Resolve this electronics paradox if you can!?

Dissipation in resistive connecting wires could be made to vanish by using superconductors say. So the answers others give relating the missing energy to Ohmic dissipation are incorrect. One has to focus on the dynamics instead. The electric field was changing in time between the initial and final static situations. This must have given rise to radiation, which carried away the energy you are missing.

Long time ago I had to do the calculation for an EM-exercises class I was teaching as a PhD student. I have forgotten the details but I remember the magnetic dipole radiation dominated and the calculation involved the second derivative wrt time of the current. The radiated energy equaled CV^2 / 4, exactly the amount " missing" ...

When you connect the discharged capacitor in parallel with the charged one, a significant current will flow, decreasing exponentially as the charge becomes equalized. This current will be limited by the resistance of the conductors and the plate material of the capacitors, dissipating energy as I^2*R = heat.

You can see this when you take a charged capacitor and short circuit it - you will see a spark and perhaps even hear something like a click when connection is made. A capacitor with a large amount of energy stored in it may even weld the shorting conductor to its terminals.

Voltage is the amount of energy required to move a charge from one place to another. In a parallel plate capacitor (the only type that I studied in my Physics II course) one plate contains only a positive charge and the other contains a negative charge. There is obviously a force of attraction between these two charges.

We know by KVL that both the voltage across the two capacitors must be the same. The energy loss comes from the fact that the negative charge has to move from one capacitor (with the positive charge attracting it) to the other one. This movement of charge decreases the overall energy available.

Well you connected a second capacitor in parallel, which now makes it equal to one 2 Farad capacitor. You can do the rest.

The energy is lost to heat and electrical sparks. And radiated as heat and EM radiation.

When you connect a charged cap to an uncharged one, you have very high currents as the caps equalize, and the current is so high that you have a spark. Energy is lost due to the high currents and the arc.

Bottom line, charge is conserved, energy is not.

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Moving charge also requires energy.

The rest of the energy was utilised to redistribute the charges.

That always happens. energy will be heating the connecting wires, which is known as Joule heat.