Chapter 4: The d-and f-Block Elements Case Study Questions | chemistry Class 12 CBSE

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A chemistry teacher demonstrated an experiment where she dissolved chromite ore (FeCr2O4) in sodium carbonate in the presence of air at high temperature. The resulting yellow solution was acidified with sulphuric acid to produce an orange solution. The orange solution was then treated with potassium chloride to yield orange crystals. She explained that chromium in these compounds can act as a powerful oxidising agent, especially in acidic medium. The students observed that when the pH was changed by adding alkali, the orange colour shifted back to yellow. She further demonstrated that the oxidation state of chromium remains +6 throughout this interconversion.

Question 1: What is the orange compound crystallised in the above experiment and what is the oxidation state of chromium in it?

The orange compound is Potassium dichromate (K2Cr2O7).
The oxidation state of chromium in K2Cr2O7 is +6.

Question 2: Write the ionic equation for the interconversion of chromate to dichromate when the solution is acidified.

2 CrO4²⁻ + 2H⁺ → Cr2O7²⁻ + H2O
When alkali is added: Cr2O7²⁻ + 2OH⁻ → 2 CrO4²⁻ + H2O

Question 3: Describe the oxidising action of potassium dichromate in acidic solution. Write the ionic equation for its reaction with iron(II) ions.

In acidic solution, K2Cr2O7 acts as a strong oxidising agent.
The half-reaction for dichromate is: Cr2O7²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O (E° = 1.33 V)
Reaction with Fe²⁺: Cr2O7²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H2O
It oxidises Fe²⁺ to Fe³⁺, iodides to iodine, sulphides to sulphur, and tin(II) to tin(IV).

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A research student was studying the magnetic properties of first-row transition metal ions. She prepared aqueous solutions of Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺ and Zn²⁺ and measured their magnetic moments. She used the spin-only formula μ = √n(n+2) BM, where n is the number of unpaired electrons. She observed that the magnetic moment increased and then decreased across the series. One ion showed zero magnetic moment. The student noted that paramagnetism arises from unpaired electrons, each having a magnetic moment due to spin angular momentum, and that for first-row transition elements, orbital angular momentum is effectively quenched.

Question 1: Which ion among Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺ and Zn²⁺ shows zero magnetic moment and why?

Zn²⁺ shows zero magnetic moment.
Zn²⁺ has the electronic configuration 3d¹⁰ — all d orbitals are completely filled with no unpaired electrons, hence μ = 0 BM.

Question 2: Calculate the magnetic moment of Mn²⁺ ion using the spin-only formula and state its electronic configuration.

Mn²⁺ has the electronic configuration 3d⁵ with 5 unpaired electrons.
μ = √5(5+2) = √35 = 5.92 BM

Question 3: Explain the concept of paramagnetism and ferromagnetism in the context of transition metals. Why do first-row transition metal compounds not require orbital contribution for magnetic moment calculation?

Paramagnetism arises from the presence of unpaired electrons; paramagnetic substances are attracted by an applied magnetic field.
Ferromagnetism is an extreme form of paramagnetism where substances are very strongly attracted.
Each unpaired electron has a magnetic moment from its spin and orbital angular momentum.
For first-row transition metal compounds, the contribution of orbital angular momentum is effectively quenched due to interaction with the crystal field/ligand environment.
Hence only the spin-only formula μ = √n(n+2) BM is used for calculation.

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During a class demonstration, a teacher prepared potassium permanganate by fusing MnO2 with KOH in the presence of air, producing a dark green intermediate. The green solution disproportionated in acidic conditions to give the deep purple permanganate. Potassium permanganate crystals were shown to be isostructural with KClO4. The teacher then showed that in acidic solution, KMnO4 oxidised oxalic acid to CO2, releasing Mn²⁺ ions. The reaction was used as a standard titration method. The students noted the intense purple colour and its use as an analytical reagent in both acidic and neutral conditions.

Question 1: What is the green intermediate formed during the preparation of KMnO4 and write the equation for its disproportionation?

The green intermediate is potassium manganate, K2MnO4.
3MnO4²⁻ + 4H⁺ → 2MnO4⁻ + MnO2 + 2H2O

Question 2: Write the ionic equation for the oxidation of oxalic acid by KMnO4 in acidic solution.

5C2O4²⁻ + 2MnO4⁻ + 16H⁺ → 2Mn²⁺ + 8H2O + 10CO2
The oxidation state of Mn changes from +7 to +2, gaining 5 electrons per Mn.

Question 3: KMnO4 behaves differently in acidic, neutral and alkaline media. Explain with examples the oxidation product of iodide in neutral/faintly alkaline solution versus acidic solution.

In acidic solution, KMnO4 reduces to Mn²⁺ (pink/colourless): 10I⁻ + 2MnO4⁻ + 16H⁺ → 2Mn²⁺ + 8H2O + 5I2
In neutral or faintly alkaline solution, KMnO4 reduces to MnO2 (brown precipitate).
Iodide is oxidised to iodate (IO3⁻) in neutral medium: 2MnO4⁻ + H2O + I⁻ → 2MnO2 + 2OH⁻ + IO3⁻
The difference is due to the hydrogen ion concentration influencing the reduction potential and final product.

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A geochemist studying rare earth elements noticed that Zirconium (Zr) and Hafnium (Hf) almost always occur together in nature and are extremely difficult to separate. Both have nearly identical atomic radii (Zr: 160 pm, Hf: 159 pm) and very similar physical and chemical properties, despite being in the same group but different periods. This anomaly puzzled early chemists. The geochemist explained that this is a direct consequence of the lanthanoid contraction — the gradual decrease in atomic and ionic radii across the lanthanoid series from La to Lu, caused by poor shielding of 4f electrons.

Question 1: What is lanthanoid contraction? Why does it occur?

Lanthanoid contraction is the gradual decrease in atomic and ionic radii from lanthanum (La) to lutetium (Lu) across the lanthanoid series.
It occurs because 4f electrons provide poor shielding of the nuclear charge for the outer electrons, so the effective nuclear charge increases across the series, pulling electrons closer and decreasing the size.

Question 2: How does lanthanoid contraction explain why Zr and Hf have almost identical radii despite being in the same group?

Normally, atomic radius increases going down a group (from Zr to Hf).
However, the filling of 4f orbitals in the lanthanoid series between 4d and 5d causes a steady decrease in size (lanthanoid contraction).
This contraction exactly compensates for the expected increase, resulting in Zr (160 pm) and Hf (159 pm) having nearly identical radii.

Question 3: List the consequences of lanthanoid contraction on the chemistry of the elements succeeding the lanthanoids.

The 5d transition series elements (3rd series) have almost the same radii as the corresponding 4d (2nd series) elements due to lanthanoid contraction.
Elements like Zr and Hf have nearly identical properties and co-occur in nature, making separation very difficult.
The 5d metals are denser and have higher melting points compared to their 4d counterparts, leading to more frequent metal-metal bonding.
The similar radii of 4d and 5d elements mean they have very similar chemical properties — far more so than expected from normal group trends.
The 3rd transition series metals tend to be 'noble' (less reactive) due to high enthalpies of atomisation.

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A student was asked to explain why transition metals show variable oxidation states, while non-transition elements like Na show only +1. She examined the electronic configuration of manganese (Z=25) and noted it could show oxidation states from +2 to +7 — the widest range in the first transition series. Her teacher pointed out that the maximum oxidation state corresponds to the total number of (n-1)d and ns electrons up to manganese. Beyond Mn, higher oxidation states become less stable and typical species are Fe²⁺/³⁺, Co²⁺/³⁺, Ni²⁺, Cu⁺/²⁺, Zn²⁺. The student also noted that in carbonyl complexes like Ni(CO)4, the oxidation state is zero.

Question 1: Why does manganese show the largest number of oxidation states (+2 to +7) in the first transition series?

Manganese (Z=25) has the electronic configuration 3d⁵4s², with the maximum number of unpaired electrons.
It can use all seven outer electrons (5 from 3d + 2 from 4s) in bonding, giving oxidation states from +2 to +7.
Elements near the middle of the series have both enough electrons to lose and available empty orbitals for higher states.

Question 2: Why does zinc show only the +2 oxidation state in all its compounds?

Zinc has the electronic configuration 3d¹⁰4s².
In the +2 state, it loses both 4s electrons, giving a stable fully-filled 3d¹⁰ configuration.
No d electrons are involved in bonding and the fully filled d¹⁰ configuration is very stable, preventing higher oxidation states.

Question 3: How does the variability in oxidation states of transition metals differ from that of non-transition elements? Explain with examples and account for the zero oxidation state in Ni(CO)4.

In transition metals, oxidation states differ by unity (e.g., V²⁺, V³⁺, V⁴⁺, V⁵⁺) because electrons are lost from closely spaced d and s orbitals.
In non-transition elements, oxidation states normally differ by two (e.g., S: 0, +2, +4, +6) due to the inert pair effect or electron pair involvement.
The variable oxidation states of transition metals arise from incomplete filling of d orbitals; the small energy gap between (n-1)d and ns levels allows variable electron loss.
In Ni(CO)4, CO is a π-acceptor ligand that accepts electron density from the metal via back-bonding; the metal needs to donate electrons, stabilising the zero oxidation state.
The availability of empty d orbitals for π-backbonding with CO stabilises the Ni(0) state.

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Chapter 4: The d-and f-Block Elements | Case Study Questions

Passage

Question 1: What is the orange compound crystallised in the above experiment and what is the oxidation state of chromium in it?

Question 2: Write the ionic equation for the interconversion of chromate to dichromate when the solution is acidified.

Question 3: Describe the oxidising action of potassium dichromate in acidic solution. Write the ionic equation for its reaction with iron(II) ions.

Passage

Question 1: Which ion among Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺ and Zn²⁺ shows zero magnetic moment and why?

Question 2: Calculate the magnetic moment of Mn²⁺ ion using the spin-only formula and state its electronic configuration.

Question 3: Explain the concept of paramagnetism and ferromagnetism in the context of transition metals. Why do first-row transition metal compounds not require orbital contribution for magnetic moment calculation?

Passage

Question 1: What is the green intermediate formed during the preparation of KMnO4 and write the equation for its disproportionation?

Question 2: Write the ionic equation for the oxidation of oxalic acid by KMnO4 in acidic solution.

Question 3: KMnO4 behaves differently in acidic, neutral and alkaline media. Explain with examples the oxidation product of iodide in neutral/faintly alkaline solution versus acidic solution.

Passage

Question 1: What is lanthanoid contraction? Why does it occur?

Question 2: How does lanthanoid contraction explain why Zr and Hf have almost identical radii despite being in the same group?

Question 3: List the consequences of lanthanoid contraction on the chemistry of the elements succeeding the lanthanoids.

Passage

Question 1: Why does manganese show the largest number of oxidation states (+2 to +7) in the first transition series?

Question 2: Why does zinc show only the +2 oxidation state in all its compounds?

Question 3: How does the variability in oxidation states of transition metals differ from that of non-transition elements? Explain with examples and account for the zero oxidation state in Ni(CO)4.

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