74HC76 integrated really use Vcc pin?

Question

I'm building a simple ring counter with two jk flip-flop. I'm using 74HC76 integrated that contains two jk.

This is the integrated pinout:

This is my schematic (very basic):

^{simulate this circuit – Schematic created using CircuitLab}

To build a working circuit it's necessary that you connect pin 5 to Vcc (5V) and pin 13 to GND. But...

The problem is that in the circuit that I'm observing on my desk the ring counter only works when pin 5 (Vcc) is unwired. It means that if I wire pin 5 to Vcc I got always 10 output. Is that possible? If no what's the matter of my circuit?

I've tried all:

• I've substituted the integrated -> same problem
• I've checked another time pin connection -> same problem
• I've changed wires -> same problem

EDIT: I initialize the first jk with 1 using a button and preset and clear pin.

Answer 1

Without a resistor in series with each LED is is likely that the corresponding Q output is being dragged down to a voltage that is not high enough to be recognized as a legitimate high level by the following J input.

Answer 2

On CMOS parts, all unused inputs must be connected, directly or through a resistor, to either Vcc or ground, whichever is required to make the part work as required. For the Preset pin where you have a switch to pull it to ground, you must also have a resistor (5K - 10K) to Vcc to ensure the preset pin is high when the switch is not taking it low.