Step 1: Identify the outermost energy level and the number of electrons. The problem states that the atom has two electrons in the outermost third energy level. This means $n=3$ is the highest principal quantum number with occupied electrons, and there are 2 electrons in this level. Step 2: Determine the orbital filling for the third energy level. Within the third energy level, orbitals fill in the order $3s$, then $3p$, then $3d$. Since there are only two electrons in this level, they will occupy the $3s$ orbital. So, the configuration for the outermost third energy level is $3s^2$. Step 3: Write the electron configuration for the inner, filled energy levels. For the third energy level to be the outermost occupied level, all preceding energy levels must be completely filled. The first energy level ($n=1$) is filled with $1s^2$. The second energy level ($n=2$) is filled with $2s^2 2p^6$. Step 4: Combine all filled orbitals to get the complete electron configuration. Combining the configurations from Step 2 and Step 3: $$1s^2 2s^2 2p^6 3s^2$$ This configuration corresponds to an atom with $2+2+6+2 = 12$ electrons, which is Magnesium (Mg). The electron configuration using s and p notations is: $$\boxed{\text{}1s^2 2s^2 2p^6 3s^2\text{}}$$ Send me the next one 📸