The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are: A. Cr2+ B. Fe2+ C. Fe3+ D…

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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)

Q jee_main_2025_29_jan_morning Preparation and Properties of Potassium Dichromate

The molar mass of the water insoluble product formed from the fusion of chromite ore

(FeCr₂O₄)

with

Na₂CO₃

in presence of

O₂

is ________

g mol⁻¹

.

Numerical Answer. Answer: 160 to 160

Solution

Related Formula

Balanced fusion reaction process description

Core Logic

Write the balanced chemical equation for the industrial preparation stage of chromate salts:

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ arrow 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

Evaluating the solubilities of the products:

$Na₂CrO₄$ $\mathrm{Na_2CrO_4}$ is highly soluble in water.
$Fe₂O₃$ $\mathrm{Fe_2O_3}$ (Iron(III) oxide) is water-insoluble.

Molar Mass of $Fe₂O₃$ $\mathrm{Fe_2O_3}$:

M = (2 \cdot 55.85) + (3 \cdot 16.0) (2 \cdot 56) + (3 \cdot 16) = 112 + 48 = 160 ~g/mol

Pattern Recognition

Transition metal oxides in high oxidation states with minimal ionic breakdown parameters reliably act as insoluble precipitates in water.

Chapter Mix

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

Q jee_main_2024_01_february_morning Oxidising Properties

In acidic medium,

K₂Cr₂O₇

shows oxidising action as represented in the half reaction

Cr₂O₇²⁻ + XH^+ + Ye^- arrow 2A + ZH₂O

X, Y, Z and A are respectively are:

A. $8, 6, 4 and Cr₂O₃$ $8, 6, 4 \text{ and } Cr_2O_3$
B. $14, 7, 6 and Cr³⁺$ $14, 7, 6 \text{ and } Cr^{3+}$
C. $8, 4, 6 and Cr₂O₃$ $8, 4, 6 \text{ and } Cr_2O_3$
D. $14, 6, 7 and Cr³⁺$ $14, 6, 7 \text{ and } Cr^{3+}$

Solution

Core Logic

The balanced half-reaction for the dichromate ion acting as an oxidising agent in an acidic medium is:

Cr₂O₇²⁻ + 14H^+ + 6e^- arrow 2Cr³⁺ + 7H₂O

Step 1: Compare with Given Equation

Comparing this with the given equation $Cr₂O₇²⁻ + XH^+ + Ye^- arrow 2A + ZH₂O$ :

$X = 14$ $X = 14$ $Y = 6$ $Y = 6$ $Z = 7$ $Z = 7$ $A = Cr³⁺$ $A = Cr^{3+}$

Pattern Recognition

In acidic medium, dichromate ( $Cr₂O₇²⁻$ $Cr_2O_7^{2-}$) always requires $14H^+$ $14H^+$ to balance $7O$ $7O$ atoms, forming $7H₂O$ $7H_2O$. Chromium reduces from +6 to +3 state, taking $6e^-$ $6e^-$ overall.

Chapter Mix

Class 12 Chemistry: The d-and f-Block Elements Class 11 Chemistry: Redox Reactions

Q73 jee_main_2024_29_january_evening Lanthanoid Oxidation States

Which of the following acts as a strong reducing agent? (Atomic number : Ce = 58, Eu = 63, Gd = 64, Lu = 71)

A. $Lu³⁺$ $\mathrm{Lu}^{3+}$
B. $Gd³⁺$ $\mathrm{Gd}^{3+}$
C. $Eu²⁺$ $\mathrm{Eu}^{2+}$
D. $Ce⁴⁺$ $\mathrm{Ce}^{4+}$

Solution

Related Formula

Electronic configuration of Eu = [Xe] 4f⁷ 6s²

Core Logic

The most common and stable oxidation state for lanthanoids is $+3$ $+3$. In the case of Europium:

Eu²⁺ = [Xe] 4f⁷

This configuration possesses a highly stable half-filled $f$ $f$-subshell. However, because the $+3$ $+3$ state is universally favored by thermodynamics in solution, $Eu²⁺$ $\text{Eu}^{2+}$ readily undergoes oxidation to lose one more electron:

Eu²⁺ arrow Eu³⁺ + 1e^-

By releasing an electron to stabilize into the $+3$ $+3$ state, it behaves as a potent reducing agent.

Step 1: Evaluation

Conversely, $Ce⁴⁺$ $\text{Ce}^{4+}$ acts as a powerful oxidizing agent to return to $+3$ $+3$, while $Lu³⁺$ $\text{Lu}^{3+}$ and $Gd³⁺$ $\text{Gd}^{3+}$ are already perfectly configured at their native stable limits.

Pattern Recognition

Europium( $II$ $II$) has a stable half-filled $f⁷$ $f^7$ configuration, yet easily loses an electron to attain the highly stable $+3$ $+3$ state typical of lanthanoids, making it a strong reducing agent.

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Solution & Explanation

Related Formula

Core Logic

Step 1: Electronic Configurations

Pattern Recognition

Chapter Mix

Reference Study Guides

More d- and f-Block Elements Previous-Year Questions — Page 6

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

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Compare with Given Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation

Pattern Recognition

Chapter Mix

More d- and f-Block Elements Questions — jee_main_2025_03_april_morning

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