Question

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:

• A. A \& B only
• B. A, B and D only
• C. A, B and C only
• D. B and D only

Hint:

For transition-metal oxoanions: oxidation state helps you decide {magnetic nature}, while number of surrounding oxygens decides {geometry}.

Answer 1

The Correct Option is B

Concept:
Magnetic nature depends on the number of unpaired electrons. Oxidation state determines electronic configuration and geometry of manganese oxoanions. Statement-wise Analysis:
(A)
{\(\mathrm{KMnO_4}\) is diamagnetic while \(\mathrm{K_2MnO_4}\) is paramagnetic.}

In \(\mathrm{MnO_4^-}\), Mn is in \(+7\) oxidation state: \(d^0\) configuration.
No unpaired electrons \(\Rightarrow\) diamagnetic.
In \(\mathrm{MnO_4^{2-}}\), Mn is in \(+6\) oxidation state: \(d^1\).
One unpaired electron \(\Rightarrow\) paramagnetic.
\(\Rightarrow\) True
(B)
{Manganate ion contains \(\mathrm{Mn^{6+}}\) while permanganate ion contains \(\mathrm{Mn^{7+}}\).}

This is directly obtained from charge calculation.
\(\Rightarrow\) True
(C)
{\(\mathrm{Mn^{2+}}\) ion on reaction with \(\mathrm{S_2O_8^{2-}}\) gives manganate ion.}

\(\mathrm{S_2O_8^{2-}}\) is a strong oxidizing agent.
It oxidizes \(\mathrm{Mn^{2+}}\) directly to permanganate (\(\mathrm{MnO_4^-}\)), not manganate.
\(\Rightarrow\) False
(D)
{Both \(\mathrm{MnO_4^-}\) and \(\mathrm{MnO_4^{2-}}\) are tetrahedral.}

Both ions have four equivalent Mn–O bonds.
Hence both possess tetrahedral geometry.
\(\Rightarrow\) True

Final Conclusion:
Correct statements are \(\boxed{A,\,B\ \text{and}\ D}\).