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$\mathrm{g \, mol^{-1}}$.

### Related Formula K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O$$K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) \rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O$$ ### Core Logic This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$) are evolved. ### Step 1: Oxidation State Calculation In chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$): Let the oxidation state of Chromium be x$x$. Oxygen is typically -2$-2$ and Chlorine is -1$-1$. x + 2(-2) + 2(-1) = 0$$x + 2(-2) + 2(-1) = 0$$ x - 4 - 2 = 0$x - 4 - 2 = 0$ x = +6$x = +6$ Thus, the oxidation state of Chromium in the product is 6$6$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Class 11 Chemistry: Practical Chemistry

### Step 1: Statement A Analysis CrO_4^2-$CrO_4^{2-}$ (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis 2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (MnO_4^2-$MnO_4^{2-}$) has manganese in the +6 oxidation state (3d^1$3d^1$). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured K_2MnO_4$K_2MnO_4$ undergoes disproportionation in neutral or acidic media to yield permanganate (MnO_4^-$MnO_4^-$) and manganese dioxide (MnO_2$MnO_2$). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements