The behavior of \(KMnO_4\) in different pH conditions can be explained as follows:
1. At pH < 7 (acidic conditions): \(KMnO_4\) is reduced to a colorless substance \(P\). Under acidic conditions, it is commonly reduced to manganese(II) ions (\(Mn^{2+}\)).
2. At pH = 7 (neutral conditions): \(KMnO_4\) forms a brown precipitate \(Q\). Under neutral conditions, it is reduced to manganese dioxide (\(MnO_2\)).
3. At pH > 7 (alkaline conditions): \(KMnO_4\) forms a green-colored solution \(R\). Under alkaline conditions, it is reduced to manganate ions (\(MnO_4^{2-}\)).
So, the relationship between \(P\), \(Q\), and \(R\) and their corresponding manganese species is as follows:
- \(P\) (colorless substance at pH < 7) corresponds to \(Mn^{2+}\). - \(Q\) (brown precipitate at pH = 7) corresponds to \(MnO_2\). - \(R\) (green-colored solution at pH > 7) corresponds to \(MnO_4^{2-}\).
Therefore, the correct answer is option (1): \(P\) corresponds to \(Mn^{2+}\), \(Q\) corresponds to \(MnO_2\), \(R\) corresponds to \(MnO_4^{2-}\). |