Q:

What destroys a LED in the reverse direction?

When a LED is included in a circuit that applies a reverse voltage that exceeds the reverse breakdown, a reverse current can flow and the LED might be destroyed.  But what is it that actually destroys the LED:  is it the reverse voltage itself, or is it the reverse current that is made to flow, or is it simply the overall power dissipation caused by the reverse current and voltage exceeding the device rating?  Or something else?
So, for instance, if I connect a 12 volt source in reverse to a LED that breaks down at 5 volts, via a resistor, the passage of reverse current will cause a voltage drop across the resistor which in turn can limit the voltage on the device to its reverse breakdown value (and thereby define the current that would flow) - rather similar to what happens in the forward direction.  Would this in itself destroy the LED as long as the total power was within the LED rating?
Normally of course, one would place a regular diode in parallel with the LED in the reverse direction to limit the LED reverse voltage to 0.7 volts or so, but there may be situations where this might not be possible or economic to do.  I am just trying to understand how much circuit design flexibility I might have to meet different requirements.
And if it is possible to expose a LED to a reverse voltage, what precautions should be taken to avoid damage, and which spec parameters are relevant?

A:

ESD appears to cause damage due to hot spots or some other localized damage. I have seen heterojunction LED failures that appear to be partial. 
Failure with DC current in reverse is probably related to power dissipation, but it might be unwise to depend on it. Breakdown could be quite high, hence the allowable current could be quite low (maybe less than 1mA). 
Safest is to follow the LED data sheet recommendation- usually 5V or so reverse is guaranteed. Many types of LEDs have much higher actual reverse voltage breakdown (perhaps 15V to 70V), but it's unwise to depend on it- the LED maker could change the chip supplier or process or purchasing could go to a different vendor. 
The typical situation where LEDs are exposed to reverse voltage is when they are operated in a multiplexed configuration- they will see up to the supply voltage in reverse. It's not really a good idea for efficiency to make the supply voltage much higher than the sum of the series LED forward voltages (often, but not always, just one LED is used). For example, you might use 5V for a single 2-3V LED array, or 12V for an array of series strings with 6-9V per string. Since the individual LEDs can take 5V each (usually guaranteed), you'd be fine in either case.
See this nice instructables gif: 

A:

Most datasheets from famous and respected manufacturers including Vishay and others, show breakdown voltage of 5V and reverse current of 10 to 50uA.  That is not true.  I just tested white, red and green leds, with a 0-30V power supply and a 1k resistor in series, measuring voltage across the led, and current in the circuit.   This is the result:  White 9V=0.4uA, 13V=1uA  /  Red 5.3V=0.3uA, 6.7V=0.5uA, 12.5V=1uA  /  Green 5.11V=0.3uA, 6.5V=0.5uA, 9.9V=1uA.   So, no led presented anything close to half a microAmper at 5V as stated the manufacturers, and only right after 10V they presented ONE microAmpere.  That is not enough to damage the component.  According to what I read, it would need a minimum of 1mA reversed to increase the depletion area.
Wagnerlip

