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

Core Logic

Step 1: Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$) reacts with an alkali like NaOH$NaOH$ to give a yellow solution of sodium chromate (Na₂CrO₄$Na_2CrO_4$).

Step 2: Sodium chromate (Na₂CrO₄$Na_2CrO_4$) reacts with hydrogen peroxide (H₂O₂$H_2O_2$) in an acidic medium (HCl$HCl$) to yield the deep blue colored chromium pentoxide (CrO₅$CrO_5$, also known as chromium(VI) oxide peroxide).

Note: NaCl$NaCl$ formation implies the overall balanced reaction uses the acid for neutralization/salt formation.

Pattern Recognition

Chromyl chloride test intermediate: Yellow solution = Na₂CrO₄$Na_2CrO_4$. Reaction of chromate with H₂O₂$H_2O_2$ in acid = Blue peroxide CrO₅$CrO_5$ (butterfly structure).

Chapter Mix

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

Core Logic

In K₂Cr₂O₇$K_2Cr_2O_7$, Chromium is in the +6$+6$ oxidation state, which means its electronic configuration is d⁰$d^0$. Since there are no d-electrons, d-d$d-d$ transitions cannot occur. The orange color is due to ligand-to-metal charge transfer (LMCT) from oxygen to chromium.

Similarly, in KMnO₄$KMnO_4$, Manganese is in the +7$+7$ oxidation state, which also corresponds to a d⁰$d^0$ configuration. Again, no d-d$d-d$ transitions are possible. The intense purple color is due to ligand-to-metal charge transfer (LMCT) from oxygen to manganese.

Step 1: Final Conclusion

Both compounds owe their colors to charge transfer transitions.

Pattern Recognition

Compounds of transition metals in their highest oxidation states (where they have d⁰$d^0$ configurations, like Cr⁺⁶$Cr^{+6}$, Mn⁺⁷$Mn^{+7}$, V⁺⁵$V^{+5}$) are deeply colored primarily due to Charge Transfer spectra, NOT d-d transitions.

Chapter Mix

Class 12 Chemistry: The d and f Block Elements

Core Logic

Step 1: Alkaline oxidative fusion of MnO₂$MnO_2$ (pyrolusite ore) with KOH$KOH$ in the presence of O₂$O_2$ (or an oxidizing agent like KNO₃$KNO_3$) yields the green-colored manganate ion (MnO₄²⁻$MnO_4^{2-}$).

So, A is MnO₄²⁻$\mathrm{MnO}_4^{2-}$.

Step 2: Electrolytic oxidation of the manganate ion (MnO₄²⁻$MnO_4^{2-}$) in an alkaline medium converts it to the purple-colored permanganate ion (MnO₄^-$MnO_4^-$).

So, B is MnO₄^-$\mathrm{MnO}_4^-$.

Pattern Recognition

Industrial preparation sequence of KMnO₄$KMnO_4$: MnO₂ fusion, KOH, O₂ MnO₄²⁻ (green) electrolytic oxidation MnO₄^- (purple)$MnO_2 \xrightarrow{\text{fusion, } KOH, O_2} MnO_4^{2-} \text{ (green)} \xrightarrow{\text{electrolytic oxidation}} MnO_4^- \text{ (purple)}$.

Chapter Mix

Class 12 Chemistry: The d and f Block Elements

Core Logic

An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question.

Step 1: Checking configurations

Cerium (Ce$Ce$, Z=58): [Xe] 4f¹ 5d¹ 6s²$[Xe] 4f^1 5d^1 6s^2$ arrow Ce⁴⁺: [Xe] 4f⁰$\rightarrow Ce^{4+}: [Xe] 4f^0$ (0 unpaired electrons arrow$\rightarrow$ Diamagnetic)

Lanthanum (La$La$, Z=57): [Xe] 4f⁰ 5d¹ 6s²$[Xe] 4f^0 5d^1 6s^2$ arrow La³⁺: [Xe] 4f⁰$\rightarrow La^{3+}: [Xe] 4f^0$ (0 unpaired electrons arrow$\rightarrow$ Diamagnetic)

Pattern Recognition

Ions with an empty f-subshell (f⁰$f^0$, e.g., La³⁺, Ce⁴⁺$La^{3+}, Ce^{4+}$) or a completely filled f-subshell (f¹⁴$f^{14}$, e.g., Lu³⁺, Yb²⁺$Lu^{3+}, Yb^{2+}$) are invariably diamagnetic.

Chapter Mix

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

Core Logic

Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost 4s$4s$ orbital.

(A) Cr$Cr$ (Z=24): [Ar] 3d⁵ 4s¹$[Ar] 3d^5 4s^1$ arrow$\rightarrow$ Cr²⁺$Cr^{2+}$: [Ar] 3d⁴$[Ar] 3d^4$ (B) Mn$Mn$ (Z=25): [Ar] 3d⁵ 4s²$[Ar] 3d^5 4s^2$ arrow$\rightarrow$ Mn^+$Mn^+$: [Ar] 3d⁵ 4s¹$[Ar] 3d^5 4s^1$ (C) Ni$Ni$ (Z=28): [Ar] 3d⁸ 4s²$[Ar] 3d^8 4s^2$ arrow$\rightarrow$ Ni²⁺$Ni^{2+}$: [Ar] 3d⁸$[Ar] 3d^8$ (D) V$V$ (Z=23): [Ar] 3d³ 4s²$[Ar] 3d^3 4s^2$ arrow$\rightarrow$ V^+$V^+$: [Ar] 3d³ 4s¹$[Ar] 3d^3 4s^1$

Step 1: Match execution

A arrow$\rightarrow$ III B arrow$\rightarrow$ II C arrow$\rightarrow$ I D arrow$\rightarrow$ IV

Chapter Mix

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