### 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

### Core Logic Step 1: Alkaline oxidative fusion of MnO_2$MnO_2$ (pyrolusite ore) with KOH$KOH$ in the presence of O_2$O_2$ (or an oxidizing agent like KNO_3$KNO_3$) yields the green-colored manganate ion (MnO_4^2-$MnO_4^{2-}$). 2mathrmMnO_2 + 4mathrmOH^- + mathrmO_2 rightarrow 2mathrmMnO_4^2- + 2mathrmH_2mathrmO$2\mathrm{MnO}_2 + 4\mathrm{OH}^- + \mathrm{O}_2 \rightarrow 2\mathrm{MnO}_4^{2-} + 2\mathrm{H}_2\mathrm{O}$ So, A is mathrmMnO_4^2-$\mathrm{MnO}_4^{2-}$. Step 2: Electrolytic oxidation of the manganate ion (MnO_4^2-$MnO_4^{2-}$) in an alkaline medium converts it to the purple-colored permanganate ion (MnO_4^-$MnO_4^-$). mathrmMnO_4^2- rightarrow mathrmMnO_4^- + mathrme^-$\mathrm{MnO}_4^{2-} \rightarrow \mathrm{MnO}_4^- + \mathrm{e}^-$ So, B is mathrmMnO_4^-$\mathrm{MnO}_4^-$. ### Pattern Recognition Industrial preparation sequence of KMnO_4$KMnO_4$: MnO_2 xrightarrowtextfusion, KOH, O_2 MnO_4^2- text (green) xrightarrowtextelectrolytic oxidation MnO_4^- text (purple)$MnO_2 \xrightarrow{\text{fusion, } KOH, O_2} MnO_4^{2-} \text{ (green)} \xrightarrow{\text{electrolytic oxidation}} MnO_4^- \text{ (purple)}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements

### Core Logic Step 1: Chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$) reacts with an alkali like NaOH$NaOH$ to give a yellow solution of sodium chromate (Na_2CrO_4$Na_2CrO_4$). CrO_2Cl_2 + 4NaOH rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 (A) + 2NaCl + 2H_2O$ Step 2: Sodium chromate (Na_2CrO_4$Na_2CrO_4$) reacts with hydrogen peroxide (H_2O_2$H_2O_2$) in an acidic medium (HCl$HCl$) to yield the deep blue colored chromium pentoxide (CrO_5$CrO_5$, also known as chromium(VI) oxide peroxide). Na_2CrO_4 + 2H_2O_2 + 2HCl rightarrow CrO_5 (B) + 2NaCl + 3H_2O$Na_2CrO_4 + 2H_2O_2 + 2HCl \rightarrow CrO_5 (B) + 2NaCl + 3H_2O$ Note: NaCl$NaCl$ formation implies the overall balanced reaction uses the acid for neutralization/salt formation. ### Pattern Recognition Chromyl chloride test intermediate: Yellow solution = Na_2CrO_4$Na_2CrO_4$. Reaction of chromate with H_2O_2$H_2O_2$ in acid = Blue peroxide CrO_5$CrO_5$ (butterfly structure). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d and f Block Elements Class 11 Chemistry: Redox Reactions