The D and F Block Elements — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "The D and F Block Elements" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "The D and F Block Elements" — 6 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 12 Chemistry — The D and F Block Elements. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
The D and F Block Elements — All 6 Questions Solved
Q1. Define transition elements. What are the characteristic properties of transition metals?
Transition Elements:
Elements with incompletely filled d-orbitals in their atoms or cations, or those with tendency to form ions by loss of electrons from d-orbitals.
They occupy the d-block of periodic table (Groups 3-12, Periods 4-6).
Characteristic Properties:
1. Variable Oxidation States:
Multiple oxidation states due to similar energies of (n-1)d and ns electrons.
Example: Manganese shows oxidation states from +2 to +7
Cr: +2, +3, +6; Mn: +2, +3, +7; Fe: +2, +3; Cu: +1, +2
2. Formation of Colored Compounds:
Colors due to d-d electronic transitions (jumping between d-orbitals).
Examples: CuSO₄ (blue), KMnO₄ (purple), K₂Cr₂O₇ (orange)
Non-transition metals form colorless compounds.
3. Catalytic Properties:
Ability to catalyze reactions due to variable oxidation states and ability to form complexes.
Examples: Fe in Haber process, Ni in hydrogenation, MnO₂ in H₂O₂ decomposition
4. Complex Formation:
Site-preference for coordination with ligands forming coordination complexes.
Due to small size and high nuclear charge.
Examples: [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺, [Cr(H₂O)₆]³⁺
5. High Melting and Boiling Points:
Strong metallic bonding due to numerous unpaired d-electrons available for bonding.
Example: Tungsten has highest melting point among metals (3695 K)
6. High Density:
Relatively high density for their atomic numbers due to d-electrons occupying same shell.
Example: Iron (7.87 g/cm³), Cobalt (8.90 g/cm³)
7. Paramagnetic Nature:
Until d-orbitals are completely filled, unpaired electrons make them paramagnetic.
Example: Iron, Cobalt, Nickel are ferromagnetic
8. Alloy Formation:
Readily form alloys with other metals improving properties.
Example: Steel (Fe + C), Bronze (Cu + Sn)
9. Electrode Potential Behavior:
Some are more reactive (Mn, Zn) while others are less reactive (Cu, Au)
Depends on ease of losing electrons
10. Hydrolysis of Cations:
Small, highly charged metal cations hydrolyze more readily.
Example: Fe³⁺ + H₂O ⇌ FeOH²⁺ + H⁺ (acidic solution)
Q2. Compare the properties of Cr and Mn. Why do they differ?
Chromium (Cr) vs Manganese (Mn):
Both are in Period 4 of d-block. Electron configurations:
Cr: [Ar] 3d⁵ 4s¹ (half-filled d-orbital)
Mn: [Ar] 3d⁵ 4s² (half-filled d-orbital + filled s-orbital)
Comparison:
Property | Chromium | Manganese
---------|----------|----------
Atomic number | 24 | 25
Electron config | 3d⁵4s¹ | 3d⁵4s²
Half-filled | Half-filled 3d | Half-filled 3d
Common oxidation states | +2, +3, +6 | +2, +3, +7
Most stable oxidation state | +3 | +2
Meeting point (K) | 2180 | 1519
Density (g/cm³) | 7.19 | 7.43
Ionization energy | Higher | Lower
Key Differences:
1. Oxidation States:
Cr: Highest is +6 (CrO₃, K₂Cr₂O₇ - orange)
Mn: Highest is +7 (KMnO₄ - purple, most powerful oxidizing agent)
Mn shows more variable oxidation states due to extra electron in 4s orbital
2. Reactivity:
Cr: Less reactive, more stable as +3 state
Mn: More reactive, readily oxidized to higher states
3. Color of Compounds:
Cr: +2 (blue), +3 (green), +6 (orange/yellow)
Mn: +2 (colorless/pale pink), +3 (brown), +7 (purple)
4. Magnetic Properties:
Cr: Antiferromagnetic at low temperature
Mn: Strongly paramagnetic, ferromagnetic below Néel temperature
5. Corrosion Resistance:
Cr: Forms passive oxide layer (Cr₂O₃), corrosion resistant
Mn: Less corrosion resistant; requires stainless steel composition
Why differences?
1. Extra electron in 4s orbital of Mn makes it more reactive
2. Cr's 3d⁵ 4s¹ configuration provides stability at +3 oxidation state
3. Mn's 3d⁵ 4s² allows easier loss of 4s electrons, facilitating higher oxidation states
4. Nuclear effective charge differs slightly (Cr has 24 protons vs Mn's 25)
5. d-orbital energy levels differ, affecting stability of various oxidation states
Q3. Discuss the chemistry of potassium permanganate (KMnO₄). What is its structure and oxidizing properties?
Potassium Permanganate (KMnO₄):
Structure:
Molecular formula: KMnO₄
Ionic compound containing K⁺ and MnO₄⁻ ions
MnO₄⁻ ion structure (tetrahedral):
1. Mn is in +7 oxidation state
2. Four oxygen atoms surround Mn (tetrahedral arrangement)
3. Oxygen atoms form covalent bonds with Mn
4. Total charge: -1 (MnO₄⁻ is tetravalent anion)
5. Mn: [Ar] 3d⁰ (all d-electrons transferred to oxygen ligands)
Physical Properties:
1. Appearance: Dark purple/violet crystals
2. Soluble in water: Yes, readily soluble
3. Melting point: 513-514 K
4. Density: 2.7 g/cm³
Oxidizing Properties:
KMnO₄ is one of strongest oxidizing agents in acidic and alkaline media.
In Acidic Medium (H₂SO₄):
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
E° = +1.51 V (strong oxidant)
Color: Purple → Colorless
Mn⁷⁺ → Mn²⁺ (7 electrons gained)
In Alkaline Medium (KOH):
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
E° = +0.59 V
Color: Purple → Brown (MnO₂ precipitate)
Mn⁷⁺ → Mn⁴⁺ (3 electrons gained)
In Neutral Medium:
MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O
Mn⁷⁺ → Mn⁴⁺
Reactions:
1. Oxidizes Fe²⁺ to Fe³⁺
2Mn⁷⁺ + 10Fe²⁺ + 16H⁺ → 2Mn²⁺ + 10Fe³⁺ + 8H₂O
2. Oxidizes oxalic acid (C₂O₄²⁻) to CO₂
5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O
3. Oxidizes alcohols to aldehydes/ketones
4. Oxidizes alkenes
Applications:
1. Analytical chemistry: Titrant in redox titration
2. Disinfection: Water purification
3. Bleaching agent
4. Oxidizing agent in organic synthesis
5. Medical: Antiseptic for wounds
Auto-oxidation:
In acidic solution, MnO₄⁻ can self-decompose to MnO₂ over time.
Must be stored in amber bottles away from light and impurities
Q4. Write the electronic configuration of Zn²⁺ and explain why zinc is not considered a typical transition metal despite being in d-block.
Zinc (Zn) Electronic Configuration:
Zn (neutral): [Ar] 3d¹⁰ 4s²
Atomic number: 30
Zn²⁺ (ion): [Ar] 3d¹⁰
Zn²⁺ loses both 4s electrons: 3d¹⁰ 4s² → 3d¹⁰ + 2e⁻
Why Zn is NOT a typical transition element:
1. Completely Filled d-orbitals:
Zn²⁺ has completely filled d-orbitals (d¹⁰), no unpaired electrons
Typical transition metals have incompletely filled d-orbitals with unpaired electrons
2. Lacks Variable Oxidation States:
Zn shows only +2 oxidation state (loses 4s² electrons)
No other stable oxidation states (unlike Fe: +2, +3 or Cr: +2, +3, +6)
3. Forms Colorless Compounds:
ZnSO₄, ZnCl₂, [Zn(H₂O)₄]²⁺ are all colorless
No d-d electronic transitions possible with filled d-orbitals
Typical transition metals form colored compounds
4. Not Paramagnetic:
Zn²⁺ has all electrons paired (d¹⁰)
Diamagenic substance
Typical transition metals are paramagnetic (unpaired d-electrons)
5. Limited Catalytic Properties:
ZnO is not a strong catalyst compared to TiO₂, Fe₂O₃, etc.
Lacking variable oxidation states reduces catalytic ability
6. Complex Formation:
Forms fewer complex ions than typical transition metals
Complexes are relatively simple and colorless
[Zn(NH₃)₄]²⁺ is colorless (no d-d transitions)
7. Electrode Potential:
Zn: E° = -0.76 V (more reactive, readily forms Zn²⁺)
Less typical transition metal behavior
Conclusion:
Zn occupies Group 12 of d-block but exhibits properties of transition metals only partially.
Often classified separately as post-transition metal or transition metal anomaly.
Similarly: Cd and Hg (in same group) also lack typical transition metal properties.
Criteria for true transition metal:
- Incompletely filled d-orbitals
- Multiple oxidation states
- Colored compounds
- Paramagnetic
Zn fails all these criteria, hence NOT a typical transition element.
Q5. Explain the lanthanide contraction and its consequences.
Lanthanide Contraction:
The phenomenon of gradual decrease in atomic and ionic radii of lanthanide elements from La to Lu, despite increase in atomic number and addition of electrons to same 4f subshell.
Cause:
As atomic number increases from La (Z=57) to Lu (Z=71):
1. Each additional electron enters the 4f orbital (inner orbital, not valence)
2. Electrons in diffuse f-orbitals provide poor shielding effect
3. Nuclear charge increases by 1 for each new electron
4. Net effect: Strong increase in effective nuclear charge (Z_eff)
5. Electrons are pulled closer to nucleus
6. Radius decreases despite adding more electrons
Ionic Radius Trend:
La³⁺ > Ce³⁺ > Pr³⁺ > Nd³⁺ > ... > Yb³⁺ > Lu³⁺
Reduction approximately 0.1-0.2 Å per element
Total contraction from La to Lu ≈ 1.5-2 Å
Comparison with transition metal contraction:
- Transition metal series: Larger radius decrease (much more gradual)
- Lanthanide series: Much sharper radius decrease (inner shell filling)
Consequences of Lanthanide Contraction:
1. Similarity of elements 65Tb to 68Er with elements 39Y and 40Zr:
Yttrium (Y) properties similar to heavier lanthanides
Allows Y to be grouped with lanthanides
Y found in lanthanide mineral deposits
2. Difficulty in separation:
Lanthanides with similar ionic radii have very similar chemical properties
Makes separation and purification extremely difficult
Requires complex ion-exchange chromatography or solvent extraction
3. Constant ionic radius of 4d transition metals:
Radii of 4d metals approximately constant
Example: Zr⁴⁺ ≈ Ti⁴⁺ (despite different atomic numbers)
La³⁺ radius ≈ Y³⁺ + 2 positions ≈ Sc³⁺ + 1 position
4. Industrial significance:
Y extracted from lanthanide ores despite being Group 3 element
Makes Y economically important with lanthanides
5. Coordination chemistry:
All lanthanides have similar ionic radii
Form similar coordination complexes
Difficult to differentiate based on size alone
6. Density trend:
Density slightly increases from La to Lu despite radius decrease
Due to increasing atomic number and mass
7. Solubility of compounds:
Compounds show parallel trends
Lanthanide hydroxides: solubility decreases from La(OH)₃ to Lu(OH)₃
Due to decreasing ionic radius (higher charge density)
Quantitative effect:
Atomic radius of La: 187 pm
Atomic radius of Lu: 175 pm
Total contraction: 12 pm
Per element: ~0.67 pm average
Comparison:
- d-block contraction (Sc to Zn): ~30 pm
- f-block contraction (La to Lu): ~12 pm
But f-block is more regular and consistent
Q6. Write the preparation and properties of potassium dichromate (K₂Cr₂O₇).
Potassium Dichromate (K₂Cr₂O₇):
Preparation:
Laboratory Method:
1. Chromite ore (FeCr₂O₄) is roasted with Na₂CO₃ at high temperature:
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2. Sodium chromate (Na₂CrO₄) solution is obtained and acidified with H₂SO₄:
2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O
(Yellow) → (Orange)
3. Cool the solution and add KCl to precipitate K₂Cr₂O₇:
Na₂CrO₇²⁻ + 2K⁺ + Cl⁻ → K₂Cr₂O₇↓ + 2NaCl
4. Filter, wash with cold water, and dry.
Alternative Method:
From potassium chromate solution by acidification:
2K₂CrO₄ + 2H⁺ + 2Cl⁻ → K₂Cr₂O₇ + 2KCl + H₂O
Structure of Cr₂O₇²⁻:
Two CrO₄ tetrahedra sharing one oxygen atom (corner-shared)
Cr atoms in +6 oxidation state
Orange-red colored ion
Physical Properties:
1. Appearance: Orange-red crystals
2. Solubility: Soluble in water, HCl, slightly soluble in ethanol
3. Melting point: 671 K
4. Density: 2.676 g/cm³
5. Molar mass: 294 g/mol
6. Hygroscopic: Readily absorbs moisture
Chemical Properties:
1. Oxidizing Power (Strong oxidizing agent):
In acidic medium:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
E° = +1.33 V
Color: Orange → Green (Cr³⁺)
2. Reaction with reducing agents:
With Fe²⁺:
6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
With C₂H₅OH:
C₂H₅OH + Cr₂O₇²⁻ + 8H⁺ → CH₃CHO + 2Cr³⁺ + 7H₂O
With I⁻:
6I⁻ + Cr₂O₇²⁻ + 14H⁺ → 3I₂ + 2Cr³⁺ + 7H₂O
3. Acid-base character:
CrO₄²⁻ ⇌ Cr₂O₇²⁻ + H₂O (equilibrium depends on pH)
In alkaline: Exists as CrO₄²⁻ (yellow)
In acidic: Exists as Cr₂O₇²⁻ (orange)
4. Thermal stability:
Stable at room temperature
Decomposes at high temperature
Applications:
1. Analytical chemistry: Titrant for redox titrations (standard solution)
2. Oxidizing agent in organic synthesis
3. Metal plating and etching
4. Photography: Bleaching agent
5. Glass and paint manufacture
6. Leather tanning
7. Indicator in redox titrations (iodometric analysis)
Precautions:
1. Toxic: Cr(VI) is carcinogenic, handle with care
2. Corrosive to skin and eyes
3. Keep away from organic materials (fire hazard)
4. Wear gloves and safety glasses when handling
The D and F Block Elements — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following elements has the highest number of unpaired electrons in its ground state atomic configuration?
- A. Fe (Z=26)
- B. Cr (Z=24)
- C. Mn (Z=25)
- D. Co (Z=27)
Answer: B. Cr (Z=24) — Chromium (Cr, Z=24) has an electronic configuration of [Ar] 3d⁵ 4s¹, resulting in 6 unpaired electrons for enhanced stability. Other options have fewer unpaired electrons: Fe (4), Mn (5), Co (3).
MCQ 2. Why do transition metals show variable oxidation states?
- A. Due to their large atomic size.
- B. Due to the involvement of (n-1)d and ns electrons in bonding.
- C. Due to their high electronegativity.
- D. Due to their low ionization enthalpies.
Answer: B. Due to the involvement of (n-1)d and ns electrons in bonding. — The energy difference between the (n-1)d and ns orbitals in transition metals is very small, allowing both sets of electrons to participate in bond formation, leading to variable oxidation states.
MCQ 3. Which of the following ions is diamagnetic?
- A. Sc³⁺ (Z=21)
- B. Mn²⁺ (Z=25)
- C. Cr³⁺ (Z=24)
- D. Ti³⁺ (Z=22)
Answer: A. Sc³⁺ (Z=21) — Sc³⁺ has an electronic configuration of [Ar] 3d⁰, meaning it has no unpaired electrons and is therefore diamagnetic. Mn²⁺ (3d⁵), Cr³⁺ (3d³), and Ti³⁺ (3d¹) all have unpaired electrons and are paramagnetic.
MCQ 4. In the preparation of potassium dichromate from chromite ore (FeCr₂O₄), what is the role of adding Na₂CO₃ and heating in air?
- A. It reduces FeCr₂O₄ to pure chromium.
- B. It oxidizes chromium to chromate and converts iron to iron(III) oxide.
- C. It extracts chromium as a soluble complex.
- D. It converts the ore into a soluble iron salt.
Answer: B. It oxidizes chromium to chromate and converts iron to iron(III) oxide. — Chromite ore (FeCr₂O₄) reacts with sodium carbonate in the presence of air (O₂) to oxidize Cr from +3 to +6 (forming soluble sodium chromate) and convert iron to insoluble Fe₂O₃.
MCQ 5. The E° value for Mn³⁺/Mn²⁺ is highly positive (+1.57 V) as compared to Cr³⁺/Cr²⁺ (-0.41 V) or Fe³⁺/Fe²⁺ (+0.77 V). What does this indicate?
- A. Mn³⁺ is more stable than Mn²⁺.
- B. Mn²⁺ is more easily oxidized than Cr²⁺ or Fe²⁺.
- C. Mn³⁺ is a stronger oxidizing agent than Cr³⁺ or Fe³⁺.
- D. Mn³⁺ is a weaker oxidizing agent than Cr³⁺ or Fe³⁺.
Answer: C. Mn³⁺ is a stronger oxidizing agent than Cr³⁺ or Fe³⁺. — A highly positive E° value indicates a strong tendency for the reduction of Mn³⁺ to Mn²⁺. This means Mn³⁺ readily accepts electrons, making it a powerful oxidizing agent compared to Cr³⁺ or Fe³⁺.
MCQ 6. Why are most transition metal compounds colored?
- A. Due to the presence of empty d-orbitals.
- B. Due to d-d transitions and charge transfer phenomena.
- C. Due to their small size.
- D. Due to their high ionization energies.
Answer: B. Due to d-d transitions and charge transfer phenomena. — Transition metal ions often have partially filled d-orbitals, allowing for d-d electronic transitions upon absorption of visible light, leading to complementary colors. Charge transfer transitions can also contribute to color.
MCQ 7. What is the main consequence of lanthanoid contraction?
- A. Increase in atomic radii across the period.
- B. Similar atomic radii of 4d and 5d transition elements.
- C. Decrease in ionization enthalpy of 5d elements.
- D. Enhanced basicity of hydroxides from La to Lu.
Answer: B. Similar atomic radii of 4d and 5d transition elements. — Lanthanoid contraction causes a steady decrease in atomic radii across the lanthanoids, which in turn leads to the 4d and 5d elements of the same group having very similar atomic radii and hence similar properties.
MCQ 8. What is the equivalent weight of KMnO₄ when it acts as an oxidizing agent in an acidic medium? (Given Molar Mass of KMnO₄ = M)
- A. M/1
- B. M/3
- C. M/5
- D. M/7
Answer: C. M/5 — In an acidic medium, KMnO₄ (MnO₄⁻) is reduced to Mn²⁺. The oxidation state of manganese changes from +7 to +2, a change of 5 electrons. Therefore, the equivalent weight is Molar Mass / 5, or M/5.
MCQ 9. Which pair of elements has anomalous electronic configurations in the d-block?
- A. V (Z=23), Ti (Z=22)
- B. Cr (Z=24), Cu (Z=29)
- C. Mn (Z=25), Fe (Z=26)
- D. Ni (Z=28), Co (Z=27)
Answer: B. Cr (Z=24), Cu (Z=29) — Chromium ([Ar] 3d⁵ 4s¹) and Copper ([Ar] 3d¹⁰ 4s¹) exhibit anomalous electronic configurations. This deviation from expected configurations occurs to achieve greater stability associated with half-filled or completely filled d-orbitals.
MCQ 10. Which statement is true for actinoids?
- A. They are all non-radioactive elements.
- B. They are generally more reactive than lanthanoids.
- C. They predominantly show a +2 oxidation state.
- D. They exhibit strong lanthanoid contraction.
Answer: B. They are generally more reactive than lanthanoids. — Actinoids are generally more reactive than lanthanoids due to their lower ionization energies and more accessible 5f electrons. All actinoids are radioactive.
MCQ 11. Which of the following statements about interstitial compounds is incorrect?
- A. They are chemically inert.
- B. They have lower melting points than the pure metals.
- C. They are very hard.
- D. They retain metallic conductivity.
Answer: B. They have lower melting points than the pure metals. — Interstitial compounds generally have higher melting points than the pure parent metals because the small non-metal atoms occupy the interstitial sites, leading to strong covalent bonding with the metal atoms.
MCQ 12. Why is Fe²⁺ readily oxidized to Fe³⁺, but Mn²⁺ is not readily oxidized to Mn³⁺?
- A. Fe²⁺ (3d⁶) loses an electron to form stable Fe³⁺ (3d⁵, half-filled), while Mn²⁺ (3d⁵, half-filled) needs to lose an electron to form less stable Mn³⁺ (3d⁴).
- B. Fe²⁺ has a higher ionization energy than Mn²⁺.
- C. Mn²⁺ forms stronger complexes than Fe²⁺.
- D. Fe³⁺ has a larger size than Fe²⁺.
Answer: A. Fe²⁺ (3d⁶) loses an electron to form stable Fe³⁺ (3d⁵, half-filled), while Mn²⁺ (3d⁵, half-filled) needs to lose an electron to form less stable Mn³⁺ (3d⁴). — Fe²⁺ (3d⁶) can achieve a stable half-filled 3d⁵ configuration by losing one electron to form Fe³⁺. In contrast, Mn²⁺ is already 3d⁵ (half-filled and stable), so losing an electron to form Mn³⁺ (3d⁴) is energetically unfavorable, making its oxidation difficult.
MCQ 13. What is the most common oxidation state for lanthanoids?
- A. +2
- B. +3
- C. +4
- D. +5
Answer: B. +3 — The most common and stable oxidation state exhibited by lanthanoids is +3, arising from the loss of two 6s electrons and one 4f electron.
MCQ 14. In the reaction: 2KMnO₄ + 3H₂SO₄ + 5(COOH)₂ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂, what is the change in the oxidation state of carbon?
- A. +2 to +3
- B. +3 to +4
- C. +2 to +4
- D. +1 to +2
Answer: B. +3 to +4 — In oxalic acid, (COOH)₂, each carbon atom has an average oxidation state of +3. In carbon dioxide, CO₂, the oxidation state of carbon is +4. Thus, carbon is oxidized from +3 to +4.
MCQ 15. Which of the following statements is a direct consequence of lanthanoid contraction?
- A. Zr and Hf have significantly different atomic radii.
- B. Nb and Ta have similar chemical properties.
- C. Mo and W have identical melting points.
- D. The stability of +2 oxidation state increases across the lanthanoids.
Answer: B. Nb and Ta have similar chemical properties. — Lanthanoid contraction leads to very similar atomic radii for elements in the same group of the 4d and 5d transition series (e.g., Nb and Ta), resulting in their similar chemical properties.
MCQ 16. Which of the following elements is generally NOT considered a transition element?
- A. Iron (Fe)
- B. Copper (Cu)
- C. Zinc (Zn)
- D. Chromium (Cr)
Answer: C. Zinc (Zn) — Zinc (Zn, Z=30) is not considered a transition element because its d-orbitals are completely filled (3d¹⁰) in its ground state and its common oxidation state (Zn²⁺ is 3d¹⁰), lacking the characteristic properties of transition metals.
MCQ 17. Calculate the spin-only magnetic moment for a Cr²⁺ ion. (Atomic number of Cr = 24)
- A. 1.73 BM
- B. 2.83 BM
- C. 3.87 BM
- D. 4.90 BM
Answer: D. 4.90 BM — The electronic configuration of Cr is [Ar] 3d⁵ 4s¹. For Cr²⁺, it is [Ar] 3d⁴, meaning there are 4 unpaired electrons. The spin-only magnetic moment μ = √n(n+2) = √4(4+2) = √24 ≈ 4.90 BM.
MCQ 18. Why do actinoids show a greater range of oxidation states compared to lanthanoids?
- A. The 5f orbitals are more diffused than 4f orbitals and participate more readily in bonding.
- B. The 5f orbitals are better shielded from nuclear charge than 4f orbitals.
- C. Actinoids have smaller atomic sizes than lanthanoids.
- D. The ionization enthalpies of actinoids are higher than lanthanoids.
Answer: A. The 5f orbitals are more diffused than 4f orbitals and participate more readily in bonding. — The 5f orbitals in actinoids are less deeply buried and more diffused compared to 4f orbitals, making their energies closer to the 6d and 7s orbitals. This allows them to participate more readily in bonding, leading to a wider range of oxidation states.
MCQ 19. What is the geometry around each chromium atom in the dichromate ion (Cr₂O₇²⁻)?
- A. Tetrahedral
- B. Octahedral
- C. Square planar
- D. Trigonal planar
Answer: A. Tetrahedral — In the dichromate ion (Cr₂O₇²⁻), each chromium atom is tetrahedrally surrounded by four oxygen atoms, with one oxygen atom serving as a bridge between the two chromium centers.
MCQ 20. Which of the following represents the correct order of atomic radii?
- A. Ti < V < Cr < Mn
- B. Fe > Co > Ni > Cu
- C. Sc > Ti > V > Cr
- D. Zn < Cu < Ni < Co
Answer: C. Sc > Ti > V > Cr — Across the 3d series, atomic radii generally decrease initially due to increasing nuclear charge, then become relatively constant, and finally increase slightly towards the end. Therefore, Sc > Ti > V > Cr shows the initial decrease.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.