The incorrect relationship in the following pairs in relation to ionisation enthalpies is :

Core Logic

In an acidic medium, both K₂Cr₂O₇$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and KMnO₄$\mathrm{KMnO}_4$ act as strong oxidizing agents and carry out the following transformations:

• A: Oxidize iodide to iodine: I^- arrow I₂$\mathrm{I}^- \rightarrow \mathrm{I}_2$
• B: Oxidize sulfide to elemental sulfur: S²⁻ arrow S$\mathrm{S}^{2-} \rightarrow \mathrm{S}$
• C: Oxidize ferrous ions to ferric ions: Fe²⁺ arrow Fe³⁺$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$

For reactions D and E:

• Iodide is oxidized to iodate (IO₃^-$\mathrm{IO}_3^-$) by KMnO₄$\mathrm{KMnO}_4$ primarily in a neutral or faintly alkaline medium, not acidic.
• Thiosulfate (S₂O₃²⁻$\mathrm{S}_2\mathrm{O}_3^{2-}$) undergoes a disproportionation/precipitation reaction in acid rather than clean oxidation to sulfate by dichromate.

Thus, statements A, B, and C are valid for both under acidic conditions.

Pattern Recognition

Sees: Shared oxidation products in an acidic environment. Shortcut: Remember that I^- arrow IO₃^-$\mathrm{I}^- \rightarrow \mathrm{IO}_3^-$ is the signature reaction for alkaline permanganate, allowing quick elimination of choices containing D.

Chapter Mix

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

Related Formula

Core Logic

For μ = 4.9 B.M.$\mu = 4.9 \text{ B.M.}$, solve for n$n$:

Thus, the metal ion must have 4 unpaired electrons.

Step 1: Electronic Configurations

A. Cr²⁺: [Ar] 3d⁴ 4$\mathrm{Cr}^{2+}: [\mathrm{Ar}] 3d^4 \implies 4$ unpaired electrons.

B. Fe²⁺: [Ar] 3d⁶ 4$\mathrm{Fe}^{2+}: [\mathrm{Ar}] 3d^6 \implies 4$ unpaired electrons.

C. Fe³⁺: [Ar] 3d⁵ 5$\mathrm{Fe}^{3+}: [\mathrm{Ar}] 3d^5 \implies 5$ unpaired electrons.

D. Co²⁺: [Ar] 3d⁷ 3$\mathrm{Co}^{2+}: [\mathrm{Ar}] 3d^7 \implies 3$ unpaired electrons.

E. Mn³⁺: [Ar] 3d⁴ 4$\mathrm{Mn}^{3+}: [\mathrm{Ar}] 3d^4 \implies 4$ unpaired electrons.

Therefore, Cr²⁺$\mathrm{Cr}^{2+}$, Fe²⁺$\mathrm{Fe}^{2+}$, and Mn³⁺$\mathrm{Mn}^{3+}$ (A, B, and E) have 4 unpaired electrons.

Pattern Recognition

Magnetic moment ~4.9 B.M. arrow n = 4$\rightarrow n = 4$ unpaired electrons. 3d⁴$3d^4$ and 3d⁶$3d^6$ high-spin ions always have n = 4$n = 4$.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Related Formula

Core Logic

• Compound A is potassium dichromate (K₂Cr₂O₇$\mathrm{K_2Cr_2O_7}$), reacting in the chromyl chloride test to release red vapors of CrO₂Cl₂$\mathrm{CrO_2Cl_2}$.
• Vapors dissolve in NaOH$\mathrm{NaOH}$ to give a yellow solution of sodium chromate, B = Na₂CrO₄$\mathrm{B} = \mathrm{Na_2CrO_4}$.
• Acidification of chromate (B) yields sodium dichromate, C = Na₂Cr₂O₇$\mathrm{C} = \mathrm{Na_2Cr_2O_7}$ (or Cr₂O₇²⁻$\mathrm{Cr_2O_7^{2-}}$).

Step 1: Structural Analysis of Dichromate Ion

The dichromate ion [O₃Cr-O-CrO₃]²⁻$\mathrm{[O_3Cr-O-CrO_3]^{2-}}$ consists of two CrO₄$\mathrm{CrO_4}$ tetrahedra sharing one oxygen corner:

• Bridging oxygen atoms: 1$1$ (Cr-O-Cr$\mathrm{Cr-O-Cr}$)
• Terminal oxygen atoms: 6$6$ (three on each chromium atom, O₃Cr...CrO₃$\mathrm{O_3Cr...CrO_3}$)

Therefore, the number of terminal oxygen atoms in compound C is 6.

Pattern Recognition

Dichromate ion structure Cr₂O₇²⁻ arrow 1$\mathrm{Cr_2O_7^{2-}} \rightarrow 1$ bridging oxygen, 6$6$ terminal oxygens.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Core Logic

Let's map the electronic configurations and count the unpaired d-orbital electrons for each option:

• For pair (1):

Both ions contain exactly 3 unpaired electrons.

• For other ions:

Pattern Recognition

D-orbital counts follow a predictable symmetry: a 3dⁿ$3d^n$ system contains the same number of unpaired electrons as a 3d¹⁰⁻ⁿ$3d^{10-n}$ system under high-spin conditions. This explains why 3d³$3d^3$ (V²⁺$V^{2+}$) and 3d⁷$3d^7$ (Co²⁺$Co^{2+}$) match perfectly with 3 unpaired electrons each.

Chapter Mix

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