Chapter 8: Transition and Inner Transition Elements
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
- Use of logarithmic tables is allowed. Calculators are not permitted.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
The highest oxidation state shown by any transition element of the 3d series is:
(A) +6(B) +7(C) +8(D) +5
-
The calculated spin-only magnetic moment for an ion with 3 unpaired electrons is:
(A) 1.73 B.M.(B) 2.83 B.M.(C) 3.87 B.M.(D) 4.90 B.M.
-
Which of the following elements is a Lanthanoid?
(A) Thorium (Th)(B) Uranium (U)(C) Cerium (Ce)(D) Radium (Ra)
-
Zinc, Cadmium, and Mercury are generally not considered true transition elements because:
(A) They have low melting points.(B) They have completely filled d-orbitals.(C) They are highly reactive.(D) They do not form complexes.
Q2. Answer the following questions in one sentence: [3 Marks]
- Write the observed electronic configuration of Copper (Z=29).
- What is the primary cause of Lanthanoid Contraction?
- Name the alloy of lanthanoids mainly used for the production of bullets and lighter flints.
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Calculate the spin-only magnetic moment of the $Ni^{2+}$ ion. (Atomic number of Ni = 28).
- Why do transition metals exhibit variable oxidation states?
- Distinguish between Lanthanoids and Actinoids. (Any 2 points).
- What are interstitial compounds? Give one example.
- Explain why the majority of transition metal ions are colored in their aqueous solutions.
SECTION C
Attempt any TWO of the following: [6 Marks]
- Describe the preparation of Potassium Permanganate ($KMnO_4$) from Pyrolusite ore ($MnO_2$).
- Explain the catalytic properties of transition metals with any two suitable examples.
- What is Lanthanoid Contraction? State any two of its consequences.
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) Explain why $Sc^{3+}$ is a colorless ion while $Ti^{3+}$ is a colored ion. (Atomic numbers: Sc = 21, Ti = 22). [2 Marks]
(b) Write the general electronic configuration of Actinoids. [2 Marks] - (a) Describe the steps involved in the preparation of Potassium Dichromate ($K_2Cr_2O_7$) from chromite ore. [3 Marks]
(b) Write the general electronic configuration of the 3d transition series. [1 Mark]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (B) +7 [1 Mark. Exhibited by Manganese (Mn)]
2. (C) 3.87 B.M. [1 Mark. $\mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87$]
3. (C) Cerium (Ce) [1 Mark for correct option]
4. (B) They have completely filled d-orbitals. [1 Mark for correct option]
Q2. Very Short Answers:
1. Electronic configuration of Copper:
The observed electronic configuration of Cu (Z=29) is $[Ar] 3d^{10} 4s^1$. [1 Mark]
2. Cause of Lanthanoid Contraction:
It is caused by the poor shielding (screening) effect of the highly diffused 4f electrons, which fails to balance the increasing effective nuclear charge. [1 Mark]
3. Alloy of Lanthanoids:
Mischmetal. [1 Mark]
SECTION B [8 Marks]
Q3. Magnetic Moment of $Ni^{2+}$:
Atomic number of Ni = 28. Configuration of Ni: $[Ar] 3d^8 4s^2$. [1/2 Mark]
Configuration of $Ni^{2+}$ (loss of two 4s electrons): $[Ar] 3d^8$. [1/2 Mark]
Filling 8 electrons in 3d subshell leaves 2 unpaired electrons ($n=2$).
$\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8}$ [1/2 Mark]
$\mu = 2.83 \text{ B.M.}$ [1/2 Mark]
Q4. Variable Oxidation States:
Transition elements exhibit variable oxidation states because the energy difference between the $(n-1)d$ orbitals and the outermost $ns$ orbitals is very small. [1 Mark]
As a result, after losing the $ns$ electrons (to form lower oxidation states like +2), they can also easily lose a variable number of $(n-1)d$ electrons to form higher oxidation states. [1 Mark]
Q5. Lanthanoids vs Actinoids:
| Lanthanoids | Actinoids |
|---|---|
| Involve the progressive filling of 4f orbitals. | Involve the progressive filling of 5f orbitals. |
| Most are non-radioactive (except Promethium). | All elements are radioactive. |
| Show limited oxidation states (mostly +3). | Show higher variable oxidation states (+3 to +7). |
[1 Mark for each point of distinction. Total 2 Marks]
Q6. Interstitial Compounds:
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. [1 Mark]
Example: Steel, Cast Iron, Titanium carbide ($TiC$). [1 Mark]
Q7. Color of Transition Metal Ions:
In aqueous solutions or complexes, the degenerate d-orbitals of transition metal ions split into two sets of different energies ($t_{2g}$ and $e_g$) due to the approach of ligands (water/anions). [1 Mark]
If the ion contains unpaired electrons in the d-subshell, an electron absorbs a specific wavelength of visible light and undergoes a d-d transition to a higher energy d-orbital. The transmitted complementary light imparts color to the solution. [1 Mark]
SECTION C [6 Marks]
Q8. Preparation of Potassium Permanganate ($KMnO_4$):
Step 1: Conversion of Pyrolusite to Potassium Manganate.
Pyrolusite ore ($MnO_2$) is fused with an alkali ($KOH$) in the presence of an oxidizing agent ($O_2$ or $KNO_3$) to form a green mass of Potassium Manganate ($K_2MnO_4$). [1 Mark]
$2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2O$
Step 2: Oxidation of Potassium Manganate to Potassium Permanganate.
The green $K_2MnO_4$ solution is oxidized by passing Chlorine gas ($Cl_2$) or Ozone ($O_3$) through it, forming the dark purple Potassium Permanganate ($KMnO_4$). [2 Marks for explanation and equation]
$2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl$
Q9. Catalytic Properties:
Transition metals act as excellent catalysts because: (1) They have variable oxidation states, allowing them to form unstable intermediate compounds with lower activation energy. (2) They provide a large surface area with free valencies for the adsorption of reactant molecules. [1 Mark]
Examples: [1 Mark each]
- Finely divided Iron (Fe): Used as a catalyst in Haber's process for the synthesis of Ammonia.
- Vanadium Pentoxide ($V_2O_5$): Used as a catalyst in the Contact process for the manufacture of Sulfuric acid.
Q10. Lanthanoid Contraction and Consequences:
Definition: The steady and continuous decrease in atomic and ionic radii of lanthanoids with the increase in atomic number (from Ce to Lu) is called lanthanoid contraction. [1 Mark]
Consequences: [1 Mark for each point = 2 Marks]
- Similar Chemical Properties: Due to almost identical ionic radii, the chemical properties of lanthanoids are extremely similar, making their separation from mixtures very difficult.
- Similarity of 2nd and 3rd transition series: The atomic radii of elements of the 3rd transition series (e.g., Hf, Ta, W) are almost identical to those of the corresponding elements of the 2nd transition series (e.g., Zr, Nb, Mo).
SECTION D [4 Marks]
Q11. (a) Color of $Sc^{3+}$ vs $Ti^{3+}$ [2 Marks] (b) Actinoid Configuration [2 Marks]
(a) Color Reasoning:
Sc ($Z=21$) is $[Ar] 3d^1 4s^2$. The $Sc^{3+}$ ion loses 3 electrons, making its configuration $[Ar] 3d^0$. Since there are no unpaired electrons in the d-orbitals, d-d transitions cannot occur. Hence, it is colorless. [1 Mark]
Ti ($Z=22$) is $[Ar] 3d^2 4s^2$. The $Ti^{3+}$ ion has the configuration $[Ar] 3d^1$. It has one unpaired electron, which undergoes d-d transitions by absorbing light in the visible region. Hence, it is colored. [1 Mark]
(b) General Electronic Configuration of Actinoids:
$[Rn] 5f^{0-14} 6d^{0-2} 7s^2$ [2 Marks]
Q12. (a) Preparation of $K_2Cr_2O_7$ [3 Marks] (b) 3d Series Configuration [1 Mark]
(a) Preparation of Potassium Dichromate:
1. Roasting of Chromite ore: Chromite ore ($FeCr_2O_4$) is roasted with sodium carbonate ($Na_2CO_3$) in the presence of air to form yellow Sodium Chromate ($Na_2CrO_4$). [1 Mark]
$4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$
2. Conversion to Sodium Dichromate: The yellow sodium chromate solution is filtered and acidified with dilute sulfuric acid ($H_2SO_4$) to form orange Sodium Dichromate ($Na_2Cr_2O_7$). [1 Mark]
$2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$
3. Conversion to Potassium Dichromate: $Na_2Cr_2O_7$ is treated with Potassium Chloride ($KCl$). Since $K_2Cr_2O_7$ is less soluble, it crystallizes out as orange crystals. [1 Mark]
$Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 \downarrow + 2NaCl$
(b) General Electronic Configuration of 3d Series:
$[Ar] 3d^{1-10} 4s^{1-2}$ [1 Mark]
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