Two volts should work pretty much with all standard bipolar junction transistors (BJTs). This is not sufficient voltage to operate a field effect transistor. A darlington BJT transistor would probably not be a good choice.
If you are trying to switch 2 volts with a transistor, remember that the forward voltage drop accross the device is about 0.7 volts.....2 volts will switch a transistor on without a problem, through a resistor (about 2K2) into the base........ Daniel.
Is 2V your source voltage (Vcc), or your drive voltage? In either case, the general answer is, yes.
As noted above, a general bipolar transistor (2N3904 / 2N3906) only requires ~0.7V on the base-emitter. Some FET's can be used with a lower drive voltage, for example the NTD3055 FET requires a gate voltage of only 2V.
If you are going to drive a relay with a bjt you should not forget that you need to have enough current available to go into the base.
Roughly I_C = I_B * beta
I usually use beta = 50 for relay drive ckts (for margin) and look at the coil resistance and voltage of the relay to get my I_C. Then calculate the I_B. From this I can size the base resistor accordingly.
R = (2V-0.7)/I_B
Using R = 2K2 you can easily drive 30mA of coil current.