Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH…

Q: Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH and KNO3 to produce

The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore (MnO2). Fusing the solid reactant directly along an alkaline base payload (KOH) combined explicitly with an oxidizing carrier (KNO3) yields the intermediate green product, **potassium manganate** (K2MnO4). Step 1 yields the +6 green compound (K2MnO4); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple +7 agent (KMnO4). Class 12 Chemistry: The d-and f-Block

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

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

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

Q jee_main_2025_29_jan_morning Melting Points of Transition Elements

The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :

A. $Fe < Mn$ $\mathrm{Fe} < \mathrm{Mn}$ , $Ru < Tc$ $\mathrm{Ru} < \mathrm{Tc}$ and $Re < Os$ $\mathrm{Re} < \mathrm{Os}$
B. $Mn < Fe, Tc < Ru$ and $Re < Os$ $\mathrm{Re} < \mathrm{Os}$
C. $Mn < Fe, Tc < Ru$ and $Os < Re$ $\mathrm{Os} < \mathrm{Re}$
D. $Fe < Mn$ $\mathrm{Fe} < \mathrm{Mn}$ , $Ru < Tc$ $\mathrm{Ru} < \mathrm{Tc}$ and $Os < Re$ $\mathrm{Os} < \mathrm{Re}$

Solution

Formulas Used

Melting point trends in $3d$ $3d$, $4d$ $4d$, and $5d$ $5d$ series transition metals depend on the extent of metallic bonding and $d$ $d$-electron participation.

Core Logic

According to NCERT transition element periodic trends:

$3d$ $3d$ Series ( $Mn$ $\mathrm{Mn}$ vs $Fe$ $\mathrm{Fe}$): Manganese ( $Mn$ $\mathrm{Mn}$, $3d⁵ 4s²$ $3d^5 4s^2$) has an abnormally low melting point compared to Iron ( $Fe$ $\mathrm{Fe}$, $3d⁶ 4s²$ $3d^6 4s^2$) because its stable, half-filled $d⁵$ $d^5$ configuration holds $d$ $d$-electrons more tightly, reducing their participation in metallic bonding $arrow Mn < Fe$ .
$4d$ $4d$ Series ( $Tc$ $\mathrm{Tc}$ vs $Ru$ $\mathrm{Ru}$): Technetium ( $Tc$ $\mathrm{Tc}$, $4d⁵ 5s²$ $4d^5 5s^2$) similarly shows a dip in melting point compared to Ruthenium ( $Ru$ $\mathrm{Ru}$, $4d⁷ 5s¹$ $4d^7 5s^1$) due to the stable $4d⁵$ $4d^5$ configuration $arrow Tc < Ru$ .
$5d$ $5d$ Series ( $Re$ $\mathrm{Re}$ vs $Os$ $\mathrm{Os}$): Rhenium ( $Re$ $\mathrm{Re}$, $5d⁵ 6s²$ $5d^5 6s^2$) has optimal interatomic interaction and a higher melting point than Osmium ( $Os$ $\mathrm{Os}$, $5d⁶ 6s²$ $5d^6 6s^2$) $arrow Os < Re$ .

Combining these trends yields: $Mn < Fe$ $\mathrm{Mn} < \mathrm{Fe}$, $Tc < Ru$ $\mathrm{Tc} < \mathrm{Ru}$, and $Os < Re$ $\mathrm{Os} < \mathrm{Re}$

Pattern Recognition

Stable half-filled $d⁵$ $d^5$ configurations in $3d$ $3d$ ( $Mn$ $\mathrm{Mn}$) and $4d$ $4d$ ( $Tc$ $\mathrm{Tc}$) restrict $d$ $d$-electron delocalization, creating characteristic dips in melting point curves compared to adjacent metals.

Correct Option: (C)

Preparation of potassium permanganate from MnO₂$\mathrm{MnO}_2$ involves two step process in which the 1st$1^{\text{st}}$ step is a reaction with KOH and KNO₃$\mathrm{KNO}_3$ to produce

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

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

Solution

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

Solution

Formulas Used

Core Logic

Pattern Recognition

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Preparation of potassium permanganate from $MnO₂$ $\mathrm{MnO}_2$ involves two step process in which the $1st$ $1^{\text{st}}$ step is a reaction with KOH and $KNO₃$ $\mathrm{KNO}_3$ to produce