Solution of \(KMnO_4\) is reduced to various products \(P, Q, R\) depending upon the pH of the solution. At pH < 7 it is reduced to colourless substance \(P\). At pH = 7 it forms a brown precipitate \(Q\). At pH > 7 it forms a green coloured solution \(R\).

\(P, Q, R\) will be :

1

The correct answer is option 1. 1

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 \(MnO_4^{2-}\).