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 5.
Q75 (2024)
Yellow compound of lead chromate gets dissolved on treatment with hot $\text{NaOH}$ solution. The product of lead formed is a :
- Tetraanionic complex with coordination number six
- Neutral complex with coordination number four
- Dianionic complex with coordination number six
- Dianionic complex with coordination number four
### Related Formula
Dissolution reaction pathway:
$$\text{PbCrO}_4 + 4\text{NaOH (hot excess)} \rightarrow \text{Na}_2[\text{Pb(OH)}_4] + \text{Na}_2\text{CrO}_4$$
### Core Logic
The reaction yields sodium tetrahydroxoplumbate(II), $[\text{Pb(OH)}_4]^{2-}$. The charge of the complex species is $-2$ (dianionic), and it binds 4 hydroxo coordination ligands, matching a coordination number of four.
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Class 12 Chemistry: Coordination Compounds
Q78 (2024)
$\text{NaCl}$ reacts with conc. $H_2SO_4$ and $K_2Cr_2O_7$ to give reddish fumes (B), which react with $\text{NaOH}$ to give yellow solution (C). (B) and (C) respectively are;
- $CrO_2Cl_2$, $Na_2CrO_4$
- $Na_2CrO_4$, $CrO_2Cl_2$
- $CrO_2Cl_2$, $KHSO_4$
- $CrO_2Cl_2$, $Na_2Cr_2O_7$
### Step 1: Production of Reddish Fumes
$$4\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{H}_2\text{SO}_4 \rightarrow 2\text{CrO}_2\text{Cl}_2\uparrow + 2\text{KHSO}_4 + 4\text{NaHSO}_4 + 3\text{H}_2\text{O}$$
Reddish brown vapors (B) are chromyl chloride ($CrO_2Cl_2$).
### Step 2: Conversion to Yellow Solution
$$\text{CrO}_2\text{Cl}_2 + 4\text{NaOH} \rightarrow \text{Na}_2\text{CrO}_4 + 2\text{NaCl} + 2\text{H}_2\text{O}$$
Yellow solution (C) corresponds to sodium chromate ($Na_2CrO_4$).
### Pattern Recognition
Chloride detection signature: $\text{Cl}^- \rightarrow \text{CrO}_2\text{Cl}_2\text{ (red-brown)} \rightarrow \text{Na}_2\text{CrO}_4\text{ (yellow chromate)}.$
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Q80 (2024)
The electronic configuration for Neodymium is:
[Atomic Number for Neodymium 60]
- $\text{[Xe]} 4f^4 6s^2$
- $\text{[Xe]} 5f^4 7s^2$
- $\text{[Xe]} 4f^6 6s^2$
- $\text{[Xe]} 4f^1 5d^1 6s^2$
### Core Logic
The noble gas configuration of Xenon ($Z=54$) provides the primary core layout. For Neodymium ($Z=60$), the 6 remaining valence electrons distribute into the inner $4\text{f}$ orbital subshell rather than filling the $5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of $\text{[Xe]} 4\text{f}^4 6\text{s}^2$.
### Pattern Recognition
Lanthanide filling sequences generally bypass $5d$ progression except for specific exceptions (La, Gd, Lu).
### Chapter Mix
Class 12 Chemistry: d-and f-Block Elements
Q1664 (2024)
$KMnO_4$ decomposes on heating at $513\mathrm{K}$ to form $O_2$ along with
- $\text{MnO}_2 \text{ \& } \text{K}_2\text{O}_2$
- $\text{K}_2\text{MnO}_4 \text{ \& } \text{Mn}$
- $\text{Mn} \text{ \& } \text{KO}_2$
- $\text{K}_2\text{MnO}_4 \text{ \& } \text{MnO}_2$
### Core Logic
Potassium permanganate ($KMnO_4$) is a strong oxidizing agent. When heated to $513\mathrm{K}$, it undergoes thermal decomposition to give potassium manganate ($K_2MnO_4$), manganese dioxide ($MnO_2$), and oxygen gas ($O_2$).
The balanced chemical equation is:
$$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$$
### Step 1: Final Identification
The products formed along with $O_2$ are $K_2MnO_4$ (green) and $MnO_2$ (black).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Q63 (2024)
In chromyl chloride test for confirmation of $Cl^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and $10\%$ $H_2O_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is
- $+6$
- $+5$
- $+10$
- $+3$
### Core Logic
The reaction sequence for the chromyl chloride test is:
$$Cl^- + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2$$
The chromyl chloride gas is then passed through a basic medium (like $NaOH$) to form a yellow solution of chromate ions:
$$CrO_2Cl_2 \xrightarrow{\text{Basic medium}} CrO_4^{2-} + Cl^-$$
Acidification of the yellow $CrO_4^{2-}$ solution followed by the addition of $H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide ($CrO_5$).
$$CrO_4^{2-} \xrightarrow[\text{yellow solution},
{1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$
### Step 1: Oxidation State Calculation
{{SOL_IMG1}}
The structure of chromium pentoxide ($CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen ($O^{2-}$) and four peroxide oxygens ($O_2^{2-}$). Therefore, there are 2 peroxo linkages.
Let the oxidation state of Chromium be $x$.
$$x + 1(-2) + 4(-1) = 0$$
$x - 2 - 4 = 0$
$x = +6$
Thus, the oxidation state of Cr in $CrO_5$ is $+6$.
### Pattern Recognition
A classic oxidation state trap. Calculating simply via formula $CrO_5$ yields $x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
