Why does atomic charge not balance in nuclear reactions?

In the discovery of neutron, the bombardment of beryllium with alpha particles is well known.

$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C + ^0_1n}$$

I don't understand how the electrons add up. As alpha particles do not have electrons, the total electrons should be 4 accounting for the electrons of beryllium, but after bombardment, there is a deficiency of 2 electrons in carbon atom. Shouldn't the $$\ce{^12_6C}$$ atom be positively charged? Like:

$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C^2+ + ^0_1n}$$

• High energy alpha particles yield high energy intermediates that, yes, may be highly charged. – Zhe Oct 13 '18 at 16:30
• @Karl While I agree with your opinion that this has a physical bent to it, I nevertheless find it difficult to accept the notion that the composition of atoms would be considered off-topic for chemistry. – Zhe Oct 13 '18 at 16:32
• In nuclear reactions we leave charge balance for other processes. There are enough electrons in the universe to neutralize the C at some point. – Jon Custer Oct 13 '18 at 16:42
• Sure enough, carbon will be positively charged, but it will have more urgent issues to worry about at the moment. – Ivan Neretin Oct 13 '18 at 18:39

There are two reasons why nuclear reactions don't consider (or often account for) charge.

The first is that charge is usually irrelevant to the reactions. If we are looking at changes in nuclei we can for all practical purposes ignore the electrons that would be associated with the nuclei under normal conditions. They are just irrelevant to the nuclear reaction taking place.

The second is that they do account for charge, but only in nuclear processes. Some radioactivity (isolated neutrons decay or any nucleus undergoing beta emission) can lead to the decay of neutrons into a proton and electron and some other minor particles. The electron clouds that would normally surround the nucleus are still irrelevant to these processes but the equation for the nuclear reaction should account for the emitted particles. When writing these equations only the nuclei (and the emitted particles) are accounted for and any net change of charge is ignored as a description of the nuclei involved implicitly describes the net nuclear charge (in the same way that many organic chemical structures don't explicitly draw every hydrogen as chemists can fill them in knowing that the total valence around carbon should add to 4).

Your right, the carbon will be charged, but it is not typically represented that way because it is a nuclear reaction and it is superfluous to consider charge since the charges of the atoms do not participate in the reaction (spectator particles).

$$\ce{^9_4Be + ^4_2He -> ^12_6C + ^0_1n }$$

Given that the carbon is in a beryllium target (maybe metal, maybe oxide) the carbon could just steal the electrons from the beryllium, you could think of it more like:

$$\ce{Be (s) -> Be^2+ +2e-}\\ \ce{^9_4Be + ^4_2He^2+ -> ^12_6C^2+ + ^0_1n ->[2e-] ^12_6C + ^0_1n}$$

But again charge is superfluous to the reaction and materials can have unbalanced charges (static electricity).