When and how is a protective diode applied to the collector of a transistor switch?
An NPN BJT will conduct significantly from collector to base (it is a junction) if the collector is brought below the base voltage by more than a few hundred mV. That could cause problems. 
Either block the voltage (series diode) or conduct it to ground. In one case the compound switch is 'open' for voltages &lt;&lt; 0V, in the other case it conducts. Depends on what you want. If you want to switch both sides of the AC you need to do something a bit different. 
You do not normally put a diode in series with an inductive load. In the case of an inductive load you need to provide a path for the current to flow when the switch turns off, in order to allow the energy stored in the inductor to dissipate, and that path is typically provided by some kind of network in parallel with the inductor- a diode, a diode with a zener in series, a bipolar TVS, an RC snubber, that sort of thing. If you put a diode in series with a pure inductive load, switching the driver off would result in the diode breaking over in reverse as the voltage rose to some very high value. In practice that might not happen- for example, you could probably safely put a 3kV diode in series with a 3V relay because the coil voltage would never get to more than a few hundred volts due to the coil distributed capacitance, but it's hard on the coil insulation. The only advantage it would have is that it would allow the coil field to collapse very quickly (oh, and the high voltage might be a useful side effect in some cases). 


Edit: Re below question about what happens with series diode- here is a simulation- the switch does not permit voltage to fall below zero, otherwise you get a relatively slow ring-down. 



Now with a parallel diode: 



1K in parallel with coil, series diode and no limitation on switch voltage: