CBSE Class 12 Chemistry The d and f Block Elements Assignment Set 01

LEVEL-1 QUESTIONS

Question 1. Explain briefly how +2 oxidation states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.
Answer: In \(\text{M}^{2+}\) ions, 3d-orbitals get occupied gradually as the atomic number increases. Since the number of empty d-orbitals decreases, the stability of cations increases from \(\text{Sc}^{2+}\) to \(\text{Mn}^{2+}\). \(\text{Mn}^{2+}\) is most stable as all d-orbitals are singly occupied.
In simple words: Half-filled d-orbitals in manganese provide extra stability.

Exam Tip: Always mention exchange energy and half-filled stability when discussing \(\text{Mn}^{2+}\).

Question 2. Explain why transition elements have many irregularities in their electronic configurations?
Answer: In the transition elements, there is a very small difference in the energy of \((n-1)\text{d}\)-orbitals and \(ns\)-orbitals. Thus, the incoming electron can occupy either of the shells. Hence, transition elements exhibit many irregularities in their electronic configurations.
In simple words: Close energy levels between d and s orbitals cause electrons to shift easily.

Exam Tip: Mention chromium and copper as classic examples of this energy proximity.

Question 3. What are different oxidation states exhibited by Lanthanides?
Answer: The common stable oxidation state of lanthanides is \(+3\). However, some members also show oxidation states of \(+2\) and \(+4\).
In simple words: Lanthanides mostly prefer the \(+3\) state due to stable noble gas-like cores combined with lost electrons.

Exam Tip: Remember that \(+4\) is shown by Cerium (\(\text{Ce}\)) and \(+2\) by Europium (\(\text{Eu}\)) to acquire stable configurations.

Question 4. How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.
Answer: The transition elements use their \((n-1)\text{d}\), \(ns\), and \(np\) orbitals, and the successive oxidation states differ by unity. For example, Mn shows all the oxidation states from \(+2\) to \(+7\). On the other hand, non-transition elements use their \(ns\), \(np\), and \(nd\) orbitals, and the successive oxidation states differ by two units, e.g., \(\text{Sn}^{2+}\), \(\text{Sn}^{4+}\), etc.
In simple words: Transition metals change oxidation states by steps of 1, while main-group elements jump by 2.

Exam Tip: Highlight the participation of inner d-orbitals in transition metals.

Question 5. Why do transition elements show variable oxidation states?
Answer: The transition elements show variable oxidation states due to the small energy difference between \((n-1)\text{d}\) and \(ns\) orbitals, as a result of which both \((n-1)\text{d}\) and \(ns\) electrons take part in bond formation.
In simple words: Both outer s and inner d electrons have similar energies, so both can participate in bonding.

Exam Tip: Use the phrase "comparable energies of \(ns\) and \((n-1)\text{d}\) orbitals" for full marks.

Question 6. Why are \(\text{Mn}^{2+}\) compounds more stable than \(\text{Fe}^{2+}\) compounds towards oxidation to \(+3\) state?
Answer: The electronic configuration of \(\text{Mn}^{2+}\) is \([\text{Ar}] 3d^5\), i.e., all five d-orbitals are singly occupied. Thus, this is a stable electronic configuration and further loss of an electron requires high energy. On the other hand, the electronic configuration of \(\text{Fe}^{2+}\) is \([\text{Ar}] 3d^6\), i.e., loss of one electron requires low energy to achieve the stable \(3d^5\) configuration.
In simple words: Manganese already has a half-filled stable d-subshell, whereas iron wants to lose an electron to reach that stable half-filled state.

Exam Tip: Always relate stability to half-filled (\(d^5\)) and completely filled (\(d^{10}\)) configurations.

Question 7. To what extent do the electronic configuration decide the stability of oxidation state in the first series of the transition elements? Illustrate your answer with an example.
Answer: In a transition series, the oxidation states which lead to exactly half-filled or completely filled orbitals are more stable. E.g., the electronic configuration of Fe is \([\text{Ar}] 3d^6 4s^2\). It shows various oxidation states, but Fe(III) (\(3d^5\)) is more stable than Fe(II) (\(3d^6\)) in certain aqueous conditions due to exchange energy.
In simple words: Configurations with half-filled or full d-orbitals dictate extra stability for specific oxidation states.

Exam Tip: Mention Hund's rule and exchange energy as the driving factors for half-filled stability.

Question 8. Which metal in the first series of transition metals exhibits \(+1\) oxidation state most frequently and why?
Answer: Copper with configuration \([\text{Ar}] 3d^{10} 4s^1\) exhibits \(+1\) oxidation state. Copper loses its \(4s^1\) electron easily and achieves a stable, completely filled configuration \(3d^{10}\) by forming \(\text{Cu}^+\).
In simple words: Copper easily loses its single 4s electron to attain a fully-filled, highly stable 10-electron d-subshell.

Question 9. What are inner transition elements?
Answer: The f-block elements in which the last electron is accommodated in the \((n-2)f\)-subshell are called inner transition elements. These include atomic numbers 58 to 71 (lanthanoids) and from 90 to 103 (actinoids).
In simple words: Elements where electrons fill the f-orbitals two shells below the outermost shell are inner transition metals.

Exam Tip: Clearly state \((n-2)f\) filling to distinguish them from regular d-block transition elements.

Question 10. The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases. Explain why?
Answer: In the 3d-transition series, as we move from Sc (21) to Mn (25), the number of unpaired electrons increases from 1 to 5, and hence paramagnetic character increases. After Mn, the pairing of electrons in the d-orbital starts, the number of unpaired electrons decreases from 4 to 0, and hence paramagnetic character decreases.
In simple words: Paramagnetism depends on unpaired electrons. They increase up to manganese and then pair up and cancel out.

Exam Tip: Use the spin-only formula \(\mu = \sqrt{n(n+2)}\) BM to support your explanation.

Question 11. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition metal.
Answer: The following points justify that the given statement is true:
(i) Ionization enthalpies of heavier transition elements are higher than the elements of the 3d series. Consequently, heavier transition elements are less reactive in comparison to 3d-elements.
(ii) Melting points of heavier transition elements are higher than 3d-elements due to stronger metal-metal bonding.
(iii) Higher oxidation states of heavier transition elements are more stable, whereas lower oxidation states are more stable in 3d-elements.
In simple words: 4d and 5d metals have stronger bonds, higher melting points, and prefer higher oxidation states compared to 3d metals.

Exam Tip: Emphasize greater effective nuclear charge and d-orbital overlap in heavier transition series.

Question 12. What are transition elements? Which d-block elements are not regarded as transition elements and why?
Answer: An element which has a partially filled \((n-1)\text{d}\)-orbital in its ground state or in any of its common oxidation states is known as a transition element. Group 12 elements, i.e., Zn, Cd, and Hg, have completely filled \((n-1)\text{d}\)-orbitals in both their atomic and ionic states, and thus these elements are not considered transition elements.
In simple words: Zinc, cadmium, and mercury have full d-orbitals even after losing electrons, so they are not true transition metals.

Exam Tip: Define transition metals strictly using the criterion of incomplete d-subshells.

Question 13. What are interstitial compounds? Why are such compounds well known for transition metal?
Answer: Compounds formed when small atoms like H, C, N, or B are trapped inside the crystal lattices of transition metals are known as interstitial compounds. These compounds are well known for transition metals because the sizes of C, N, O, and B atoms are very similar to the sizes of the interstitial voids present in transition metal crystal lattices.
In simple words: Small atoms fit snugly into the empty spaces between large transition metal atoms.

Exam Tip: Note that interstitial compounds are hard, rigid, and retain metallic conductivity while having high melting points.

Question 14. (a) What is actinide contraction? What effect does it have on the chemistry of the elements which follow actinides?
(b) Name an important alloy, which contains some of the lanthanide metals. Mention its uses.
Answer: (a) The size of actinoids and their trivalent ions steadily decreases from Ac to Lw. This is known as actinide contraction.
Cause: It is due to the poor shielding effect of 5f electrons.
Consequences of actinide contraction:
(i) Basic strength of hydroxides decreases from \(\text{Ac(OH)}_3\) to \(\text{Lw(OH)}_3\).
(ii) Due to extremely similar chemical properties, actinoids are extremely difficult to separate.
(b) An important alloy containing lanthanoid metals is mischmetal, which contains 95% lanthanide metal and 5% Fe along with traces of S, C, Ca, and Al. It is used in Mg-based alloys to produce bullets, shells, and lighter flints.
In simple words: Poor shielding by f-electrons causes atomic shrinkage across the series, and mischmetal is a useful lanthanide alloy for flints.

Exam Tip: Always link actinide and lanthanide contractions to poor shielding by f-orbitals.

LEVEL-2 QUESTIONS

Question 1. Give reasons for the following:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalysts.
(v) Transition metals have a strong tendency to form complexes.
Answer:
(i) Transition metals and many of their compounds show paramagnetic behaviour due to the presence of unpaired electrons in their \((n-1)\text{d}\) orbitals.
(ii) The enthalpies of atomisation of transition metals are high because of a large number of unpaired electrons in their atoms, leading to strong interatomic interactions and robust metallic bonding.
(iii) Transition metals generally form coloured compounds due to the presence of unpaired electrons in \((n-1)\text{d}\) orbitals, enabling d-d electronic transitions upon absorbing visible light.
(iv) Transition metals act as good catalysts due to their partially filled \((n-1)\text{d}\) orbitals, variable oxidation states, and ability to provide a large surface area with active adsorption sites.
(v) Transition metals have a strong tendency to form complexes due to their small ionic sizes, high ionic charges, and the presence of vacant d-orbitals of suitable energy to accept electron pairs from ligands.
In simple words: Unpaired d-electrons and vacant orbitals make transition metals versatile, magnetic, colourful, and catalytic.

Exam Tip: Memorize these five fundamental properties as they form the core of d-block descriptive chemistry questions.

Question 2. Give reasons for the following:
(i) Fe has higher melting point than Cu.
(ii) \([\text{Ti(H}_2\text{O)}_6]^{3+}\) is coloured while \([\text{Sc(H}_2\text{O)}_6]^{3+}\) is colourless.
(iii) The 4d and 5d series of transition metals have more frequent metal-metal bonding in their compounds than do the 3d metals.
(iv) Transition metals sometimes exhibit very low oxidation states such as \(+1\) and \(0\).
(v) Hg is not considered a transition metal.
Answer:
(i) Fe (\(3d^6 4s^2\)) has four unpaired electrons in the 3d-subshell, whereas Cu (\(3d^{10} 4s^1\)) has only one unpaired electron. Hence, metallic bonding is much stronger in Fe than in Cu.
(ii) The oxidation state of Ti in \([\text{Ti(H}_2\text{O)}_6]^{3+}\) is \(+3\) with a \(3d^1\) configuration, containing one unpaired electron, making it coloured. In contrast, Sc in \([\text{Sc(H}_2\text{O)}_6]^{3+}\) is in the \(+3\) state with a \(3d^0\) configuration, having no unpaired electrons, hence it is colourless.
(iii) In the same group of d-block elements, 4d and 5d transition elements are larger in size than 3d elements, allowing their valence orbitals to overlap more effectively, forming metal-metal bonds more frequently.
(iv) \(+1\) oxidation state is shown by elements like Cu because after losing one electron, it achieves a stable configuration. Zero oxidation state is shown in metal carbonyls because \(\pi\)-electrons donated by CO molecules are accepted into empty metal d-orbitals.
(v) The characteristic properties of transition metals are due to partially filled d-orbitals. Hg has a completely filled \(3d^{10}\) (or \(5d^{10}\)) configuration; as a result, it does not exhibit typical transition metal properties.
In simple words: Differences in unpaired electrons, d-orbital occupancy, and spatial size govern these specific metallic behaviors.

Exam Tip: Always quote electronic configurations explicitly when explaining colour and melting point differences.

Question 3. (a) Write electronic configuration of element having atomic number 101.
(b) Which element shows maximum oxidation state in 3d transition series?
(c) What is mischmetal?
(d) Explain why \(\text{Cu}^+\) ion is not stable in aqueous solution?
(e) Name the transition metal which is well known to exhibit \(+4\) oxidation state?
Answer:
(a) \([\text{Rn}] 5f^{13} 6d^0 7s^2\)
(b) Manganese (Mn), which shows a \(+7\) oxidation state in \(\text{KMnO}_4\).
(c) Mischmetal is an important alloy containing 95% lanthanide metal and 5% iron, along with traces of sulfur, carbon, calcium, and aluminum. It is used in magnesium-based alloys to produce bullets, shells, and lighter flints.
(d) Water is a good complexing agent, and the hydration enthalpy of \(\text{Cu}^{2+}\) is much higher than that of \(\text{Cu}^+\), which more than compensates for the second ionization enthalpy, causing \(\text{Cu}^+\) to disproportionate into \(\text{Cu}^{2+}\) and Cu in aqueous solution.
(e) Cerium (\(Z = 58\)) is well known for exhibiting the \(+4\) oxidation state (though strictly an f-block element, it is frequently cited in this context; among d-block transition metals, titanium and zirconium show stable \(+4\) states).
In simple words: Hydration energy drives copper(I) to disproportionate into copper(II) in water, while manganese reaches the peak oxidation state of \(+7\).

Exam Tip: For \(\text{Cu}^+\) stability questions, always mention hydration energy as the driving force.

Question 4. (a) Write the preparation of potassium dichromate from iron chromite. What happens when potassium dichromate reacts with (i) Hydrogen sulphide (ii) \(\text{FeSO}_4\)?
(b) Why do Zr and Hf exhibit almost similar properties?
(c) Why is \(\text{La(OH)}_3\) a stronger base than \(\text{Lu(OH)}_3\)?
Answer:
(a) Preparation takes place in three steps:
(i) Conversion of chromite ore to sodium chromate:
\(4\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2\)
(ii) Conversion of sodium chromate to 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) Conversion of sodium dichromate to potassium dichromate:
\(\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}\)
Reactions with potassium dichromate:
(i) With Hydrogen sulphide:
\(\text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{S}\)
(ii) With \(\text{FeSO}_4$\):
\(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}\)
(b) Zr and Hf exhibit almost similar properties due to the lanthanide contraction, which results in nearly identical atomic and ionic radii.
(c) Due to lanthanide contraction, the size of \(\text{La}^{3+}\) is larger than that of \(\text{Lu}^{3+}\). Consequently, the \(\text{Lu-O}\) bond is much stronger than the \(\text{La-O}\) bond, making \(\text{La(OH)}_3\) dissociate more easily to release hydroxide ions, rendering it a stronger base.
In simple words: Lanthanide contraction makes zirconium and hafnium twins in size and properties, and larger lanthanum ions form weaker bonds with hydroxide, releasing \(\text{OH}^-\) more readily.

Exam Tip: Practice the three-step industrial conversion of chromite ore to potassium dichromate thoroughly.

Question 5. Give reasons for the following:
(i) Transition metals have high enthalpy of hydration.
(ii) Zn, Cd and Hg are not regarded as transition metal.
(iii) d-block elements exhibit a large number of oxidation states than f-block elements.
(iv) The second and third members in each group of transition element have similar atomic radii.
(v) \(\text{K}_2[\text{PtCl}_6]\) is a well known compound whereas the corresponding Ni compound is not known.
Answer:
(i) Transition metal ions are smaller in size and carry a higher nuclear charge, therefore they attract water molecules strongly, resulting in high enthalpy of hydration.
(ii) Group 12 elements, i.e., Zn, Cd, and Hg, have completely filled \((n-1)\text{d}\)-orbitals in their ground and ionic states; thus they are not considered transition metals.
(iii) The energy difference between \((n-1)\text{d}\) and \(ns\) orbitals in d-block elements is very small, allowing electrons from both subshells to participate in bonding. In contrast, \((n-2)f\) orbitals in f-block elements lie deep inside and are not readily available for bonding.
(iv) The second and third members in each group of transition elements have similar atomic radii due to the lanthanide contraction, which offsets the expected size increase down the group.
(v) The \(+4\) oxidation state is stable for platinum, making \(\text{K}_2[\text{PtCl}_6]\) stable. However, the \(+4\) oxidation state for nickel is extremely difficult to achieve because the sum of its first four ionization energies is exceptionally high, making the corresponding nickel(IV) compound non-existent.
In simple words: Small size, vacant d-orbitals, and lanthanide contraction explain these distinct chemical trends.

Exam Tip: Reference ionization energy sums when explaining why high oxidation states are unstable in lighter group members like nickel.

Question 6. What is meant by disproportionation? Give two examples.
Answer: Those reactions in which the same substance undergoes oxidation as well as reduction simultaneously are called disproportionation reactions.
Examples:
(i) \(2\text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\)
(ii) \(3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
In simple words: A disproportionation reaction is one where a single chemical species is both oxidized and reduced at the same time.

Exam Tip: Always verify that the element in question changes to both a higher and a lower oxidation state.

Question 7. Complete the following reactions:
(i) \(\text{MnO}_4^- + \text{H}^+ + \text{Fe}^{2+} \rightarrow \dots+ \dots+ \dots\)
(ii) \(\text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} + \text{H}^+ \rightarrow \dots+ \dots+ \dots\)
(iii) \(\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \rightarrow \dots+ \dots+ \dots\)}
Answer:
(i) \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)
(ii) \(2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)
(iii) \(2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^-\)
In simple words: Permanganate acts as a strong oxidizing agent in acidic and neutral/faintly alkaline media, converting into manganese(II) or manganese dioxide.

Question 9. Decide giving reason which one of the following pairs exhibits the property indicated:
(i) \(\text{Sc}^{3+}\) or \(\text{Cr}^{3+}\) exhibits paramagnetism
(ii) V or Mn exhibits more number of oxidation states
(iii) \(\text{V}^{4+}\) or \(\text{V}^{5+}\) exhibits colour
Exam Tip: Always write out the noble gas core electronic configuration when comparing magnetic and optical properties.

Question 10. (a) Describe the general trends in the following properties of the first series of the transition elements:
(i) Stability of \(+2\)-oxidation state
(ii) Formation of oxometal ions
(b) Write steps involved in the preparation of \(\text{KMnO}_4\) from \(\text{K}_2\text{MnO}_4\).
Answer:
(a) (i) The elements of the first transition series show a decreasing tendency to form divalent cations as we move from left to right in the series. This trend is due to a general increase in the sum of first and second ionization energies. The greater stability of \(\text{Mn}^{2+}\) is due to the half-filled \(d^5\) configuration, and that of zinc is due to the \(d^{10}\) configuration.
(ii) All metals except Sc form basic oxides of the type MO. The highest oxidation number in all oxides coincides with the group number, attained from \(\text{Sc}_2\text{O}_3\) to \(\text{Mn}_2\text{O}_7\). The formation of oxoanions is driven by the high electronegativity and small size of the oxygen atom.
(b) Preparation of \(\text{KMnO}_4\) from \(\text{K}_2\text{MnO}_4\) takes place in two steps:
(i) Conversion of pyrolusite ore into potassium manganate:
\(2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}\)
(ii) Oxidation of potassium manganate to potassium permanganate:
\(3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
In simple words: Divalent stability decreases across the period except at half/full shells, and manganate is oxidized to permanganate to produce \(\text{KMnO}_4\).

Exam Tip: Note that potassium manganate disproportionates in acidic or neutral solutions to yield permanganate and manganese dioxide.

Question 11. (a) Write the steps involved in the preparation of \(\text{K}_2\text{Cr}_2\text{O}_7\) from chromite ore.
(b) What is the effect of pH on dichromate ion solution?
Answer:
(a) The preparation takes place in three steps:
(i) Conversion of chromite ore to sodium chromate:
\(4\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2\)
(ii) Conversion of sodium chromate to 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) Conversion of sodium dichromate to potassium dichromate:
\(\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}\)
(b) Dichromate ion is orange in acidic solution (\(\text{pH} < 7\)) and turns yellow in basic solution (\(\text{pH} > 7\)). This is due to the interconversion of dichromate and chromate ions:
\(2\text{CrO}_4^{2-} \text{ (yellow)} + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} \text{ (orange)} + \text{H}_2\text{O}\)
\(\text{Cr}_2\text{O}_7^{2-} \text{ (orange)} + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} \text{ (yellow)} + \text{H}_2\text{O}\)
In simple words: Acid turns chromate into orange dichromate, while base turns dichromate back into yellow chromate.

Exam Tip: Memorize the colour change: Yellow in alkali, Orange in acid.

Question 12. (a) What is lanthanide contraction? What effect does it have on the chemistry of the elements, which follow lanthanoids?
(b) The chemistry of actinoid elements is not so much smooth as that of lanthanoids. Justify these statements by giving some examples from the oxidation state of these elements.
Answer:
(a) The steady decrease in atomic and ionic radii of lanthanoids and their trivalent ions from La to Lu is known as lanthanide contraction. Its main effect on post-lanthanide elements (like Zr and Hf) is that they possess nearly identical atomic radii, making them notoriously difficult to separate.
(b) Lanthanoids show a limited number of oxidation states, principally \(+3\), because of a large energy gap between 4f, 5d, and 6s subshells. In contrast, actinoids show a wide range of oxidation states (such as \(+4, +5, +6, +7\)) alongside \(+3\) because the energy difference between 5f, 6d, and 7s orbitals is very small, making their chemistry more complex and varied.
In simple words: Poor shielding in lanthanoids causes atomic contraction, while actinoids exhibit diverse oxidation states due to closely spaced energy levels.

Exam Tip: Emphasize the energy gap difference between 4f/5d and 5f/6d subshells.

Question 13. Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) Of the \(d^4\) species, \(\text{Cr}^{2+}\) is strongly reducing while manganese(III) is strongly oxidizing.
Answer:
(i) Oxides of transition metals in lower oxidation states are generally basic because the metal ion retains more electron density, whereas those in higher oxidation states are acidic due to high charge density and covalent character. E.g., \(\text{MnO}\) is basic, \(\text{Mn}_3\text{O}_4\) is amphoteric, and \(\text{Mn}_2\text{O}_7\) is acidic.
(ii) A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are the most electronegative and oxidizing elements, capable of stabilizing high oxidation states and unpairing metal d-electrons.
(iii) Redox properties depend on \(E^0\) values. Since the \(E^0\) value for \(\text{Cr}^{3+}/\text{Cr}^{2+}\) is negative, \(\text{Cr}^{2+}\) readily loses an electron to become \(\text{Cr}^{3+}\) (acting as a reducing agent). Conversely, the \(E^0\) value for \(\text{Mn}^{3+}/\text{Mn}^{2+}\) is positive, making \(\text{Mn}^{3+}\) easily reduced to the stable half-filled \(\text{Mn}^{2+}\) state (acting as a strong oxidizing agent).
In simple words: Low oxidation oxides are basic, high oxidation oxides are acidic, and electrode potentials dictate whether a \(d^4\) ion acts as a reducer or oxidant.

Exam Tip: Link acidic character of metal oxides directly to increasing oxidation number and charge density.

Question 14. For \(\text{M}^{2+}/\text{M}\) and \(\text{M}^{3+}/\text{M}^{2+}\) systems, the \(E^0\) values for some metals are as follows:
\(\text{Cr}^{2+}/\text{Cr} = -0.9\text{V}, \quad \text{Cr}^{3+}/\text{Cr}^{2+} = -0.4\text{V}\)
\(\text{Mn}^{2+}/\text{Mn} = -1.2\text{V}, \quad \text{Mn}^{3+}/\text{Mn}^{2+} = +1.5\text{V}\)
\(\text{Fe}^{2+}/\text{Fe} = -0.4\text{V}, \quad \text{Fe}^{3+}/\text{Fe}^{2+} = +0.8\text{V}\)
Use this data to comment upon:
(i) the stability of \(\text{Fe}^{3+}\) in acid solution as compared to \(\text{Cr}^{3+}\) or \(\text{Mn}^{3+}\) and
(ii) the ease with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.
Answer:
(i) The \(E^0\) for \(\text{Cr}^{3+}/\text{Cr}^{2+}\) is \(-0.4\text{V}\) (negative), meaning \(\text{Cr}^{3+}\) cannot be easily reduced to \(\text{Cr}^{2+}\) (making \(\text{Cr}^{3+}\) very stable). The \(E^0\) for \(\text{Mn}^{3+}/\text{Mn}^{2+}\) is \(+1.5\text{V}\) (strongly positive), meaning \(\text{Mn}^{3+}\) is easily reduced to \(\text{Mn}^{2+}\). The \(E^0\) for \(\text{Fe}^{3+}/\text{Fe}^{2+}\) is \(+0.8\text{V}\), indicating moderate stability. Thus, stability order is: \(\text{Mn}^{3+} < \text{Fe}^{3+} < \text{Cr}^{3+}\).
(ii) The oxidation potentials for \(\text{M}/\text{M}^{2+}\) are \(+0.9\text{V}\) for Cr, \(+1.2\text{V}\) for Mn, and \(+0.4\text{V}\) for Fe. Thus, the ease of getting oxidized follows the order: \(\text{Mn} > \text{Cr} > \text{Fe}\).
In simple words: Negative reduction potentials indicate stable ions and easy metal oxidation, letting us rank metal reactivities.

Exam Tip: Remember that a more positive reduction potential means the species is a stronger oxidizing agent and harder to oxidize as a metal.

Question 15. Account for the following statements:
(i) Cobalt(II) is stable in aqueous solution, but in the presence of strong ligands and air, it can be oxidized to Co(III).
(ii) The \(d^1\) configuration is very unstable in ions.
(iii) One among the lanthanides, Ce(III) can be easily oxidized to Ce(IV).
Answer:
(i) Strong ligands force cobalt(II) to pair its electrons, causing crystal field stabilization energy (CFSE) to overcome ionization energy, enabling loss of one more electron from the 3d-subshell and promoting \(d^2sp^3\) hybridization.
(ii) Ions with a \(d^1\) configuration tend to lose their single remaining d-electron to achieve a stable, lower noble gas core configuration.
(iii) The electronic configuration of Ce is \([\text{Xe}] 4f^1 5d^1 6s^2\). There is very little energy difference between 4f, 5d, and 6s orbitals, allowing Ce to utilize electrons present in these orbitals easily to attain a stable empty f-orbital configuration (\(f^0\)) corresponding to the \(+4\) oxidation state.
In simple words: Ligand field stabilization, noble gas configurations, and vacant f-orbital stability explain these specific redox behaviours.

Exam Tip: Highlight crystal field stabilization energy (CFSE) when explaining cobalt oxidation in complex ions.

Question 16. Compare the chemistry of actinides with that of the lanthanoids with special reference to:
(i) electronic configuration
(ii) atomic and ionic sizes
(iii) oxidation state
(iv) chemical reactivity
Answer:

In simple words: Lanthanoids and actinoids both prefer \(+3\) states, but actinoids show broader oxidation states and higher reactivity due to lower orbital energy gaps.

Exam Tip: Present comparisons in clear tabular format to score maximum marks in comparative chemistry questions.

LEVEL-3 QUESTIONS

Question 1. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess reacts with \(\text{NH}_3\) to give an explosive compound (C). Identify compounds A, B, and C.
Answer:
\(A = \text{MnO}_2\) (Manganese dioxide, brown/black solid)
\(B = \text{Cl}_2\) (Chlorine gas)
\(C = \text{NCl}_3\) (Nitrogen trichloride, explosive compound)
Chemical Equations:
\(\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\)
\(\text{NH}_3 + 3\text{Cl}_2 \rightarrow \text{NCl}_3 + 3\text{HCl}\)
In simple words: Manganese dioxide reacts with hydrochloric acid to release chlorine gas, which then reacts with ammonia to form explosive nitrogen trichloride.

Exam Tip: Remember that \(\text{MnO}_2\) acts as an oxidizing agent in acidic media to convert chloride ions into elemental chlorine gas.

Question 2. A mixed oxide of iron and chromium \(\text{FeOCr}_2\text{O}_3\) is fused with sodium carbonate in the presence of air to form a yellow colored compound (A). On acidification the compound (A) forms an orange colored compound (B), which is a strong oxidizing agent.
a. Identify compound (A) and (B)
b. Write a balanced chemical equation.
Exam Tip: Practice chromite ore oxidation steps as they frequently appear in board exams.

Question 3. A green solution of potassium manganate turns purple when \(\text{CO}_2\) gas is passed through the solution. Explain why?
Answer: \(\text{CO}_2\) reacts with water to form carbonic acid as given below:
\(\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\)
In acidic medium, manganate ion (\(\text{MnO}_4^{2-}\)) undergoes disproportionation to form permanganate ions (\(\text{MnO}_4^-\)), which imparts a deep purple colour to the solution:
\(3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
In simple words: Carbon dioxide makes the solution acidic, triggering potassium manganate to disproportionate into purple permanganate and brown manganese dioxide.

Exam Tip: This is a classic example of acid-induced disproportionation of manganate(VI).

Question 4. Draw structure of the following:
a) \(\text{CrO}_4^{2-}\)
b) \(\text{Cr}_2\text{O}_7^{2-}\)
c) \(\text{MnO}_4^{2-}\)
d) \(\text{MnO}_4^-\)}
Answer:
a) \(\text{CrO}_4^{2-}\): Tetrahedral structure with four \(\text{Cr=O}\) bonds.
b) \(\text{Cr}_2\text{O}_7^{2-}\): Two tetrahedral \(\text{CrO}_4\) units sharing one corner oxygen atom forming a \(\text{Cr-O-Cr}\) bridge with an angle of \(126^\circ\).
c) \(\text{MnO}_4^{2-}\): Tetrahedral green manganate ion structure.
d) \(\text{MnO}_4^-\): Tetrahedral deep purple permanganate ion structure with \(p\pi-d\pi\) bonding.
In simple words: All these oxoanions adopt tetrahedral geometries with multiple bonds between the transition metal and oxygen atoms.

Exam Tip: Practice drawing the bridged dichromate structure with correct bond angles (\(126^\circ\)).