d-and f-Block Elements Previous Year Questions — JEE Main Chemistry
44 past-year d-and f-Block Elements questions from JEE Main (Chemistry).
Q39 (2025)
Given below are two statements:
Statement I: $\mathrm{CrO}_3$ is a stronger oxidizing agent than $\mathrm{MoO}_3$.
Statement II: $\mathrm{Cr(VI)}$ is more stable than $\mathrm{Mo(VI)}$.
In the light of the above statements, choose the correct answer from the options given below :
- Statement I is false but Statement II is true
- Statement I is true but Statement II is false
- Both Statement I and Statement II are true
- Both Statement I and Statement II are false
### Related Formula
In transition metal groups:
- Stability of higher oxidation states increases down the group:
$$\text{Stability: } \mathrm{Cr(VI)} < \mathrm{Mo(VI)} < \mathrm{W(VI)}$$
- Oxidizing power is inversely proportional to the stability of the high oxidation state.
### Core Logic
Statement I Analysis:
- Since $\mathrm{Cr(VI)}$ is less stable than $\mathrm{Mo(VI)}$, chromium is easily reduced from $+6$ to $+3$, making $\mathrm{CrO}_3$ a much stronger oxidizing agent than $\mathrm{MoO}_3$. Statement I is True.
### Step 1: Analyze Statement II
- Statement II asserts that $\mathrm{Cr(VI)}$ is more stable than $\mathrm{Mo(VI)}$. As we go down a transition metal group, the higher oxidation states become increasingly stable due to better shielding of the core electrons and relativistic effects. Hence, $\mathrm{Mo(VI)}$ is more stable than $\mathrm{Cr(VI)}$. Statement II is False.
### Step 2: Conclusion
Therefore, Statement I is True but Statement II is False, matching Option (2).
### Pattern Recognition
For d-block elements, higher oxidation states are more stable down the group (e.g., $\mathrm{Mo(VI)}$ and $\mathrm{W(VI)}$ are very stable and non-oxidizing, whereas $\mathrm{Cr(VI)}$ is unstable and strongly oxidizing). This is the exact opposite of p-block elements where the inert pair effect makes lower oxidation states more stable down the group.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q49 (2025)
Among, $\mathrm{Sc}$, $\mathrm{Mn}$, $\mathrm{Co}$ and $\mathrm{Cu}$, identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ________ BM (in nearest integer).
### Related Formula
Spin-only magnetic moment ($\mu$) is given by:
$$\mu = \sqrt{n(n+2)}\mathrm{~BM}$$
where $n$ is the number of unpaired d-electrons.
### Core Logic
Enthalpies of atomization of the given 3d transition elements (in $\mathrm{kJ/mol}$):
- Scandium ($\mathrm{Sc}$): $326$
- Manganese ($\mathrm{Mn}$): $281$
- Cobalt ($\mathrm{Co}$): $425$
- Copper ($\mathrm{Cu}$): $339$
Thus, Cobalt ($\mathrm{Co}$) has the highest enthalpy of atomization.
### Step 1: Determine unpaired electrons in $\mathrm{Co}^{2+}$
Electronic configuration of Cobalt ($Z=27$):
$$\mathrm{Co}: [\mathrm{Ar}] 3d^7 4s^2$$
For divalent Cobalt ion ($\mathrm{Co}^{2+}$):
$$\mathrm{Co}^{2+}: [\mathrm{Ar}] 3d^7$$
In the d-subshell (five orbitals):
- Three orbitals are paired, and three are unpaired ($n=3$).
### Step 2: Calculate spin-only magnetic moment
$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87\mathrm{~BM}$$
Rounding to the nearest integer gives $4$.
### Pattern Recognition
Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, $\mathrm{Mn}$ ($3d^5 4s^2$) is an anomaly with an exceptionally low value ($281\mathrm{~kJ/mol}$) due to its highly stable half-filled $d^5$ subshell configuration which reduces electron delocalization in metallic bonding.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q36 (2025)
The number of valence electrons present in the metal among $\mathrm{Cr}$, $\mathrm{Co}$, $\mathrm{Fe}$ and $\mathrm{Ni}$ which has the lowest enthalpy of atomisation is:
- 8
- 9
- 6
- 10
### Core Logic
Let's look at the enthalpy of atomisation values for the given 3d transition metals:
- **Chromium ($\mathrm{Cr}$)**: $397 \text{ kJ mol}^{-1}$
- **Iron ($\mathrm{Fe}$)**: $416 \text{ kJ mol}^{-1}$
- **Cobalt ($\mathrm{Co}$)**: $425 \text{ kJ mol}^{-1}$
- **Nickel ($\mathrm{Ni}$)**: $430 \text{ kJ mol}^{-1}$
Among the choices, **Chromium ($\mathrm{Cr}$)** has the lowest enthalpy of atomisation ($397 \text{ kJ mol}^{-1}$), due to a highly stable half-filled d-subshell configuration which leads to weaker metallic bonding relative to the other metals listed.
The valence electronic configuration of $\mathrm{Cr}$ is:
$$\mathrm{Cr} = [\mathrm{Ar}] 3\mathrm{d}^5 4\mathrm{s}^1$$
Total valence electrons $= 5 + 1 = 6$.
### Pattern Recognition
In transition metals, manganese ($Mn$) has the absolute lowest enthalpy of atomisation in the 3d series because of its completely half-filled $d^5$ and completely filled $s^2$ stability. Since $Mn$ is not in the list, Chromium ($"Cr"$) is next, having $6$ valence electrons.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Q39 (2025)
The first transition series metal 'M' has the highest enthalpy of atomisation in its series. One of its aquated ions $(\mathbf{M}^{n+})$ exists in green colour. The nature of the oxide formed by the above M ion is:
- $\text{neutral}$
- $\text{acidic}$
- $\text{basic}$
- $\text{amphoteric}$
### Core Logic
1. In the $3\mathrm{d}$ transition series, **Vanadium ($\mathrm{V}$)** has the highest enthalpy of atomisation ($515 \text{ kJ mol}^{-1}$).
2. One of its aquated ions, $\mathrm{V}^{3+}\mathrm{(aq)}$ [specifically $[\mathrm{V(H_2O)_6}]^{3+}$], has a characteristic **green colour**.
3. The corresponding oxide for this state is $\mathrm{V}_2\mathrm{O}_3$ (Vanadium(III) oxide).
4. Metal oxides in lower oxidation states ($+2, +3$) are typically **basic** in nature, while intermediate states like $\mathrm{V}_2\mathrm{O}_4$ are amphoteric, and high states like $\mathrm{V}_2\mathrm{O}_5$ are acidic. Therefore, $\mathrm{V}_2\mathrm{O}_3$ is purely a basic oxide.
### Pattern Recognition
Vanadium ($V$) stands out with high atomisation enthalpy and characteristic oxidation states. Lower oxides of transition metals are always basic, higher oxidation state oxides are acidic.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Q586 (2025)
The correct decreasing order of spin-only magnetic moment values (BM) of $\text{Cu}^+$, $\text{Cu}^{2+}$, $\text{Cr}^{2+}$, and $\text{Cr}^{3+}$ ions is:
- $\text{Cu}^+ > \text{Cu}^{2+} > \text{Cr}^{3+} > \text{Cr}^{2+}$
- $\text{Cu}^{2+} > \text{Cu}^+ > \text{Cr}^{2+} > \text{Cr}^{3+}$
- $\text{Cr}^{2+} > \text{Cr}^{3+} > \text{Cu}^{2+} > \text{Cu}^+$
- $\text{Cr}^{3+} > \text{Cr}^{2+} > \text{Cu}^+ > \text{Cu}^{2+}$
### Related Formula
Spin-only magnetic moment equation:
$$\mu = \sqrt{n(n+2)} \quad \text{BM}$$
where $n$ is the exact count of unpaired d-shell electrons.
### Execution
Let us compute the unpaired electron distribution for each transition metal ion:
1. **$\text{Cu}^+$**: Electronic configuration is $[\text{Ar}]3d^{10}$. All electrons are paired up.
$$n = 0 \implies \mu = 0 \text{ BM}$$
2. **$\text{Cu}^{2+}$**: Electronic configuration is $[\text{Ar}]3d^9$. Has one unpaired hole.
$$n = 1 \implies \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$$
3. **$\text{Cr}^{3+}$**: Electronic configuration is $[\text{Ar}]3d^3$. Has three unpaired parallel spins.
$$n = 3 \implies \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$$
4. **$\text{Cr}^{2+}$**: Electronic configuration is $[\text{Ar}]3d^4$. Has four unpaired spins.
$$n = 4 \implies \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$$
Arranging these values in decreasing structural order:
$$\mu(\text{Cr}^{2+}) > \mu(\text{Cr}^{3+}) > \mu(\text{Cu}^{2+}) > \mu(\text{Cu}^+)$$
### Pattern Recognition
The value of the spin-only magnetic moment scales monotonically with the number of unpaired electrons ($n$). More unpaired electrons directly translate to a higher magnetic moment, bypassing any tedious square-root calculations during testing.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
