There are two band gaps in your case, Elder. For spin-up and spin-down. But You can also run restricted Hartree-Fock calculation (if the number of electrons is even) for paired electrons with spin zero and receive one gap. Actually, You must explore every spin state separatly.

Good luck!

I ran a restricted calculation (using the B3LYP funcional), but I realized that the charge and spin multiplicity of my structure (e.g when I doped with Cobalt) were 0 and 2, respectively. So I got these results:

NBasis: 1720

HOMO: 379     378

Groups: 0

Alpha MO         eV        Symmetry  Beta MO           eV        Symmetry

1720  L+1340  267.3   A                1720 L+1341   267.31 A

1719  L+1339  99.54   A                1719 L+1340   99.55    A

         .                      .        .                            .                  .         .

         .                      .        .                            .                  .         . 

381  L+1          -2.56      A                 381 L+2           -2.38    A

380  LUMO     -2.97      A                 380 L+1           -2.8       A

379  HOMO    -4.24      A                 379 LUMO       -3.08    A

378  H-1          -4.45      A                 378 HOMO      -4.22    A

377  H-2          -4.86      A                 377 H-1            -4.61    A

And I do not know what I have to choice to be the band gap energy of the structure: If 1.27 eV (alpha MO) or 1.14 eV (beta MO) 

You have one more Alpha electron. It is not spin zero state. So, You have again two gaps, but they are essentially the same. In solid state language You have half-filled zone. In your case there is no gap in the semiconductor sense.

So, Do I have to rebuild the structure and run the calculation again? Because I have another structure that have the same properties: Charge 0 and spin multiplicity 2 (doping using Iridium). When I did these structures I did not understand why this unpaired alpha electron appearance.