The spin only magnetic moment (mu) value (B.M.) of the compound with strongest oxidising power among Mn2O3, TiO and VO…

Year	2026	2025	2024	Total
Questions	10	18	17	45

Q47 jee_main_2025_04_april_morning Chemical Properties of KMnO4

KMnO₄

acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of

X + Y

is ______.

Numerical Answer. Answer: 10 to 10

Solution

Core Logic

Let's resolve both components step by step:

Finding X: In an acidic medium, the permanganate ion ( $KMnO₄$ $KMnO_4$, where $Mn$ $Mn$ is in the $+7$ $+7$ state) is reduced to the divalent manganese cation ( $Mn²⁺$ $Mn^{2+}$, state $+2$ $+2$):

$X = 7 - 2 = 5$ $X = 7 - 2 = 5$

Finding Y: During qualitative salt analysis, the acetate ion reacts with neutral ferric chloride to produce a characteristic blood-red coordination solution. Boiling this solution throws down a brown-red precipitate of basic ferric acetate, $[Fe(OH)₂(CH₃COO)]$ $[Fe(OH)_2(CH_3COO)]$. In this complex, Iron retains its $+3$ $+3$ oxidation state:

Fe³⁺ [Ar] 3d⁵ 4s⁰ Number of d-electrons (Y) = 5

Summing the values yields:

X + Y = 5 + 5 = 10

Pattern Recognition

This problem elegantly links standard redox transitions with qualitative inorganic salt tests. Remember that throughout the basic ferric acetate precipitation test, Iron remains steadily in its ferric $+3$ $+3$ ( $d⁵$ $d^5$) core configuration.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements Inorganic Qualitative Analysis

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

The spin only magnetic moment (μ)$(\mu)$ value (B.M.) of the compound with strongest oxidising power among Mn₂O₃$Mn_{2}O_{3}$, TiO and VO is ______ B.M. (Nearest integer).

Solution & Explanation

Related Formula

Core Logic

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

Step 2: Rounding to Nearest Integer

Pattern Recognition

Chapter Mix

Reference Study Guides

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

Solution

Core Logic

Pattern Recognition

Chapter Mix

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

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The spin only magnetic moment $(μ)$ $(\mu)$ value (B.M.) of the compound with strongest oxidising power among $Mn₂O₃$ $Mn_{2}O_{3}$, TiO and VO is ______ B.M. (Nearest integer).