The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of…

Q: The number of valence electrons present in the metal among Cr, Co, Fe and Ni which has the lowest enthalpy of atomisation is:

Let's look at the enthalpy of atomisation values for the given 3d transition metals: - **Chromium (Cr)**: 397 kJ mol-1 - **Iron (Fe)**: 416 kJ mol-1 - **Cobalt (Co)**: 425 kJ mol-1 - **Nickel (Ni)**: 430 kJ mol-1 Among the choices, **Chromium (Cr)** has the lowest enthalpy of atomisation (397 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 is: Total valenc

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Q35 jee_main_2025_24_jan_evening Magnetic Properties of Transition Metals

Match List-I with List-II.

List-I (Transition metal ion)	List-II (Spin only magnetic moment (B.M.))
(A) $Ti³⁺$ $\mathrm{Ti}^{3+}$	(I) $3.87$ $3.87$
(B) $V²⁺$ $\mathrm{V}^{2+}$	(II) $0.00$ $0.00$
(C) $Ni²⁺$ $\mathrm{Ni}^{2+}$	(III) $1.73$ $1.73$
(D) $Sc³⁺$ $\mathrm{Sc}^{3+}$	(IV) $2.84$ $2.84$

Choose the correct answer from the options given below :

A. \text{(A)-(III), (B)-(I), (C)-(II), (D)-(IV)}
B. \text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}
C. \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
D. \text{(A)-(II), (B)-(IV), (C)-(I), (D)-(III)}

Solution

Related Formula

μ = \sqrt(n(n+2)) B.M.

where $n$ $n$ represents the number of unpaired electrons.

Core Logic

Let's calculate the number of unpaired d-electrons ( $n$ $n$) and the resulting spin-only magnetic moment for each transition metal ion:

(A) $Ti³⁺$ $\mathrm{Ti}^{3+}$:

Electronic configuration $= [Ar] 3d¹ arrow n = 1$ $= [Ar] 3d^1 \rightarrow n = 1$

μ = \sqrt(1(1+2)) = \sqrt(3) \approx 1.73 B.M. arrow (III)

(B) $V²⁺$ $\mathrm{V}^{2+}$:

Electronic configuration $= [Ar] 3d³ arrow n = 3$ $= [Ar] 3d^3 \rightarrow n = 3$

μ = \sqrt(3(3+2)) = \sqrt(15) \approx 3.87 B.M. arrow (I)

(C) $Ni²⁺$ $\mathrm{Ni}^{2+}$:

Electronic configuration $= [Ar] 3d⁸$ $= [Ar] 3d^8$. The $3d$ $3d$ subshell has 3 paired orbitals and 2 unpaired orbitals $arrow n = 2$ $\rightarrow n = 2$

μ = \sqrt(2(2+2)) = \sqrt(8) \approx 2.84 B.M. arrow (IV)

(D) $Sc³⁺$ $\mathrm{Sc}^{3+}$:

Electronic configuration $= [Ar] 3d⁰ arrow n = 0$ $= [Ar] 3d^0 \rightarrow n = 0$

μ = 0.00 B.M. arrow (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 ( $n$ $n$). For example, a value of $3.87 B.M.$ $3.87\text{ B.M.}$ means there are exactly $3$ $3$ unpaired electrons.

Chapter Mix

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

Q31 jee_main_2025_24_jan_morning Lanthanoids Oxidation States

Which of the following ions is the strongest oxidizing agent? [Atomic Number of

Ce=58

,

Eu=63

,

Tb=65,

Lu=71

]

A. $Lu³⁺$ $Lu^{3+}$
B. $Eu²⁺$ $Eu^{2+}$
C. $Tb⁴⁺$ $Tb^{4+}$
D. $Ce³⁺$ $Ce^{3+}$

Solution

Core Logic

The most common and chemically robust oxidation state for lanthanoid elements is $+3$ $+3$. Consequently, ions existing in unstable $+4$ $+4$ oxidation states exhibit a pronounced thermodynamic driving force to capture electrons and revert to the $+3$ $+3$ form.

Among the options, $Tb⁴⁺$ $Tb^{4+}$ acts as a potent oxidizing agent due to this stability drive.

Pattern Recognition

$Ln⁴⁺$ $Ln^{4+}$ forms naturally act as electron grabbers to sink back into the thermodynamic sweet spot of $+3$ $+3$ states.

Chapter Mix

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

Q37 jee_main_2025_24_jan_morning Preparation and Properties of Potassium Permanganate

Preparation of potassium permanganate from

MnO₂

involves two step process in which the

1st

step is a reaction with KOH and

KNO₃

to produce

A. $K₄[Mn(OH)₆]$
B. $K₃MnO₄$ $\mathrm{K}_3\mathrm{MnO}_4$
C. $KMnO₄$ $\mathrm{KMnO}_4$
D. $K₂MnO₄$ $\mathrm{K}_2\mathrm{MnO}_4$

Solution

Related Formula

2MnO₂ + 4KOH + O₂ KNO₃ 2K₂MnO₄ + 2H₂O

Core Logic

The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore ( $MnO₂$ $MnO_2$). Fusing the solid reactant directly along an alkaline base payload ( $KOH$ $KOH$) combined explicitly with an oxidizing carrier ( $KNO₃$ $KNO_3$) yields the intermediate green product, potassium manganate ( $K₂MnO₄$ $K_2MnO_4$).

Pattern Recognition

Step 1 yields the $+6$ $+6$ green compound ( $K₂MnO₄$ $K_2MnO_4$); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple $+7$ $+7$ agent ( $KMnO₄$ $KMnO_4$).

Chapter Mix

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

Q27 jee_main_2025_28_jan_evening Oxides and Oxoanions of Transition Metals

The amphoteric oxide among

V₂O₃

,

V₂O₄

and

V₂O₅

upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

A. $+3$ $+3$
B. $+7$ $+7$
C. $+5$ $+5$
D. $+4$ $+4$

Solution

Related Formula

Oxidation state equation for an oxoanion $VO₄³⁻$ $VO_4^{3-}$:

$x + 4(-2) = -3$ $x + 4(-2) = -3$

Core Logic

Among the given oxides of Vanadium:

$V₂O₃$ $V_2O_3$ is basic.
$V₂O₄$ $V_2O_4$ is less basic / amphoteric.
$V₂O₅$ $V_2O_5$ is predominantly amphoteric (reacts with both acids and alkalies).

When $V₂O₅$ $V_2O_5$ reacts with an alkali, it forms the orthovanadate ion ( $VO₄³⁻$ $VO_4^{3-}$).

Step 1: Finding the Oxidation State

In $VO₄³⁻$ $VO_4^{3-}$ ion:

x - 8 = -3 x = +5

Thus, the oxidation state of Vanadium in the resulting oxide anion is $+5$ $+5$.

Pattern Recognition

As the oxidation state of a transition metal increases, its oxide shifts from basic to amphoteric to acidic. $V₂O₅$ $V_2O_5$ has the highest oxidation state ( $+5$ $+5$) here and dissolves in alkali to retain its $+5$ $+5$ oxidation state in $VO₄³⁻$ $VO_4^{3-}$.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Q46 jee_main_2025_28_jan_evening Magnetic Properties and Oxidation States

The spin only magnetic moment

(μ)

value (B.M.) of the compound with strongest oxidising power among

Mn₂O₃

, TiO and VO is ______ B.M. (Nearest integer).

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

Spin-only magnetic moment expression:

μ = \sqrt(n(n+2)) B.M.

Core Logic

Evaluating the oxidation states and stability profiles:

In $TiO$ $TiO$: $Ti²⁺$ $Ti^{2+}$
In $VO$ $VO$: $V²⁺$ $V^{2+}$
In $Mn₂O₃$ $Mn_2O_3$: $Mn³⁺$ $Mn^{3+}$

$Mn³⁺$ $Mn^{3+}$ possesses a very high reduction potential ( $E^°Mn³⁺/Mn²⁺ = +1.57 V$ ), making it an exceptionally strong oxidizing agent because it easily gains an electron to form stable $Mn²⁺$ $Mn^{2+}$ ( $d⁵$ $d^5$ configuration).

Step 1: Calculate the Magnetic Moment of Mn(III)

Electronic configuration of $Mn³⁺$ $Mn^{3+}$:

Mn³⁺ = [Ar]3d⁴ n = 4 unpaired electrons

Calculating the spin-only magnetic moment:

μ = \sqrt(4(4+2)) = \sqrt(24) \approx 4.89 B.M.

Step 2: Rounding to Nearest Integer

Rounding $4.89 B.M.$ $4.89\mathrm{\ B.M.}$ to the nearest integer gives $5$ $5$.

Pattern Recognition

High reduction potentials are strongly tied to manganese in its $+3$ $+3$ oxidation state. To quickly estimate magnetic moments, remember that a system with $n$ $n$ unpaired electrons always results in a value of ' $n.something$ $n.\text{something}$' B.M. Thus, $4$ $4$ unpaired electrons $arrow 4.89 B.M.$ $\rightarrow 4.89\mathrm{\ B.M.}$, which rounds up to $5$ $5$.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

The number of valence electrons present in the metal among Cr$\mathrm{Cr}$, Co$\mathrm{Co}$, Fe$\mathrm{Fe}$ and Ni$\mathrm{Ni}$ which has the lowest enthalpy of atomisation is:

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More d- and f-Block Elements Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Oxidation State

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Magnetic Moment of Mn(III)

Step 2: Rounding to Nearest Integer

Pattern Recognition

Chapter Mix

More d- and f-Block Elements Questions — jee_main_2025_07_april_morning

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The number of valence electrons present in the metal among $Cr$ $\mathrm{Cr}$, $Co$ $\mathrm{Co}$, $Fe$ $\mathrm{Fe}$ and $Ni$ $\mathrm{Ni}$ which has the lowest enthalpy of atomisation is: