This is an old thread but just in case anyone comes across it like I did. pi.FileName needs to be set to the file name (and possibly full path to file ) of the executable you want to use to open your file. The below code works for me to open a video file with VLC.
var path = files[currentIndex].fileName;
var pi = new ProcessStartInfo(path)
{
Arguments = Path.GetFileName(path),
UseShellExecute = true,
WorkingDirectory = Path.GetDirectoryName(path),
FileName = "C:\\Program Files (x86)\\VideoLAN\\VLC\\vlc.exe",
Verb = "OPEN"
};
Process.Start(pi)
Tigran's answer works but will use windows' default application to open your file, so using ProcessStartInfo may be useful if you want to open the file with an application that is not the default.