Chapter 8: Transition and Inner Transition Elements
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The general electronic configuration of lanthanoids is $[Xe] 4f^{1-14} 5d^{0-1} 6s^2$.
Answer: The spin-only formula is $\mu = \sqrt{n(n + 2)} \text{ B.M.}$, where $n$ is the number of unpaired electrons and B.M. stands for Bohr Magneton.
Answer: Zinc ($Z=30$) has a completely filled d-subshell ($3d^{10}$) in its ground state as well as in its most common oxidation state ($Zn^{2+}$). Since a transition element requires a partially filled d-orbital, Zinc is not considered a true transition element.
Answer: The observed electronic configuration of Chromium is $[Ar] 3d^5 4s^1$. (One electron shifts from the 4s to the 3d orbital to achieve the extra stability of a half-filled $d^5$ configuration).
2 Mark Questions (Short Answer-I)
Answer:
Lanthanoid Contraction: The steady and gradual decrease in the atomic and ionic radii of the lanthanoid elements (from Cerium to Lutetium) with an increase in atomic number is called lanthanoid contraction.
Cause: As the atomic number increases across the lanthanoid series, new electrons are added to the inner 4f subshell. The 4f electrons have a highly diffused shape and provide very poor shielding (screening) effect. Consequently, the effective nuclear charge pulling the valence shell electrons inwards increases steadily, causing the atomic size to shrink.
Answer:
| Lanthanoids | Actinoids |
|---|---|
| 1. Involve the progressive filling of 4f orbitals. | 1. Involve the progressive filling of 5f orbitals. |
| 2. Except Promethium, they are non-radioactive in nature. | 2. All actinoids are highly radioactive elements. |
| 3. Do not have a strong tendency to form complex compounds. | 3. Have a much stronger tendency to form complex compounds. |
| 4. Show limited oxidation states (mostly +3, sometimes +2, +4). | 4. Show a wider range of oxidation states (+3, +4, +5, +6, +7). |
Answer:
Alloys are solid solutions of two or more metals. Transition metals form alloys very easily because their atomic radii are very similar (they do not differ by more than 15%). Because of this similarity in size, atoms of one transition metal can easily take the positions of the atoms of another transition metal in its crystal lattice without distorting the structure significantly.
3 Mark Questions (Short Answer-II / Numericals)
Answer:
Reason for Color:
In aqueous solutions or in solid compounds, the five degenerate d-orbitals of transition metal ions split into two sets of different energies ($t_{2g}$ and $e_g$ sets) due to the approach of ligands (water molecules/anions). If the metal ion has incompletely filled d-orbitals (unpaired electrons), an electron from a lower energy d-orbital can absorb a specific wavelength of light from the visible spectrum and jump to a higher energy d-orbital. This is called a d-d transition. The transmitted or reflected light gives the complementary color to the solution.
Why $Sc^{3+}$ is colorless and $Ti^{3+}$ is colored:
- $Sc^{3+}$ ($Z=21$): The electronic configuration of $Sc$ is $[Ar] 3d^1 4s^2$. The $Sc^{3+}$ ion is formed by losing 3 electrons, making its configuration $[Ar] 3d^0$. Since there are no electrons in the d-orbitals, d-d transitions cannot occur. Hence, it is colorless.
- $Ti^{3+}$ ($Z=22$): The electronic configuration of $Ti$ is $[Ar] 3d^2 4s^2$. The $Ti^{3+}$ ion has the configuration $[Ar] 3d^1$. It possesses one unpaired electron in the 3d subshell, which undergoes d-d transition by absorbing light in the visible region. Hence, it is colored (purple).
Answer:
Definition:
Interstitial compounds are those which are 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. They are non-stoichiometric in nature (e.g., steel, cast iron, $TiC, VH_{0.56}$).
Characteristics (Write any two):
- They have very high melting points, which are often higher than those of the pure metals.
- They are extremely hard. Some borides approach the hardness of diamond.
- They retain metallic conductivity.
- They are generally chemically inert.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark numerical combined with a 2-mark reasoning question).
(b) Explain why transition metals act as good catalysts. [2 Marks] March 2017, March 2019, Oct 2022
Answer (a): Numerical on Magnetic Moment
Step 1: Write electronic configurations
Atomic number of Cobalt (Co) = 27.
Ground state electronic configuration of Co: $[Ar] 3d^7 4s^2$
Electronic configuration of $Co^{2+}$ ion (loss of two 4s electrons): $[Ar] 3d^7$
Step 2: Determine number of unpaired electrons ($n$)
Filling 7 electrons in the five degenerate 3d orbitals (according to Hund's rule):
$\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow$
There are 3 unpaired electrons. So, $n = 3$.
Step 3: Calculate magnetic moment ($\mu$)
$$\mu = \sqrt{n(n + 2)} \text{ B.M.}$$
$$\mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15}$$
$$\mu = 3.87 \text{ B.M.}$$
The spin-only magnetic moment of $Co^{2+}$ is 3.87 B.M.
Answer (b): Catalytic Properties
Transition metals and their compounds are widely used as commercial catalysts (e.g., $V_2O_5$, finely divided Iron, Nickel). They act as good catalysts due to the following reasons:
- Variable Oxidation States: Transition metals have the ability to adopt multiple oxidation states. This allows them to form unstable intermediate compounds with reactants, providing an alternative reaction pathway with a lower activation energy.
- Large Surface Area: In their finely divided state, they provide a large solid surface area with free valencies. This allows reactant molecules to be easily adsorbed onto their surface, bringing them closer together and facilitating the reaction.
(b) Transition elements exhibit variable oxidation states. Why? [2 Marks] Oct 2013, March 2020
Answer (a): Preparation of $KMnO_4$
The preparation of Potassium Permanganate involves two main steps:
Step 1: Conversion of Pyrolusite to Potassium Manganate
Finely powdered pyrolusite ore ($MnO_2$) is fused with an alkali metal hydroxide like Potassium hydroxide ($KOH$) in the presence of air or an oxidizing agent (like $KNO_3$). A dark green mass of Potassium Manganate ($K_2MnO_4$) is formed.
$2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 (\text{green}) + 2H_2O$
Step 2: Oxidation of Potassium Manganate to Potassium Permanganate
The green solution of $K_2MnO_4$ is extracted with water and then oxidized either electrolytically or by passing chlorine or ozone gas through it. This converts the green manganate ion into the dark purple permanganate ion.
$2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 (\text{dark purple}) + 2KCl$
Answer (b): Reasoning for Variable Oxidation States
In transition elements, the energy difference between the valence $ns$ orbital and the inner $(n-1)d$ orbital is very small. When transition metals form compounds, they readily lose the $ns$ electrons to exhibit lower oxidation states (like +2). However, because the energy gap is so small, they can also easily lose a variable number of $(n-1)d$ electrons to form higher oxidation states. Since both $s$ and $d$ electrons can participate in bonding, transition elements exhibit a wide variety of oxidation states.
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