1. Pin 2 must be taken LOW for it to activate the 555. The circuit shows a positive voltage being applied to pin 2. This will do nothing.
2. Connecting pin 2 to pin 4 and leaving them "open" as shown in the diagram above is very dangerous. These are fairly high-impedance pins and the consequences of leaving them open will be unpredictable.
3. Pin 2 cannot be left "open." The 10u will initially charge to 2/3 rail voltage and the voltage will be detected by pin 6. Pin 7 will then discharge the 10u and wait for a low to be detected by pin 2. Pin 2 will actually have no voltage on it but the pin requires a very small current (about 500 nano-amp) to activate the chip. If a static charge delivers this current, the chip will cycle. The outcome is unpredictable.
4. Pin 4 cannot be left "open." For pin 4 to reset the chip, it must be taken below 0.7v and supplied a current of 100uA. If it is left open you cannot guarantee the chip will operate.