The molar mass of the water insoluble product formed from the fusion of chromite ore mathrm(FeCr_2O_4)$\mathrm{(FeCr_2O_4)}$ with mathrmNa_2mathrmCO_3$\mathrm{Na}_2\mathrm{CO}_3$ in presence of mathrmO_2$\mathrm{O}_2$ is ________ mathrmg \, mol^-1$\mathrm{g \, mol^{-1}}$.

### Core Logic The noble gas configuration of Xenon (Z=54$Z=54$) provides the primary core layout. For Neodymium (Z=60$Z=60$), the 6 remaining valence electrons distribute into the inner 4textf$4\text{f}$ orbital subshell rather than filling the 5textd$5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of text[Xe] 4textf^4 6texts^2$\text{[Xe]} 4\text{f}^4 6\text{s}^2$. ### Pattern Recognition Lanthanide filling sequences generally bypass 5d$5d$ progression except for specific exceptions (La, Gd, Lu). ### Chapter Mix Class 12 Chemistry: d-and f-Block Elements

### Core Logic The reaction sequence for the chromyl chloride test is: Cl^- + K_2Cr_2O_7 + H_2SO_4 rightarrow CrO_2Cl_2$$Cl^- + K_2Cr_2O_7 + H_2SO_4 \rightarrow CrO_2Cl_2$$ The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions: CrO_2Cl_2 xrightarrowtextBasic medium CrO_4^2- + Cl^-$$CrO_2Cl_2 \xrightarrow{\text{Basic medium}} CrO_4^{2-} + Cl^-$$ Acidification of the yellow CrO_4^2-$CrO_4^{2-}$ solution followed by the addition of H_2O_2$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO_5$CrO_5$). CrO_4^2- xrightarrow[textyellow solution, 1. textAcidification CrO_5 text (blue compound)$$CrO_4^{2-} \xrightarrow[\text{yellow solution}, {1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$ ### Step 1: Oxidation State Calculation

The structure of chromium pentoxide (CrO_5$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O^2-$O^{2-}$) and four peroxide oxygens (O_2^2-$O_2^{2-}$). Therefore, there are 2 peroxo linkages. Let the oxidation state of Chromium be x$x$. x + 1(-2) + 4(-1) = 0$$x + 1(-2) + 4(-1) = 0$$ x - 2 - 4 = 0$x - 2 - 4 = 0$ x = +6$x = +6$ Thus, the oxidation state of Cr in CrO_5$CrO_5$ is +6$+6$. ### Pattern Recognition A classic oxidation state trap. Calculating simply via formula CrO_5$CrO_5$ yields x - 10 = 0 implies x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

### Core Logic Potassium permanganate (KMnO_4$KMnO_4$) is a strong oxidizing agent. When heated to 513mathrmK$513\mathrm{K}$, it undergoes thermal decomposition to give potassium manganate (K_2MnO_4$K_2MnO_4$), manganese dioxide (MnO_2$MnO_2$), and oxygen gas (O_2$O_2$). The balanced chemical equation is: 2KMnO_4 xrightarrowDelta K_2MnO_4 + MnO_2 + O_2$$2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2$$ ### Step 1: Final Identification The products formed along with O_2$O_2$ are K_2MnO_4$K_2MnO_4$ (green) and MnO_2$MnO_2$ (black). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements

### Core Logic The behavior of the permanganate ion (MnO_4^-$MnO_4^-$) varies with the pH of the medium. In a faintly alkaline or neutral medium, MnO_4^-$MnO_4^-$ oxidizes iodide (I^-$I^-$) completely to iodate (IO_3^-$IO_3^-$) while getting reduced to manganese dioxide (MnO_2$MnO_2$). The balanced ionic equation is: 2MnO_4^- + H_2O + I^- rightarrow 2MnO_2 + 2OH^- + IO_3^-$$2MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-$$ ### Pattern Recognition Rule of thumb for I^-$I^-$ oxidation by KMnO_4$KMnO_4$: In acidic medium: I^- rightarrow I_2$I^- \rightarrow I_2$ In alkaline/neutral medium: I^- rightarrow IO_3^-$I^- \rightarrow IO_3^-$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: d and f Block Elements

### Core Logic In K_2Cr_2O_7$K_2Cr_2O_7$, Chromium is in the +6$+6$ oxidation state, which means its electronic configuration is d^0$d^0$. 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 KMnO_4$KMnO_4$, Manganese is in the +7$+7$ oxidation state, which also corresponds to a d^0$d^0$ 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 d^0$d^0$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements