Hi Jose and welcome to Electronics Point
Your flip-flop is powering up in the activated state. This could be due to leakage currents, or stray capacitance - through the breadboard, for example.
Try shorting the base and emitter together on one transistor (it doesn't matter which). This will force that transistor OFF, and the other transistor should turn OFF as well, since the first transistor should not be passing any significant collector current. The LED should turn OFF while the transistor's base is shorted to its emitter.
Now remove the short, disconnecting your shorting wire from the base first. If the LED turns ON again, leakage current is the reason.
That circuit works, and you've drawn it right, but for best reliability, you should have a resistor between the base and the emitter of each transistor. For a 5V supply, and 22k resistors for R1 and R2, these resistors should be around 4k7. They ensure that each transistor will not turn on due to small leakage currents from the other transistor; the other transistor needs to pass at least 130 µA and pull its collector to within 1.5V of its emitter, before the transistor will start to conduct.
If you're wondering how I calculated those numbers, here's the explanation.
Each transistor will only start to conduct when its base-emitter voltage reaches about 0.6V. The 4k7 resistor across each base-emitter junction needs a certain amount of current to be flowing through it to cause a voltage drop of 0.6V across it. This current can be calculated using Ohm's Law:
I = V / R
= 0.6V / 4700 ohms
= 0.00013 amps (roughly)
= 130 µA (roughly)
So the other transistor needs to be passing at least 130 µA to cause enough voltage drop across the 4k7 resistor to make the transistor start to conduct.
The 22k resistor and the 4k7 resistor form a voltage divider. The total voltage that must appear across the series combination of the 22k resistor and the 4k7 resistor, in order to get 0.6V across the 4k7 resistor can be calculated in various ways. One way is by calculating the voltage across the 22k resistor at 130 µA. Using Ohm's Law:
V = I R
= 0.00013 * 22000
= 2.9V (roughly)
Now add the 0.6V that's across the 4k7 resistor (because voltages in series add together) and you get 3.5V total across both resistors.
Subtract 3.5V from the 5V supply rail and you get 1.5V between the collector and emitter of the other transistor. So the other transistor needs to pull its collector to within 1.5V of its emitter in order to produce enough voltage across the 4k7 resistor to make the other transistor start to conduct.
