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 3.
Q35 (2025)
Match List-I with List-II.
<div class="rankbit-table-container"><table class="rankbit-table"><thead><tr><th>List-I (Transition metal ion)</th><th>List-II (Spin only magnetic moment (B.M.))</th></tr></thead><tbody><tr><td>(A) $\mathrm{Ti}^{3+}$</td><td>(I) $3.87$</td></tr><tr><td>(B) $\mathrm{V}^{2+}$</td><td>(II) $0.00$</td></tr><tr><td>(C) $\mathrm{Ni}^{2+}$</td><td>(III) $1.73$</td></tr><tr><td>(D) $\mathrm{Sc}^{3+}$</td><td>(IV) $2.84$</td></tr></tbody></table></div>
Choose the correct answer from the options given below :
- \text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}
- \text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}
- \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
- \text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}
### Related Formula
$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$
where $n$ represents the number of unpaired electrons.
### Core Logic
Let's calculate the number of unpaired d-electrons ($n$) and the resulting spin-only magnetic moment for each transition metal ion:
* (A) $\mathrm{Ti}^{3+}$:
Electronic configuration $= [Ar] 3d^1 \rightarrow n = 1$
$$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ B.M.} \rightarrow \text{(III)}$$
* (B) $\mathrm{V}^{2+}$:
Electronic configuration $= [Ar] 3d^3 \rightarrow n = 3$
$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\text{ B.M.} \rightarrow \text{(I)}$$
* (C) $\mathrm{Ni}^{2+}$:
Electronic configuration $= [Ar] 3d^8$. The $3d$ subshell has 3 paired orbitals and 2 unpaired orbitals $\rightarrow n = 2$
$$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84\text{ B.M.} \rightarrow \text{(IV)}$$
* (D) $\mathrm{Sc}^{3+}$:
Electronic configuration $= [Ar] 3d^0 \rightarrow n = 0$
$$\mu = 0.00\text{ B.M.} \rightarrow \text{(II)}$$
Matching these values yields the sequence: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
### Pattern Recognition
Shortcut: The digit before the decimal point in a spin-only magnetic moment matches the number of unpaired electrons ($n$). For example, a value of $3.87\text{ B.M.}$ means there are exactly $3$ unpaired electrons.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q31 (2025)
Which of the following ions is the strongest oxidizing agent?
[Atomic Number of $Ce=58$, $Eu=63$, $Tb=65,$ $Lu=71$]
- $Lu^{3+}$
- $Eu^{2+}$
- $Tb^{4+}$
- $Ce^{3+}$
### Core Logic
The most common and chemically robust oxidation state for lanthanoid elements is $+3$. Consequently, ions existing in unstable $+4$ oxidation states exhibit a pronounced thermodynamic driving force to capture electrons and revert to the $+3$ form.
Among the options, $Tb^{4+}$ acts as a potent oxidizing agent due to this stability drive.
### Pattern Recognition
$Ln^{4+}$ forms naturally act as electron grabbers to sink back into the thermodynamic sweet spot of $+3$ states.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q37 (2025)
Preparation of potassium permanganate from $\mathrm{MnO}_2$ involves two step process in which the $1^{\text{st}}$ step is a reaction with KOH and $\mathrm{KNO}_3$ to produce
- $\mathrm{K}_{4}[\mathrm{Mn}(\mathrm{OH})_{6}]$
- $\mathrm{K}_3\mathrm{MnO}_4$
- $\mathrm{KMnO}_4$
- $\mathrm{K}_2\mathrm{MnO}_4$
### Related Formula
$$2MnO_2 + 4KOH + O_2 \xrightarrow{KNO_3} 2K_2MnO_4 + 2H_2O$$
### Core Logic
The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore ($MnO_2$). Fusing the solid reactant directly along an alkaline base payload ($KOH$) combined explicitly with an oxidizing carrier ($KNO_3$) yields the intermediate green product, **potassium manganate** ($K_2MnO_4$).
### Pattern Recognition
Step 1 yields the $+6$ green compound ($K_2MnO_4$); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple $+7$ agent ($KMnO_4$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q27 (2025)
The amphoteric oxide among $V_{2}O_{3}$, $V_{2}O_{4}$ and $V_{2}O_{5}$ upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :
- $+3$
- $+7$
- $+5$
- $+4$
### Related Formula
Oxidation state equation for an oxoanion $VO_4^{3-}$:
$x + 4(-2) = -3$
### Core Logic
Among the given oxides of Vanadium:
- $V_2O_3$ is basic.
- $V_2O_4$ is less basic / amphoteric.
- $V_2O_5$ is predominantly amphoteric (reacts with both acids and alkalies).
When $V_2O_5$ reacts with an alkali, it forms the orthovanadate ion ($VO_4^{3-}$).
### Step 1: Finding the Oxidation State
In $VO_4^{3-}$ ion:
$$x - 8 = -3 \implies x = +5$$
Thus, the oxidation state of Vanadium in the resulting oxide anion is $+5$.
### Pattern Recognition
As the oxidation state of a transition metal increases, its oxide shifts from basic to amphoteric to acidic. $V_2O_5$ has the highest oxidation state ($+5$) here and dissolves in alkali to retain its $+5$ oxidation state in $VO_4^{3-}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Q46 (2025)
The spin only magnetic moment $(\mu)$ value (B.M.) of the compound with strongest oxidising power among $Mn_{2}O_{3}$, TiO and VO is ______ B.M. (Nearest integer).
### Related Formula
Spin-only magnetic moment expression:
$$\mu = \sqrt{n(n+2)}\mathrm{\ B.M.}$$
### Core Logic
Evaluating the oxidation states and stability profiles:
- In $TiO$: $Ti^{2+}$
- In $VO$: $V^{2+}$
- In $Mn_2O_3$: $Mn^{3+}$
$Mn^{3+}$ possesses a very high reduction potential ($E^\circ_{Mn^{3+}/Mn^{2+}} = +1.57\mathrm{\ V}$), making it an exceptionally strong oxidizing agent because it easily gains an electron to form stable $Mn^{2+}$ ($d^5$ configuration).
### Step 1: Calculate the Magnetic Moment of Mn(III)
Electronic configuration of $Mn^{3+}$:
$$Mn^{3+} = [Ar]3d^4 \implies n = 4\text{ unpaired electrons}$$
Calculating the spin-only magnetic moment:
$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.89\mathrm{\ B.M.}$$
### Step 2: Rounding to Nearest Integer
Rounding $4.89\mathrm{\ B.M.}$ to the nearest integer gives $5$.
### Pattern Recognition
High reduction potentials are strongly tied to manganese in its $+3$ oxidation state. To quickly estimate magnetic moments, remember that a system with $n$ unpaired electrons always results in a value of '$n.\text{something}$' B.M. Thus, $4$ unpaired electrons $\rightarrow 4.89\mathrm{\ B.M.}$, which rounds up to $5$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
