Consider the following reactions: The number of terminal 'O' present in the compound 'C' is _____

Q: Consider the following reactions: The number of terminal 'O' present in the compound 'C' is ______

### Core Logic Let us identify the sequential chemical components via the chromyl chloride test pathway: 1. Reactant **A** represents a dichromate salt such as K2Cr2O7. Heating it with a metal chloride and concentrated sulfuric acid generates deep red chromyl chloride vapors (CrO2Cl2). 2. Passing these vapors into sodium hydroxide dissolves them, producing yellow sodium chromate compound **B** (Na2CrO4). 3. Acidifying the chromate solution dimerizes it into orange sodium dichromate compound **C*

Year	2026	2025	2024	Total
Questions	10	17	17	44

Q39 jee_main_2025_03_april_evening Oxidation States and Stability of Transition Metals

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 :

A. Statement I is false but Statement II is true
B. Statement I is true but Statement II is false
C. Both Statement I and Statement II are true
D. Both Statement I and Statement II are false

Solution

Related Formula

In transition metal groups:

Stability of higher oxidation states increases down the group:

Stability: Cr(VI) < Mo(VI) < W(VI)

Oxidizing power is inversely proportional to the stability of the high oxidation state.

Core Logic

Statement I Analysis:

Since $Cr(VI)$ $\mathrm{Cr(VI)}$ is less stable than $Mo(VI)$ $\mathrm{Mo(VI)}$, chromium is easily reduced from $+6$ $+6$ to $+3$ $+3$, making $CrO₃$ $\mathrm{CrO}_3$ a much stronger oxidizing agent than $MoO₃$ $\mathrm{MoO}_3$. Statement I is True.

Step 1: Analyze Statement II

Statement II asserts that $Cr(VI)$ $\mathrm{Cr(VI)}$ is more stable than $Mo(VI)$ $\mathrm{Mo(VI)}$. 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, $Mo(VI)$ $\mathrm{Mo(VI)}$ is more stable than $Cr(VI)$ $\mathrm{Cr(VI)}$. 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., $Mo(VI)$ $\mathrm{Mo(VI)}$ and $W(VI)$ $\mathrm{W(VI)}$ are very stable and non-oxidizing, whereas $Cr(VI)$ $\mathrm{Cr(VI)}$ 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.

Chapter Mix

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

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

Core Logic

Step 1: Structural Analysis of Dichromate

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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

Solution

Related Formula

Core Logic

Step 1: Analyze Statement II

Step 2: Conclusion

Pattern Recognition

Chapter Mix

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 $Co²⁺$ $\mathrm{Co}^{2+}$