From what I understand, split receptacles are allowed to share a neutral since the two hots come from opposite sides of the panel and are 180 deg out of phase. Therefore, for purely resistive loads, the current in the common neutral should never exceed the wire rating.

Question: What about reactive loads? If capacitive and inductive loads are present, it is theoretically possible that the current in the common neutral would exceed the rating of the wire.

From what I understand, split receptacles are allowed to share a neutral since the two hots come from opposite sides of the panel and are 180 deg out of phase. Therefore, for purely resistive loads, the current in the common neutral should never exceed the wire rating.

Question: What about reactive loads? If capacitive and inductive loads are present, it is theoretically possible that the current in the common neutral would exceed the rating of the wire.

How does the code deal with this?

if you were able to google that much, why not the rest?

From what I understand, split receptacles are allowed to share a neutral since the two hots come from opposite sides of the panel and are 180 deg out of phase. Therefore, for purely resistive loads, the current in the common neutral should never exceed the wire rating.

Question: What about reactive loads? If capacitive and inductive loads are present, it is theoretically possible that the current in the common neutral would exceed the rating of the wire.

How does the code deal with this?

It doesn't. The example you make is only theoretical. It is solved with simple trig functions. If the voltage on first phase is sin(x) and say you put a pure capacitive load on it the current will follow sin(x+pi/2), where the
+pi/2 is the current lead of a capacitor. On the other phase you put an inductive load on it the current will be sin(x-pi/2-pi), the -pi/2 is the lag of an inductor and -pi is the alternate phase (180 degrees out of phase).

We know sin(x+pi/2) = sin(x-3pi/2) therefore the magnitude of the current is exactly 2x the value of either branch.

Again the odds of a highly (i.e. near pure inductor) inductive load already are nil in a home, especially one at exactly the branch circuit current rating. Similarly highly capacitive and at the full branch circuit current is also low. The combined probability of these two loads then being on opposite phases of a single split duplex... come on, not going to happen.....