Still bad idea. No LED have an exackt Vf. It is all a question of tolerances. Some may have 4.8V - in which case they will get 0.2V too much resulting in a huge current. Some may have 5.2V, in which case they will be very weak.

When using a LED, you must always have a supply voltage that have enough wiggle room for the resistor.

With a 6V supply and diodes that varies between 4.8 and 5.2V, the voltage over the resistor will vary between 0.8 and 1.2V. That represents a current difference (and light difference) of 1.2/0.8 = 1.5, i.e. some diodes may get 50% more current than other diodes.

With a 7V supply, the resistor may get between 1.8 and 2.2V, which gives a potential variance between the diodes of 2.2/1.8 = 1.22.

With a 8V supply, you may get 3.2/2.8 = 1.14.

The more voltage you put over the resistor, the more even your current will be, and the less variance depending on production tolerances of the diodes. For best result, you can replace the resistor with a constant-current drive. Because it will actively regulate the current by dynamically adjusting its internal resistance, you will not need as many volts for the regulation.

Back to your original question. As noted, the more voltage over the resistor, the more even current. You are walking in the other direction, trying to reduce the voltage over the resistor to 0V, and the resistance of the resistor to 0 ohm. For an ideal diode, that would mean an infinite current through a 5.2Vf diode.