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Problem: Explain why Boron (a second-period element) can only form the molecule BF4(-) with a maximum of 4 bonds, while Aluminum (right below Boron in the third period) can form AlF6(3-) with 6 bonds around it. Step 1: Identify the period of each element. Boron is in the second period, and Aluminum is in the third period. Step 2: Check the electron housing capacity (orbitals). Boron only has 2s and 2p orbitals available in its valence shell, meaning it can hold a maximum of 8 electrons (octet rule, 4 pairs of electrons). Step 3: Analyze Aluminum. Aluminum is in the third period, so it has vacant 3d orbitals available in addition to its 3s and 3p orbitals. Step 4: Relate this to the molecules. In BF4(-), Boron has 4 pairs of electrons around it and cannot accept any more because its small backpack is full. In AlF6(3-), Aluminum uses its vacant d-orbitals to expand its octet and accommodate 6 pairs (12 electrons) of electrons from six Fluorine atoms. Conclusion: Boron shows anomalous behavior by strictly obeying the octet rule due to a lack of d-orbitals, unlike Aluminum.