Electrons, energy levels and atomic orbitals — practice questions

1. Identify the number of electrons: A neutral silicon atom has a proton number (Z) of 14, so it has 14 electrons.
2. Fill orbitals according to the Aufbau principle:
   • The 1s orbital takes 2 electrons: $1s^2$. (12 electrons remaining)
   • The 2s orbital takes 2 electrons: $1s^2 2s^2$. (10 electrons remaining)
   • The 2p sub-shell takes 6 electrons: $1s^2 2s^2 2p^6$. (4 electrons remaining)
   • The 3s orbital takes 2 electrons: $1s^2 2s^2 2p^6 3s^2$. (2 electrons remaining)
   • The 3p sub-shell takes the final 2 electrons: $1s^2 2s^2 2p^6 3s^2 3p^2$.

Full electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^2$. 3. 'Electrons in boxes' for valence electrons (n=3):

• The valence shell is the outermost shell, n=3. It contains the 3s and 3p sub-shells.
• The 3s orbital is filled with two electrons of opposite spin.
• For the 3p orbitals, Hund's rule applies. The two electrons will occupy separate p orbitals with parallel spins.

3s 3p [↑↓] [↑ ][↑ ][ ]

1. Write the configuration for the neutral Fe atom (26 electrons): Following the filling order (...3p, 4s, 3d): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$. It is often helpful to write this with shells grouped: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$.
2. Determine which electrons to remove for Fe²⁺: We need to remove 2 electrons. The highest principal quantum shell is n=4. Therefore, we remove the two electrons from the 4s orbital.
3. Write the final configuration for Fe²⁺: Removing the $4s^2$ electrons leaves: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$. This can also be written using noble gas notation as $[Ar] 3d^6$.