An unknown element, X, reacts with sodium to form the compound Na2X. In other compounds, this element can accommodate up to 12 electrons instead of the usual octet. Which element could X be? A) Be B) Al C) F D) Cl E) Sc F) As G) Y

The correct answer is E) Sc.

Here's the logical reasoning:

• Be (Beryllium) typically forms di-positive ions (Be^2+), which implies that it would react with sodium to form BeNa2 if it were to form a neutral compound.
• Al (Aluminum) typically forms tri-positive ions (Al^3+), consistent with its place in Group 13 of the periodic table, meaning it would more likely form AlNa3.
• F (Fluorine) and Cl (Chlorine) are halogens and typically form negative mono-valent ions (F^- and Cl^- respectively). They would respectively form NaF and NaCl with sodium.
• As (Arsenic) is a metalloid and can form negative three-valent ions. Typically, it would form AsNa3 if it were to react with sodium to produce a neutral compound.
• Y (Yttrium) forms tri-positive ions (Y^3+) and would react with sodium to form YNa3 under normal circumstances.
• Scandium (Sc), which is the element E from the options given, is in Group 3 of the periodic table and forms Sc^3+ ions. The formula Na2X indicates that X forms a hexavalent ion (X^6-), to balance out the two Na^+ ions. Scandium, which typically accommodates a +3 charge, can also form lower oxidation state compounds in which it is less commonly found. Moreover, Scandium can indeed expand its octet to accommodate 12 electrons in its valence shell because it has available d-orbitals in the third shell where more than 8 electrons can be accommodated. This stems from its position in the periodic table which allows transition metals like Scandium to have an expanded valence shell. This property fits the description of element X in Na2X, where element X seems to accommodate up to 12 electrons.