Electronic – Fail on using 2 Transistor as a Switch

It looks like you need a current limiting resistor between your output pin and the transistor base (assuming you didn't omit it on purpose, for brevity). Without the resistor, when you set the output pin to HIGH, you are causing a short-circuit to ground. That may damage the pin, if it hasn't done so already.

I would guess that a 10K resistor would do it. That's what I use with my ATmegas and ATminis. But check your MCU datasheet for appropriate values.

It's quite a common mistake to think that the base-emitter junction already has some sort of current limiting property, but it doesn't. It has very low impedance. I usually think of it as a plain wire. That helps me avoid these kinds of mistakes. Same applies to collector-emitter junction.

By the way, your collector-emitter path has no current limiting resistor or load. Watch out! 

Your basic problem is PNP1/NPN2. With the bases tied together, you are getting maximum current flow through the two emitter-base junctions from +12 to ground. Any real transistors in this configuration will die instantly.

If you must use BJTs and want to keep your circuit ideas as much as possible, replicate NPN1/PNP2, and use a fifth transistor to invert the base drive signal between the right and left sides of the bridge. Also, you'll need to either use a 12-volt coil voltage or provide a 6th NPN as a level shifter. But be aware that this will not work well, since the NPNs will not get enough base voltage to turn on hard, and you won't get very good performance.

A better idea is something like

^{simulate this circuit – Schematic created using CircuitLab}

You will have noticed, with some frustration, that none of the resistors have values. There's a very good reason for that: you haven't specified your load current requirements. At any rate, Q1 - Q4 are power transistors, able to handle the load current, while Q5 - Q9 are smaller signal transistors able to handle about 1/10 the load current.