d-and f-Block Elements Previous Year Questions — JEE Main Chemistry
44 past-year d-and f-Block Elements questions from JEE Main (Chemistry) — page 6.
Q80 (2024)
In alkaline medium. $MnO_4^-$ oxidises $I^-$ to
- $IO_4^-$
- $IO^-$
- $I_2$
- $IO_3^-$
### Core Logic
The behavior of the permanganate ion ($MnO_4^-$) varies with the pH of the medium.
In a faintly alkaline or neutral medium, $MnO_4^-$ oxidizes iodide ($I^-$) completely to iodate ($IO_3^-$) while getting reduced to manganese dioxide ($MnO_2$).
The balanced ionic equation is:
$$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$$
### Pattern Recognition
Rule of thumb for $I^-$ oxidation by $KMnO_4$:
In acidic medium: $I^- \rightarrow I_2$
In alkaline/neutral medium: $I^- \rightarrow IO_3^-$
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q1758 (2024)
A and B formed in the following reactions are:
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- $\text{A = Na}_2\text{CrO}_4\text{, B = CrO}_5$
- $\text{A = Na}_2\text{Cr}_2\text{O}_4\text{, B = CrO}_4$
- $\text{A = Na}_2\text{Cr}_2\text{O}_7\text{, B = CrO}_3$
- $\text{A = Na}_2\text{Cr}_2\text{O}_7\text{, B = CrO}_5$
### Core Logic
Step 1: Chromyl chloride ($CrO_2Cl_2$) reacts with an alkali like $NaOH$ to give a yellow solution of sodium chromate ($Na_2CrO_4$).
$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O$$
Step 2: Sodium chromate ($Na_2CrO_4$) reacts with hydrogen peroxide ($H_2O_2$) in an acidic medium ($HCl$) to yield the deep blue colored chromium pentoxide ($CrO_5$, also known as chromium(VI) oxide peroxide).
$$Na_2CrO_4 + 2H_2O_2 + 2HCl \rightarrow CrO_5 (B) + 2NaCl + 3H_2O$$
Note: $NaCl$ formation implies the overall balanced reaction uses the acid for neutralization/salt formation.
### Pattern Recognition
Chromyl chloride test intermediate: Yellow solution = $Na_2CrO_4$.
Reaction of chromate with $H_2O_2$ in acid = Blue peroxide $CrO_5$ (butterfly structure).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
Class 11 Chemistry: Redox Reactions
Q70 (2024)
The orange colour of $K_2Cr_2O_7$ and purple colour of $KMnO_4$ is due to
- $\text{Charge transfer transition in both.}$
- $\text{d} \rightarrow \text{d transition in KMnO}_4 \text{ and charge transfer transitions in K}_2\text{Cr}_2\text{O}_7$
- $\text{d} \rightarrow \text{d transition in K}_2\text{Cr}_2\text{O}_7 \text{ and charge transfer transitions in KMnO}_4.$
- $\text{d} \rightarrow \text{d transition in both.}$
### Core Logic
In $K_2Cr_2O_7$, Chromium is in the $+6$ oxidation state, which means its electronic configuration is $d^0$. Since there are no d-electrons, $d-d$ transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.
Similarly, in $KMnO_4$, Manganese is in the $+7$ oxidation state, which also corresponds to a $d^0$ configuration. Again, no $d-d$ transitions are possible. The intense purple color is due to ligand-to-metal charge transfer (LMCT) from oxygen to manganese.
### Step 1: Final Conclusion
Both compounds owe their colors to charge transfer transitions.
### Pattern Recognition
Compounds of transition metals in their highest oxidation states (where they have $d^0$ configurations, like $Cr^{+6}$, $Mn^{+7}$, $V^{+5}$) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
Q71 (2024)
Alkaline oxidative fusion of $MnO_2$ gives "A" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are:
- $\text{Mn}_2\text{O}_7 \text{ and MnO}_4^-$
- $\text{MnO}_4^{2-} \text{ and MnO}_4^-$
- $\text{Mn}_2\text{O}_3 \text{ and MnO}_4^{2-}$
- $\text{MnO}_4^{2-} \text{ and Mn}_2\text{O}_7$
### Core Logic
Step 1: Alkaline oxidative fusion of $MnO_2$ (pyrolusite ore) with $KOH$ in the presence of $O_2$ (or an oxidizing agent like $KNO_3$) yields the green-colored manganate ion ($MnO_4^{2-}$).
$$2\mathrm{MnO}_2 + 4\mathrm{OH}^- + \mathrm{O}_2 \rightarrow 2\mathrm{MnO}_4^{2-} + 2\mathrm{H}_2\mathrm{O}$$
So, A is $\mathrm{MnO}_4^{2-}$.
Step 2: Electrolytic oxidation of the manganate ion ($MnO_4^{2-}$) in an alkaline medium converts it to the purple-colored permanganate ion ($MnO_4^-$).
$$\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^- + \mathrm{e}^-$$
So, B is $\mathrm{MnO}_4^-$.
### Pattern Recognition
Industrial preparation sequence of $KMnO_4$: $MnO_2 \xrightarrow{\text{fusion, } KOH, O_2} MnO_4^{2-} \text{ (green)} \xrightarrow{\text{electrolytic oxidation}} MnO_4^- \text{ (purple)}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
Q66 (2024)
Diamagnetic Lanthanoid ions are:
- $Nd^{3+}\text{ and }Eu^{3+}$
- $La^{3+}\text{ and }Ce^{4+}$
- $Nd^{3+}\text{ and }Ce^{4+}$
- $Lu^{3+}\text{ and }Eu^{3+}$
### Core Logic
An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons).
Let's write the electronic configuration for the elements in question.
### Step 1: Checking configurations
Cerium ($Ce$, Z=58): $[Xe] 4f^1 5d^1 6s^2$
$\rightarrow Ce^{4+}: [Xe] 4f^0$ (0 unpaired electrons $\rightarrow$ Diamagnetic)
Lanthanum ($La$, Z=57): $[Xe] 4f^0 5d^1 6s^2$
$\rightarrow La^{3+}: [Xe] 4f^0$ (0 unpaired electrons $\rightarrow$ Diamagnetic)
### Pattern Recognition
Ions with an empty f-subshell ($f^0$, e.g., $La^{3+}, Ce^{4+}$) or a completely filled f-subshell ($f^{14}$, e.g., $Lu^{3+}, Yb^{2+}$) are invariably diamagnetic.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
