Transition & Inner Transition Elements
d and f Block Elements: Properties, Lanthanoid Contraction & Board PYQs
1. Introduction to Transition Elements (d-Block)
The elements belonging to groups 3 to 12 of the periodic table are called d-block elements. They are called transition elements because their properties represent a transition between the highly reactive s-block metals and the less reactive p-block elements.
Definition: A transition element is defined as an element which has a partially filled (incompletely filled) d-subshell in its atomic state or in any of its common oxidation states.
Exception: Zinc (Zn), Cadmium (Cd), and Mercury (Hg) have completely filled d-orbitals ($d^{10}$) in their ground state as well as in their common oxidation states (e.g., $Zn^{2+}$ is $3d^{10}$). Hence, they are d-block elements but not considered typical transition elements.
2. General Electronic Configuration
The general outer electronic configuration of d-block elements is:
$(n-1)d^{1-10} \, ns^{1-2}$
The 3d Series (Scandium to Zinc)
The filling of electrons takes place in the 3d subshell. General configuration: $[Ar] 3d^{1-10} 4s^{1-2}$.
Anomalous Electronic Configurations: Chromium and Copper
- Chromium (Cr, Z=24): Expected: $[Ar] 3d^4 4s^2$.
Actual: $[Ar] 3d^5 4s^1$.
Reason: Shifting one electron from 4s to 3d gives a half-filled d-subshell ($d^5$), which provides extra stability due to symmetrical distribution of electrons and higher exchange energy. - Copper (Cu, Z=29): Expected: $[Ar] 3d^9 4s^2$.
Actual: $[Ar] 3d^{10} 4s^1$.
Reason: Shifting one electron gives a completely filled d-subshell ($d^{10}$), providing extra thermodynamic stability.
3. Properties of d-Block Elements
A. Variable Oxidation States
Transition elements exhibit a large number of oxidation states because the energy difference between the $(n-1)d$ and $ns$ orbitals is very small. Both $ns$ and $(n-1)d$ electrons can participate in bond formation.
- Scandium (Sc) shows only +3.
- Manganese (Mn) shows the maximum number of oxidation states (+2 to +7) in the 3d series because it has the maximum number of unpaired electrons ($3d^5 4s^2$).
B. Magnetic Properties
Substances are classified based on their behavior in a magnetic field:
- Paramagnetic: Attracted by a magnetic field. Arises due to the presence of unpaired electrons. Most transition metal ions are paramagnetic.
- Diamagnetic: Repelled by a magnetic field. Arises when all electrons are paired (e.g., $Zn^{2+}, Sc^{3+}, Cu^+$).
Calculation of Magnetic Moment ($\mu$)
The "spin-only" formula for magnetic moment is:
$$\mu = \sqrt{n(n + 2)} \text{ B.M.}$$
Where $n$ = number of unpaired electrons. Unit is Bohr Magneton (B.M.).
C. Color of Transition Metal Ions
Most transition metal compounds are colored in their solid state or in aqueous solution. This is due to d-d transitions.
When ligands approach the transition metal ion, the degenerate d-orbitals split into two sets of different energies ($t_{2g}$ and $e_g$). An electron from a lower energy d-orbital can absorb a specific wavelength from visible light and jump to a higher energy d-orbital. The transmitted/reflected light gives the complementary color to the compound.
Note: Ions with $d^0$ (e.g., $Sc^{3+}, Ti^{4+}$) or $d^{10}$ (e.g., $Zn^{2+}, Cu^+$) configurations are colorless because d-d transitions are not possible.
D. Catalytic Properties
Many transition metals and their compounds act as excellent catalysts (e.g., $V_2O_5$ in Contact process, finely divided Fe in Haber process). This is due to:
- Their ability to adopt multiple (variable) oxidation states to form unstable intermediate compounds.
- They provide a large surface area with free valencies for the adsorption of reactant molecules.
E. Formation of Interstitial Compounds
Transition metals form interstitial compounds when small atoms like H, C, N, or B are trapped inside the empty spaces (interstices) of their crystal lattices. (e.g., Steel, Cast Iron).
Properties: They are typically very hard, have high melting points, and retain metallic conductivity. They are chemically inert.
4. Important Compounds of Transition Metals
Potassium Permanganate ($KMnO_4$)
It is a dark purple crystalline solid and a powerful oxidizing agent.
Preparation from Pyrolusite ($MnO_2$):
- Conversion of $MnO_2$ to Potassium Manganate ($K_2MnO_4$):
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 (\text{green}) + 2H_2O$ - Oxidation of $K_2MnO_4$ to $KMnO_4$ (Electrolytic or by Chlorine):
$2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 (\text{purple}) + 2KCl$
Potassium Dichromate ($K_2Cr_2O_7$)
It is an orange crystalline solid, widely used as an oxidizing agent in volumetric analysis.
Effect of pH on Chromate/Dichromate:
Chromate ($CrO_4^{2-}$, yellow) and dichromate ($Cr_2O_7^{2-}$, orange) ions are interconvertible in aqueous solution depending on the pH.
$2CrO_4^{2-} (\text{Yellow}) + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (\text{Orange}) + H_2O$
$Cr_2O_7^{2-} (\text{Orange}) + 2OH^- \rightleftharpoons 2CrO_4^{2-} (\text{Yellow}) + H_2O$
5. Inner Transition Elements (f-Block)
Elements in which the last electron enters the anti-penultimate $(n-2)f$ orbital are called f-block or inner transition elements. They consist of two series: Lanthanoids (4f series) and Actinoids (5f series).
A. Lanthanoids (Atomic number 58 to 71)
- General Configuration: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
- Oxidation State: The most common and stable oxidation state is +3. Some elements show +2 (e.g., $Eu^{2+}, Yb^{2+}$) or +4 (e.g., $Ce^{4+}$) to achieve stable $f^0, f^7,$ or $f^{14}$ configurations.
Lanthanoid Contraction
Definition: The steady and gradual decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number (from Cerium to Lutetium) is known as lanthanoid contraction.
Cause:
As the atomic number increases, the positive nuclear charge increases. The new electrons enter the inner 4f subshell. The shielding (screening) effect of 4f electrons is very poor due to the diffused shape of f-orbitals. Therefore, the effective nuclear charge pulling the valence electrons inward increases steadily, shrinking the atomic size.
Consequences (Effects):
- Similarity in radii of 2nd and 3rd transition series: The radii of elements of the 5d series (e.g., Hf, Ta) are almost identical to those of the corresponding 4d series (e.g., Zr, Nb) due to lanthanoid contraction. Thus, Zr and Hf have identical sizes and chemical properties, making their separation difficult.
- Decrease in basic strength of hydroxides: As the size of $Ln^{3+}$ ion decreases from $La^{3+}$ to $Lu^{3+}$, the covalent character of the $Ln-OH$ bond increases (Fajans' rule). Hence, basic strength decreases from $La(OH)_3$ (most basic) to $Lu(OH)_3$ (least basic).
B. Actinoids (Atomic number 90 to 103)
These elements involve the filling of 5f orbitals. General configuration: $[Rn] 5f^{1-14} 6d^{0-1} 7s^2$.
- All actinoids are radioactive.
- They exhibit a large number of oxidation states (+3, +4, +5, +6, +7) because the energy difference between 5f, 6d, and 7s orbitals is very small.
- Actinoid Contraction: Similar to lanthanoid contraction, but the contraction is greater from element to element because 5f electrons have even poorer shielding effect than 4f electrons.
6. Solved Textbook Numericals & Reasoning
Problem 1: Magnetic Moment Calculation
Calculate the spin-only magnetic moment of $Fe^{2+}$ ion. (Atomic number of Fe = 26).
Solution:
1. Electronic configuration of Fe ($Z=26$): $[Ar] 3d^6 4s^2$
2. Electronic configuration of $Fe^{2+}$ (loss of two 4s electrons): $[Ar] 3d^6$
3. Arrangement in 3d subshell (5 degenerate orbitals):
$\uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow$
4. Number of unpaired electrons ($n$) = 4
5. Formula: $\mu = \sqrt{n(n + 2)}$ B.M.
$\mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} = 4.89 \text{ B.M.}$
Answer: The magnetic moment of $Fe^{2+}$ is 4.89 B.M.
Problem 2: Coloring properties
Why are $Sc^{3+}$ salts colorless, whereas $Ti^{3+}$ salts are colored? (Z: Sc=21, Ti=22)
Reason:
Electronic configuration of Sc ($Z=21$) is $[Ar] 3d^1 4s^2$. The $Sc^{3+}$ ion is formed by losing 3 electrons, so its configuration is $[Ar] 3d^0$. Since it has no electrons in the d-orbitals, d-d transition is not possible. Hence, $Sc^{3+}$ salts are colorless.
Electronic configuration of Ti ($Z=22$) is $[Ar] 3d^2 4s^2$. The $Ti^{3+}$ ion has the configuration $[Ar] 3d^1$. It has one unpaired electron in the d-orbital, which can undergo a d-d transition by absorbing light from the visible region. Hence, $Ti^{3+}$ salts are colored (purple).
7. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Write the general electronic configuration of Lanthanoids. (Oct 2013, March 2018)
Answer: $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
Q2. Why is zinc not considered a transition element? (March 2016, Oct 2020)
Answer: Zinc ($Z=30$) has a completely filled d-subshell ($3d^{10}$) in its ground state as well as in its common oxidation state ($Zn^{2+}$). Hence, it lacks a partially filled d-orbital required to be a transition element.
Q3. Write the spin-only formula for magnetic moment. (March 2017)
Answer: $\mu = \sqrt{n(n + 2)} \text{ B.M.}$ where $n$ is the number of unpaired electrons.
2 Mark Questions (SA-I)
Q4. What is Lanthanoid contraction? State any one of its causes. (March 2015, Oct 2019, March 2022)
Lanthanoid Contraction: The steady and gradual decrease in atomic and ionic radii of lanthanoids with an increase in atomic number is called lanthanoid contraction.
Cause: It is caused by the imperfect/poor shielding effect of the 4f electrons. As atomic number increases, the increased nuclear charge strongly attracts the outer electrons inwards because the diffuse 4f orbitals fail to shield them effectively.
Q5. Why do transition metals form alloys easily? (March 2014, March 2021)
Transition metals have very similar atomic radii (they do not differ by more than 15%). Because of this similar size, atoms of one transition metal can easily substitute or replace the atoms of another transition metal in its crystal lattice, forming solid solutions known as alloys.
3 Mark Questions (SA-II)
Q6. Explain why transition elements show variable oxidation states. Name the transition element which shows maximum oxidation state in 3d series. (Oct 2014, March 2019, March 2023)
Reason: In transition elements, the energy difference between the outer $ns$ orbital and the inner $(n-1)d$ orbital is very small. Therefore, after the $ns$ electrons are lost to form lower oxidation states, the $(n-1)d$ electrons can also be easily lost or shared to form higher oxidation states. Since a variable number of $(n-1)d$ electrons can take part in bonding, they exhibit variable oxidation states.
Element with max oxidation state: Manganese (Mn) shows the maximum oxidation state of +7 in the 3d series (e.g., in $KMnO_4$) because it has the maximum number of unpaired electrons ($3d^5 4s^2$) available for bonding.
Q7. What are interstitial compounds? Write any two characteristics of interstitial compounds. (Oct 2016, March 2020)
Definition: Compounds formed when small non-metallic atoms like Hydrogen (H), Carbon (C), Nitrogen (N), or Boron (B) get trapped inside the empty spaces (interstices) of the crystal lattice of transition metals are called interstitial compounds.
Characteristics (any two):
- They have high melting points, higher than those of pure metals.
- They are very hard (some borides approach the hardness of diamond).
- They retain metallic conductivity.
- They are generally chemically inert.
📚 Also Read
Lecture Notes🏆 Complete Maharashtra HSC Class 12 Chemistry Preparation
Prepare for the Maharashtra HSC Class 12 Chemistry Board Exam with chapter-wise revision notes, important questions, PYQs, formula sheets, mock tests, quick revision resources and exam-oriented study material. Everything you need to score high in one comprehensive learning hub.
🚀 Explore the Complete Maharashtra HSC Chemistry Hub
No comments:
Post a Comment