Consider the following statements:

[(A)] $\mathrm{KMnO_4}$ is diamagnetic while $\mathrm{K_2MnO_4}$ is paramagnetic.
[(B)] Manganate ion contains $\mathrm{Mn^{6+}}$ while permanganate ion contains $\mathrm{Mn^{7+}}$.
[(C)] $\mathrm{Mn^{2+}}$ ion on reaction with $\mathrm{S_2O_8^{2-}}$ ions gives manganate ion.
[(D)] Both $\mathrm{MnO_4^-}$ and $\mathrm{MnO_4^{2-}}$ are tetrahedral.
Correct statements are:

Question

Statement-I : The correct increasing order of bond length among the following is $O_2^+ < O_2 < O_2^- < O_2^{2-}$.
Statement-II : The correct order of number of unpaired electrons is $O_2^{2-} < O_2^+ < O_2^- < O_2$.

Answer 1

Let's analyze each statement one by one to determine which are correct:

Statement A: \mathrm{KMnO_4} is diamagnetic while \mathrm{K_2MnO_4} is paramagnetic.
Explanation: Potassium permanganate \mathrm{(KMnO_4)} has the Mn in the +7 oxidation state, which results in no unpaired electrons as the electron configuration is \mathrm{3d^0}. Therefore, it is diamagnetic. In contrast, potassium manganate \mathrm{(K_2MnO_4)} has the Mn in the +6 oxidation state with an electron configuration of \mathrm{3d^1}, meaning it has one unpaired electron, making it paramagnetic. Hence, this statement is correct.
Statement B: Manganate ion contains \mathrm{Mn^{6+}} while permanganate ion contains \mathrm{Mn^{7+}}.
Explanation: In the manganate ion \mathrm{(MnO_4^{2-})}, manganese is in the +6 oxidation state, and in the permanganate ion \mathrm{(MnO_4^-}), it is in the +7 oxidation state. Thus, this statement is correct.
Statement C: \mathrm{Mn^{2+}} ion on reaction with \mathrm{S_2O_8^{2-}} ions gives manganate ion.
Explanation: Persulfate ions \mathrm{S_2O_8^{2-}} are strong oxidizing agents, and they can oxidize \mathrm{Mn^{2+}} to \mathrm{MnO_4^-} ions (permanganate), not manganate (\mathrm{MnO_4^{2-}}). Thus, this statement is incorrect.
Statement D: Both \mathrm{MnO_4^-} and \mathrm{MnO_4^{2-}} are tetrahedral.
Explanation: Both the permanganate (\mathrm{MnO_4^-}) and manganate ions (\mathrm{MnO_4^{2-}}) have a tetrahedral geometry due to the presence of d
^2
sp
^3
hybridization. Thus, the statement is correct.

Conclusion: The correct statements are A, B, and D. Therefore, the correct answer is "A, B and D only".

Answer 2

Let's analyze each statement one by one to determine which are correct:

Statement A: \mathrm{KMnO_4} is diamagnetic while \mathrm{K_2MnO_4} is paramagnetic.
Explanation: Potassium permanganate \mathrm{(KMnO_4)} has the Mn in the +7 oxidation state, which results in no unpaired electrons as the electron configuration is \mathrm{3d^0}. Therefore, it is diamagnetic. In contrast, potassium manganate \mathrm{(K_2MnO_4)} has the Mn in the +6 oxidation state with an electron configuration of \mathrm{3d^1}, meaning it has one unpaired electron, making it paramagnetic. Hence, this statement is correct.
Statement B: Manganate ion contains \mathrm{Mn^{6+}} while permanganate ion contains \mathrm{Mn^{7+}}.
Explanation: In the manganate ion \mathrm{(MnO_4^{2-})}, manganese is in the +6 oxidation state, and in the permanganate ion \mathrm{(MnO_4^-}), it is in the +7 oxidation state. Thus, this statement is correct.
Statement C: \mathrm{Mn^{2+}} ion on reaction with \mathrm{S_2O_8^{2-}} ions gives manganate ion.
Explanation: Persulfate ions \mathrm{S_2O_8^{2-}} are strong oxidizing agents, and they can oxidize \mathrm{Mn^{2+}} to \mathrm{MnO_4^-} ions (permanganate), not manganate (\mathrm{MnO_4^{2-}}). Thus, this statement is incorrect.
Statement D: Both \mathrm{MnO_4^-} and \mathrm{MnO_4^{2-}} are tetrahedral.
Explanation: Both the permanganate (\mathrm{MnO_4^-}) and manganate ions (\mathrm{MnO_4^{2-}}) have a tetrahedral geometry due to the presence of d
^2
sp
^3
hybridization. Thus, the statement is correct.

Conclusion: The correct statements are A, B, and D. Therefore, the correct answer is "A, B and D only".

Show Hint

The Correct Option is B

Solution and Explanation

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