Given below are two statements: Statement I: CrO3 is a stronger oxidizing agent than MoO3. Statement II: Cr(VI) is more stable than Mo(VI). In the light of the above statements, choose the correct answer from the options given below :

In transition metal groups: - Stability of higher oxidation states increases down the group: - Oxidizing power is inversely proportional to the stability of the high oxidation state. Statement I Analysis: - Since Cr(VI) is less stable than Mo(VI), chromium is easily reduced from +6 to +3, making CrO3 a much stronger oxidizing agent than MoO3. Statement I is True. - Statement II asserts that Cr(VI) is more stable than 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, Mo(VI) is more stable than Cr(VI). Statement II is False. Therefore, Statement I is True but Statement II is False, matching Option (2). For d-block elements, higher oxidation states are more stable down the group (e.g., Mo(VI) and W(VI) are very stable and non-oxidizing, whereas 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. Class 12 Chemistry: The d-and f-Block Elements

Given below are two statements: Statement I: CrO3 is a stronger oxidizing agent than MoO3. Statement II: Cr(VI) is more stable than Mo(VI). In the light of the above statements, choose the correct ans

In transition metal groups: - Stability of higher oxidation states increases down the group: - Oxidizing power is inversely proportional to the stability of the high oxidation state. Statement I Analysis: - Since Cr(VI) is less stable than Mo(VI), chromium is easily reduced from +6 to +3, making CrO3 a much stronger oxidizing agent than MoO3. Statement I is True. - Statement II asserts that Cr(VI) is more stable than Mo(VI). As we go down a transition metal group, the higher oxidation states bec

Given below are two statements: Statement I: CrO3 is a stronger oxidizing agent than MoO3. Statement II: Cr(VI) is more stable than Mo(VI). In the light of the above statements, choose the correct answer from the options given below :

In transition metal groups: - Stability of higher oxidation states increases down the group: - Oxidizing power is inversely proportional to the stability of the high oxidation state. Statement I Analysis: - Since Cr(VI) is less stable than Mo(VI), chromium is easily reduced from +6 to +3, making CrO3 a much stronger oxidizing agent than MoO3. Statement I is True. - Statement II asserts that Cr(VI) is more stable than 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, Mo(VI) is more stable than Cr(VI). Statement II is False. Therefore, Statement I is True but Statement II is False, matching Option (2). For d-block elements, higher oxidation states are more stable down the group (e.g., Mo(VI) and W(VI) are very stable and non-oxidizing, whereas 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. Class 12 Chemistry: The d-and f-Block Elements

Given below are two statements: Statement I: CrO3 is a stronger oxidizing agent than MoO3. Statement II: Cr(VI) is more stable than Mo(VI). In the light of the above statements, choose the correct ans

In transition metal groups: - Stability of higher oxidation states increases down the group: - Oxidizing power is inversely proportional to the stability of the high oxidation state. Statement I Analysis: - Since Cr(VI) is less stable than Mo(VI), chromium is easily reduced from +6 to +3, making CrO3 a much stronger oxidizing agent than MoO3. Statement I is True. - Statement II asserts that Cr(VI) is more stable than Mo(VI). As we go down a transition metal group, the higher oxidation states bec

Given below are two statements: Statement I: CrO3 is a stronger oxidizing agent than MoO3. Statement II: Cr(VI) is more…

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More The d-and f-Block Elements Previous-Year Questions — Page 3

Q49 jee_main_2025_03_april_evening Enthalpy of Atomisation and Magnetic Properties

Among,

Sc

Mn

Co

and

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).

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Spin-only magnetic moment ( $μ$ $\mu$) is given by:

μ = \sqrt(n(n+2))~BM

where $n$ $n$ is the number of unpaired d-electrons.

Core Logic

Enthalpies of atomization of the given 3d transition elements (in $kJ/mol$ $\mathrm{kJ/mol}$):

Scandium ( $Sc$ $\mathrm{Sc}$): $326$ $326$
Manganese ( $Mn$ $\mathrm{Mn}$): $281$ $281$
Cobalt ( $Co$ $\mathrm{Co}$): $425$ $425$
Copper ( $Cu$ $\mathrm{Cu}$): $339$ $339$

Thus, Cobalt ( $Co$ $\mathrm{Co}$) has the highest enthalpy of atomization.

Step 1: Determine unpaired electrons in $Co²⁺$ $\mathrm{Co}^{2+}$

Electronic configuration of Cobalt ( $Z=27$ $Z=27$):

Co: [Ar] 3d⁷ 4s²

For divalent Cobalt ion ( $Co²⁺$ $\mathrm{Co}^{2+}$):

Co²⁺: [Ar] 3d⁷

In the d-subshell (five orbitals):

Three orbitals are paired, and three are unpaired ( $n=3$ $n=3$).

Step 2: Calculate spin-only magnetic moment

μ = \sqrt(3(3+2)) = \sqrt(15) \approx 3.87~BM

Rounding to the nearest integer gives $4$ $4$.

Pattern Recognition

Enthalpy of atomization generally peaks near the middle of transition series due to maximum metallic bonding. However, $Mn$ $\mathrm{Mn}$ ( $3d⁵ 4s²$ $3d^5 4s^2$) is an anomaly with an exceptionally low value ( $281~kJ/mol$ $281\mathrm{~kJ/mol}$) due to its highly stable half-filled $d⁵$ $d^5$ subshell configuration which reduces electron delocalization in metallic bonding.

Chapter Mix

Class 12 Chemistry: The d-and f-Block Elements

Q36 jee_main_2025_07_april_morning Enthalpy of Atomisation

The number of valence electrons present in the metal among

Cr

Co

Fe

and

Ni

which has the lowest enthalpy of atomisation is:

A. 8
B. 9
C. 6
D. 10

Solution

Core Logic

Let's look at the enthalpy of atomisation values for the given 3d transition metals:

Chromium ( $Cr$ $\mathrm{Cr}$): $397 kJ mol⁻¹$ $397 \text{ kJ mol}^{-1}$
Iron ( $Fe$ $\mathrm{Fe}$): $416 kJ mol⁻¹$ $416 \text{ kJ mol}^{-1}$
Cobalt ( $Co$ $\mathrm{Co}$): $425 kJ mol⁻¹$ $425 \text{ kJ mol}^{-1}$
Nickel ( $Ni$ $\mathrm{Ni}$): $430 kJ mol⁻¹$ $430 \text{ kJ mol}^{-1}$

Among the choices, Chromium ( $Cr$ $\mathrm{Cr}$) has the lowest enthalpy of atomisation ( $397 kJ mol⁻¹$ $397 \text{ kJ mol}^{-1}$), 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 $Cr$ $\mathrm{Cr}$ is:

Cr = [Ar] 3d⁵ 4s¹

Total valence electrons $= 5 + 1 = 6$ $= 5 + 1 = 6$.

Pattern Recognition

In transition metals, manganese ( $Mn$ $Mn$) has the absolute lowest enthalpy of atomisation in the 3d series because of its completely half-filled $d⁵$ $d^5$ and completely filled $s²$ $s^2$ stability. Since $Mn$ $Mn$ is not in the list, Chromium ( $"Cr"$ $"Cr"$) is next, having $6$ $6$ valence electrons.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Q39 jee_main_2025_07_april_morning Properties of Oxides

The first transition series metal 'M' has the highest enthalpy of atomisation in its series. One of its aquated ions

(Mⁿ⁺)

exists in green colour. The nature of the oxide formed by the above M ion is:

A. $neutral$ $\text{neutral}$
B. $acidic$ $\text{acidic}$
C. $basic$ $\text{basic}$
D. $amphoteric$ $\text{amphoteric}$

Solution

Core Logic

In the $3d$ $3\mathrm{d}$ transition series, Vanadium ( $V$ $\mathrm{V}$) has the highest enthalpy of atomisation ( $515 kJ mol⁻¹$ $515 \text{ kJ mol}^{-1}$).
One of its aquated ions, $V³⁺(aq)$ $\mathrm{V}^{3+}\mathrm{(aq)}$ [specifically $[V(H₂O)₆]³⁺$ $[\mathrm{V(H_2O)_6}]^{3+}$], has a characteristic green colour.
The corresponding oxide for this state is $V₂O₃$ $\mathrm{V}_2\mathrm{O}_3$ (Vanadium(III) oxide).
Metal oxides in lower oxidation states ( $+2, +3$ $+2, +3$) are typically basic in nature, while intermediate states like $V₂O₄$ $\mathrm{V}_2\mathrm{O}_4$ are amphoteric, and high states like $V₂O₅$ $\mathrm{V}_2\mathrm{O}_5$ are acidic. Therefore, $V₂O₃$ $\mathrm{V}_2\mathrm{O}_3$ is purely a basic oxide.

Pattern Recognition

Vanadium ( $V$ $V$) stands out with high atomisation enthalpy and characteristic oxidation states. Lower oxides of transition metals are always basic, higher oxidation state oxides are acidic.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Q jee_main_2025_08_april_evening Magnetic Properties

The correct decreasing order of spin-only magnetic moment values (BM) of

Cu^+

Cu²⁺

Cr²⁺

, and

Cr³⁺

ions is:

A. $Cu^+ > Cu²⁺ > Cr³⁺ > Cr²⁺$
B. $Cu²⁺ > Cu^+ > Cr²⁺ > Cr³⁺$
C. $Cr²⁺ > Cr³⁺ > Cu²⁺ > Cu^+$
D. $Cr³⁺ > Cr²⁺ > Cu^+ > Cu²⁺$

Solution

Related Formula

Spin-only magnetic moment equation:

μ = \sqrt(n(n+2)) BM

where $n$ $n$ is the exact count of unpaired d-shell electrons.

Execution

Let us compute the unpaired electron distribution for each transition metal ion:

$Cu^+$ $\text{Cu}^+$: Electronic configuration is $[Ar]3d¹⁰$ $[\text{Ar}]3d^{10}$. All electrons are paired up.

n = 0 μ = 0 BM

$Cu²⁺$ $\text{Cu}^{2+}$: Electronic configuration is $[Ar]3d⁹$ $[\text{Ar}]3d^9$. Has one unpaired hole.

n = 1 μ = \sqrt(1(1+2)) = \sqrt(3) \approx 1.73 BM

$Cr³⁺$ $\text{Cr}^{3+}$: Electronic configuration is $[Ar]3d³$ $[\text{Ar}]3d^3$. Has three unpaired parallel spins.

n = 3 μ = \sqrt(3(3+2)) = \sqrt(15) \approx 3.87 BM

$Cr²⁺$ $\text{Cr}^{2+}$: Electronic configuration is $[Ar]3d⁴$ $[\text{Ar}]3d^4$. Has four unpaired spins.

n = 4 μ = \sqrt(4(4+2)) = \sqrt(24) \approx 4.90 BM

Arranging these values in decreasing structural order:

μ(Cr²⁺) > μ(Cr³⁺) > μ(Cu²⁺) > μ(Cu^+)

Pattern Recognition

The value of the spin-only magnetic moment scales monotonically with the number of unpaired electrons ( $n$ $n$). More unpaired electrons directly translate to a higher magnetic moment, bypassing any tedious square-root calculations during testing.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Solution & Explanation

Related Formula

Core Logic

Step 1: Analyze Statement II

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More The d-and f-Block Elements Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Determine unpaired electrons in Co²⁺$\mathrm{Co}^{2+}$

Step 2: Calculate spin-only magnetic moment

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

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Step 1: Determine unpaired electrons in $Co²⁺$ $\mathrm{Co}^{2+}$