Consider the following statements about manganate and permanganate ions. Identify the correct statements. A. The geometry of both manganate and permanganate ions is tetrahedral. B. The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively. C. Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product. D. Manganate ion is paramagnetic and permanganate ion is diamagnetic. E. Acidified permanganate ion reduces oxalate, nitrite and iodide ions.

Question

Structures of four disaccharides are given below. Among the given disaccharides, the non-reducing sugar is:

Answer 1

To determine the correct statements about manganate and permanganate ions, let's analyze each given statement:

Statement A: The geometry of both manganate and permanganate ions is tetrahedral.
This statement is true because both MnO₄^2- (manganate ion) and MnO₄^- (permanganate ion) have tetrahedral geometries. The manganese atom is at the center surrounded by four oxygen atoms in a tetrahedral arrangement.
Statement B: The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively.
This statement is false because the oxidation state of Mn in manganate (MnO₄^2-) is +6, and in permanganate (MnO₄^-) is +7.
Statement C: Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product.
This statement is false as the oxidation by peroxodisulphate usually leads to the formation of manganate (VII) ion or permanganate ion, not manganate ion, because peroxodisulphate is a very strong oxidizing agent.
Statement D: Manganate ion is paramagnetic and permanganate ion is diamagnetic.
This statement is false because manganate ion is actually paramagnetic due to unpaired electrons in its d orbitals, while permanganate ion is diamagnetic since it has no unpaired electrons.
Statement E: Acidified permanganate ion reduces oxalate, nitrite, and iodide ions.
This statement is true. In an acidic medium, permanganate ion (MnO₄^-) is a strong oxidizing agent and can oxidize oxalate (C₂O₄^2-), nitrite (NO₂^-), and iodide (I^-) ions.

Thus, the correct statements are A and E. Therefore, the correct answer is A, B and E Only.

Answer 2

To determine the correct statements about manganate and permanganate ions, let's analyze each given statement:

Statement A: The geometry of both manganate and permanganate ions is tetrahedral.
This statement is true because both MnO₄^2- (manganate ion) and MnO₄^- (permanganate ion) have tetrahedral geometries. The manganese atom is at the center surrounded by four oxygen atoms in a tetrahedral arrangement.
Statement B: The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively.
This statement is false because the oxidation state of Mn in manganate (MnO₄^2-) is +6, and in permanganate (MnO₄^-) is +7.
Statement C: Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product.
This statement is false as the oxidation by peroxodisulphate usually leads to the formation of manganate (VII) ion or permanganate ion, not manganate ion, because peroxodisulphate is a very strong oxidizing agent.
Statement D: Manganate ion is paramagnetic and permanganate ion is diamagnetic.
This statement is false because manganate ion is actually paramagnetic due to unpaired electrons in its d orbitals, while permanganate ion is diamagnetic since it has no unpaired electrons.
Statement E: Acidified permanganate ion reduces oxalate, nitrite, and iodide ions.
This statement is true. In an acidic medium, permanganate ion (MnO₄^-) is a strong oxidizing agent and can oxidize oxalate (C₂O₄^2-), nitrite (NO₂^-), and iodide (I^-) ions.

Thus, the correct statements are A and E. Therefore, the correct answer is A, B and E Only.

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