d and f Block Elements Part 3

Concept:Manganese in its highest oxidation state forms a covalent oxide with oxygen using multiple bonds to stabilise the high charge. The structure of $\mathrm{Mn}_2\mathrm{O}_7$ contains one bridging oxygen.Explanation:For statement A: In $\mathrm{Mn}_2\mathrm{O}_7$, let Mn oxidation state be $x$. Then $2x + 7(-2) = 0$ gives $x = +7$. Manganese can lose all seven valence electrons ($3d^5 4s^2$), so +7 is its highest oxidation state. Statement A is correct.For statement B: In high oxidation states, Mn strongly attracts electrons and forms multiple bonds (e.g., $\mathrm{Mn}= \mathrm{O}$) with oxygen. These bonds delocalise electron density and stabilise the $\mathrm{Mn}^{7+}$ centre. Statement B is correct.For statement C: $\mathrm{Mn}_2\mathrm{O}_7$ is a dark green liquid with a low melting point (−20 °C). These are properties of a covalent molecular compound, not an ionic solid. The high charge density of $\mathrm{Mn}^{7+}$ increases covalent character. So $\mathrm{Mn}_2\mathrm{O}_7$ is covalent, not ionic. Statement C is incorrect.For statement D: The structure of $\mathrm{Mn}_2\mathrm{O}_7$ is $\mathrm{O}_3\mathrm{Mn} - \mathrm{O} - \mathrm{MnO}_3$, consisting of two $\mathrm{MnO}_4$ tetrahedra sharing one common oxygen atom. That common oxygen is the bridging oxygen. Statement D is correct.Hence, statements A, B, and D are correct.Answer:Option B (A, B and D Only).