The mean lifetime of the first excited state of B12 at 0.95 MeV was measured by the Doppler shift attenuation technique using the B11(d, p)B12 reaction to populate the state. The result is τ=(3.4+/-1)×10-13 sec. Sum rules are invoked to show that the deexcitation of this state to the Jπ=1+ B12 ground state cannot be predominantly E2 so that the spin parity of the B12 first excited state cannot be 3+. The angular distribution of the ground-state decay from this level was measured at a deuteron energy of 0.8 MeV. The result rules out a zero-spin assignment. It is concluded that the available experimental evidence indicates 2+ for the spin parity of the B12 first excited state.