$ configuration. However, because the common stable state is +3, $Eu^{2+}$ tends to lose one electron to become $Eu^{3+}$, acting as a strong reducing agent.
Mischmetall is a highly pyrophoric alloy composed primarily of lanthanoid metals ($\approx 95\%$), Iron ($\approx 5\%$), and traces of S, C, Ca, and Al.
Use: It sparks easily when struck, so it is widely used in magnesium-based alloys to produce bullets, shells, and lighter flints.
Like lanthanoids, actinoids experience a steady decrease in size with increasing atomic number. This is the Actinoid Contraction.
It is stronger (greater decrease from element to element) because the 5f electrons provide even poorer shielding than 4f electrons, allowing the effective nuclear charge to compress the atom more drastically.
In actinoids, the 5f, 6d, and 7s energy levels are of comparable energies (very close to each other). Consequently, electrons from all three shells can participate in bonding, leading to a much wider variety of oxidation states (up to +7) compared to lanthanoids where 4f is buried deeper.
- Radioactivity: Except Promethium, all lanthanoids are non-radioactive. All actinoids are radioactive.
- Complex Formation: Lanthanoids have less tendency to form complexes. Actinoids have a much stronger tendency to form complex compounds due to higher charge density.
Chapter 4 Mastered!
You have just conquered the 50 most critical subjective questions for Class 12 Chemistry, Chapter 4: d- and f-Block Elements. Transition to the next chapter with confidence!
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