Board Previous Year Questions

How to Score Full Marks in the Board Exam

A complete bank of board-style questions with model answers for The d- and f-Block Elements. For "account for" questions, give the NCERT-canonical reasons: half/fully-filled stability, small electronegative ligands stabilising high oxidation states, and poor 4f shielding for the lanthanoid contraction. For magnetic-moment numericals, write $\mu = \sqrt{n(n+2)}$ BM, find the ion's d-configuration, count $n$, then substitute.

1-Mark Questions (Definitions & Direct)

Q1. Write the general electronic configuration of d-block elements. Answer: $(n-1)d^{1-10}\,ns^{1-2}$.

Q2. Write the electronic configuration of chromium (Z = 24). Answer: $[\text{Ar}]\,3d^5 4s^1$ — a half-filled 3d subshell is more stable.

Q3. What is the highest oxidation state shown in the 3d series and by which element? Answer: +7, shown by manganese (in MnO$_4^-$).

Q4. Write the spin-only magnetic moment of an ion with 5 unpaired electrons. Answer: $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$ BM.

1-Mark Questions (continued)

Q5. What is the lanthanoid contraction? Answer: The steady decrease in atomic and ionic radii of the lanthanoids from La to Lu, caused by the poor shielding of the nucleus by the 4f electrons.

Q6. Name the oxoanion in which manganese shows the +7 state. Answer: The permanganate ion, MnO$_4^-$.

Q7. Why is Zn$^{2+}$ colourless? Answer: Zn$^{2+}$ has a completely filled $3d^{10}$ configuration, so no d-d transition is possible.

Q8. Write the configuration of copper (Z = 29). Answer: $[\text{Ar}]\,3d^{10} 4s^1$ — a fully-filled 3d subshell is more stable.

2-Mark Reasoning Questions

Q9. Why is Cr$^{2+}$ a stronger reducing agent than Fe$^{2+}$? Answer: Cr$^{2+}$ ($d^4$) is readily oxidised to the very stable Cr$^{3+}$ ($d^3$, which is stable in an octahedral field), so Cr$^{2+}$ acts as a reducing agent. Fe$^{2+}$ ($d^6$) is oxidised to Fe$^{3+}$ ($d^5$); although $d^5$ is stable, the change for Cr is more favourable, making Cr$^{2+}$ the stronger reducing agent.

Q10. Why do transition metals show variable oxidation states? Answer: Because the $ns$ and $(n-1)d$ electrons have similar energies, both sets can take part in bonding. An element may lose only its $ns$ electrons (lower states) or additionally some $d$ electrons (higher states), giving a range of oxidation states differing by 1.

2-Mark Reasoning Questions (continued)

Q11. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride? Answer: Oxygen and fluorine are small and highly electronegative, so they form strong bonds that can stabilise the metal in its highest oxidation state (e.g. Mn in MnO$_4^-$ and Mn$_2$O$_7$, Cr in CrO$_4^{2-}$).

Q12. Why are Mn$^{2+}$ compounds more stable to oxidation than Fe$^{2+}$ compounds? Answer: Mn$^{2+}$ has a $3d^5$ (stable half-filled) configuration and resists oxidation to Mn$^{3+}$ ($d^4$). Fe$^{2+}$ is $3d^6$ and is readily oxidised to the stable Fe$^{3+}$ ($d^5$). Hence Mn$^{2+}$ is more stable toward oxidation.

Q13. Why do transition metals form coloured compounds? Answer: Their ions have partially filled d subshells. In a ligand field the d orbitals split, and a d electron can absorb visible light to jump from the lower to the higher set (a d-d transition). The colour observed is complementary to the light absorbed.

2-3 Mark Reasoning Questions

Q14. Why do transition metals and their compounds act as good catalysts? Answer: (i) They show variable oxidation states, so they can readily accept and donate electrons, forming intermediates and providing an alternative low-energy pathway. (ii) Finely divided metals provide a large surface for adsorption of reactants, weakening their bonds. Examples: Fe in the Haber process, V$_2$O$_5$ in the Contact process.

Q15. What is the lanthanoid contraction and what are its two consequences? Answer: It is the steady decrease in size across the lanthanoid series due to poor 4f shielding. Consequences: (i) the second (4d) and third (5d) transition series have nearly the same atomic radii (e.g. Zr and Hf), making them hard to separate; (ii) the basicity of the lanthanoid hydroxides decreases from La(OH)$_3$ to Lu(OH)$_3$.

Q16. Why are the enthalpies of atomisation of transition metals high? Answer: Transition metals have many unpaired d electrons that participate in strong metallic (and some covalent) bonding between atoms. Breaking these strong bonds to form gaseous atoms requires a large amount of energy, so the enthalpies of atomisation (and melting points) are high.

3-Mark Numericals & Reactions

Q17. Calculate the spin-only magnetic moment of Fe$^{2+}$ (Z = 26). Answer: Fe$^{2+}$ is $3d^6$ with 4 unpaired electrons. $\mu = \sqrt{4(4+2)} = \sqrt{24} = \mathbf{4.90\ BM}$.

Q18. Calculate the spin-only magnetic moment of the Mn$^{2+}$ ion. Answer: Mn$^{2+}$ is $3d^5$ with 5 unpaired electrons. $\mu = \sqrt{5(5+2)} = \sqrt{35} = \mathbf{5.92\ BM}$.

Q19. Write the ionic equation for the oxidation of Fe$^{2+}$ by acidified KMnO$_4$. Answer: $5\text{Fe}^{2+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$.

3-Mark Numericals & Reactions (continued)

Q20. Write the ionic equation for the oxidation of iodide ions by acidified potassium dichromate. Answer: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}$.

Q21. The molar mass of K$_2$Cr$_2$O$_7$ is 294 g mol$^{-1}$. Calculate its equivalent mass as an oxidising agent in acidic medium. Answer: Acidified dichromate gains 6 electrons per formula unit, so equivalent mass $= \dfrac{294}{6} = \mathbf{49\ g\ equiv^{-1}}$.

Q22. The molar mass of KMnO$_4$ is 158 g mol$^{-1}$. Calculate its equivalent mass in acidic medium. Answer: Acidified permanganate gains 5 electrons, so equivalent mass $= \dfrac{158}{5} = \mathbf{31.6\ g\ equiv^{-1}}$.

Q23. Write the configuration of Fe$^{3+}$ and state its number of unpaired electrons. Answer: Fe is $[\text{Ar}]\,3d^6 4s^2$; Fe$^{3+}$ = $[\text{Ar}]\,3d^5$ with 5 unpaired electrons (a stable half-filled subshell).

3-Mark Reactions & Configurations

Q24. What happens when an acidified solution of potassium dichromate is treated with an alkali? Write the equation. Answer: The orange dichromate is converted to yellow chromate: $\text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O}$. (Adding acid reverses it to orange dichromate.)

Q25. Write the half-reaction for the oxidising action of acidified permanganate, stating how many electrons are gained. Answer: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ — 5 electrons are gained (Mn goes from +7 to +2).

3-Mark Reasoning

Q26. Why is the second ionisation enthalpy of copper unusually high? Answer: Cu$^+$ has a $[\text{Ar}]\,3d^{10}$ configuration — a stable fully-filled subshell. Removing a second electron disrupts this stable $d^{10}$, requiring a large amount of energy, so IE$_2$(Cu) is high.

Q27. Why are the radii of the second and third transition series almost the same? Answer: Because of the lanthanoid contraction. The 14 lanthanoids between the 4d and 5d series cause a size reduction (poor 4f shielding) that offsets the expected increase from 4d to 5d, so the congeners (e.g. Zr/Hf, Nb/Ta) are nearly the same size.

Q28. What is the action of acidified KMnO$_4$ on oxalate ions? Write the equation. Answer: Oxalate is oxidised to CO$_2$: $5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$.

3-Mark Reasoning (continued)

Q29. Why is the E0(M2+/M) value for copper positive, unlike the other 3d metals? Answer: Copper's high enthalpy of atomisation and high ionisation enthalpies are not compensated by its hydration enthalpy. As a result $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V (positive), which means copper is not oxidised by H$^+$ and does not liberate hydrogen from dilute acids.

Q30. Why is Cu$^+$ ion unstable in aqueous solution? Answer: Cu$^+$ disproportionates: $2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}$. The high hydration enthalpy of Cu$^{2+}$ more than compensates for its higher second ionisation enthalpy, making Cu$^{2+}$ + Cu more stable than 2 Cu$^+$ in water.

Q31. Both Cr$^{2+}$ and Mn$^{3+}$ are $d^4$. Why is Cr$^{2+}$ reducing while Mn$^{3+}$ is oxidising? Answer: Cr$^{2+}$ ($d^4$) is oxidised to the stable Cr$^{3+}$ ($d^3$), so it is a reducing agent. Mn$^{3+}$ ($d^4$) is reduced to the stable half-filled Mn$^{2+}$ ($d^5$), so it is an oxidising agent. Each moves toward the more stable configuration.

5-Mark / Long-Answer Questions

Q32. Describe the preparation of potassium dichromate from chromite ore. Answer: (i) Chromite ore (FeCr$_2$O$_4$) is fused with sodium carbonate in the presence of air to give sodium chromate: $4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$. (ii) The yellow sodium chromate solution is acidified to give sodium dichromate: $2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O}$. (iii) Treatment with KCl gives orange crystals of potassium dichromate (less soluble): $\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}$.

Q33. Explain the trends in the chemistry of the lanthanoids: electronic configuration, oxidation states and the lanthanoid contraction. Answer: Lanthanoids have the general configuration $[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^2$. Their characteristic and most stable oxidation state is +3, though a few show +2 (e.g. Eu$^{2+}$) or +4 (e.g. Ce$^{4+}$) where a stable empty, half-filled or full 4f results. Across the series the atomic and ionic radii decrease steadily (the lanthanoid contraction) because the 4f electrons shield poorly, allowing the effective nuclear charge to increase.

5-Mark / Long-Answer Questions (continued)

Q34. Account for the following: (a) transition metals form interstitial compounds; (b) transition metals form alloys; (c) Zn, Cd and Hg are not regarded as transition metals. Answer: (a) Small atoms (H, C, N, B) occupy the interstices of the metal lattice, forming hard, high-melting interstitial compounds that retain metallic conductivity. (b) Transition metals have similar atomic radii, so one metal can substitute for another in the lattice, forming substitutional alloys (e.g. brass, steel). (c) Zn, Cd and Hg have a completely filled $d^{10}$ configuration in their atoms and common +2 ions, so they have no partially filled d subshell and are not typical transition metals.

Q35. Compare the lanthanoids and actinoids with respect to oxidation states, radioactivity and contraction. Answer: The lanthanoids fill the 4f subshell and are dominated by the +3 oxidation state; most are non-radioactive. The actinoids fill the 5f subshell and show a much greater range of oxidation states (up to +6 or +7), because the 5f, 6d and 7s orbitals are close in energy; all actinoids are radioactive and those beyond uranium are synthetic. Both show a contraction, but the actinoid contraction is greater per element (5f shields even more poorly than 4f).

Q36. Why do the transition elements exhibit characteristic magnetic properties? Calculate the spin-only moment of Co$^{2+}$ (Z = 27). Answer: Most transition-metal ions have unpaired d electrons, each contributing a magnetic moment from its spin; such ions are paramagnetic. The moment is given by the spin-only formula $\mu = \sqrt{n(n+2)}$ BM. For Co$^{2+}$ ($[\text{Ar}]\,3d^7$), there are 3 unpaired electrons, so $\mu = \sqrt{3(3+2)} = \sqrt{15} = \mathbf{3.87\ BM}$.