I have a resistor pack 5 pins through-hole with a value of 4.7 k ohms. I want to connect it with the DS1307 8-pin IC on breadboard to communicate with mcu because I2C requires pull-up resistors for the SDA and clock pins.

Since I only need three pins - one for supply and the other two for SDA and SCL - do I need to cut off the other unused pins of the resistors pack?

 

All of the resistors are common on one end, which presumably you're connecting to +5 volts. The other side of one resistor connects to SCL. Another to SDA.

The other two resistors aren't connected to anything. There's no circuit through them. Why would they matter?

 

All of the resistors are common on one end, which presumably you're connecting to +5 volts. The other side of one resistor connects to SCL. Another to SDA.

The other two resistors aren't connected to anything. There's no circuit through them. Why would they matter?

Please find the attached picture illustrating the current situation regarding the DS1307 socket and the 4-pin headers. As you can see, there is a hole on the board designated for pull-up resistors. However, considering that only three pins of the resistor package are required (one for VCC, and the others for SDA and SCL), it seems that the SQW pin is unnecessary for pull-up resistor functionality.

Therefore, I propose to cut the SQW pin and the other unused pin of the resistor package to optimize the configuration.

 

No, dont cut them off as there is a good chance you'll crack the substrate. There's no harm in pulling up SQW even if unused, and its almost certain that somewhere in your unfolding design you'll need a further pullup.