The correct order of basic strength of given aniline is:

The correct answer is option 2. m-nitroaniline > p-nitroaniline > o-nitroaniline.

In order to give the decreasing order of the basicity of the aniline, we must first know about the \(pK_a\) value and the inductive effect present in the given aniline groups such as o-nitroaniline, p-nitroaniline and m-nitroaniline.

The \(pK_a\) value of o-nitroaniline, p-nitroaniline and m-nitroaniline is \(-0.3, 1.0\) and \(2.5\) respectively. First thing is that we have to see the group present in the Nitroaniline compound. The \(−NO_2\) is the group that is present in the nitroaniline. \(−NO_2\) group is the electron-withdrawing group. As the electron-withdrawing group is present, there will be \(−I\) effect present. In the o-nitroaniline and p-nitroaniline compound, delocalization of the lone pair is observed. Therefore, both \(-I\) effect and \(-M\) effect is observed. But compared to p-nitroaniline, o-nitroaniline will be having a close inductive effect. Therefore, o-nitroaniline is less basic than p-nitroaniline compounds. In m-nitroaniline only inductive effect is observed and therefore, it is more basic compared to others.

Therefore, the decreasing order of basic strength of given aniline is: m-nitroaniline > p-nitroaniline > o-nitroaniline