KMnO_4$KMnO_4$ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y$X + Y$ is ______.

### 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