Study the table below and use it to answer the questions that follow: a) Complete the table to show the formulae of the oxides of A, E, G, and L. b) Select an oxide that reacts with both sodium hydrox

• Element A: Atomic number 11 (Sodium, Na). Group 1, valency +1.
• Element E: Atomic number 13 (Aluminum, Al). Group 13, valency +3.
• Element G: Atomic number 14 (Silicon, Si). Group 14, valency +4.
• Element L: Atomic number 16 (Sulfur, S). Group 16, common valencies +4 or +6.

Oxygen has a valency of -2.

• Oxide of A: $\mathrm{Na_2O}$
• Oxide of E: $\mathrm{Al_2O_3}$
• Oxide of G: $\mathrm{SiO_2}$
• Oxide of L: $\mathrm{SO_2}$ (sulfur dioxide) or $\mathrm{SO_3}$ (sulfur trioxide). $\mathrm{SO_2}$ is a common simple molecular oxide. Answer:
• Oxide of A: $\boxed{\mathrm{Na_2O}}$
• Oxide of E: $\boxed{\mathrm{Al_2O_3}}$
• Oxide of G: $\boxed{\mathrm{SiO_2}}$
• Oxide of L: $\boxed{\mathrm{SO_2}}$

Answer: Oxide of E ($\mathrm{Al_2O_3}$)

• Element E: Atomic number 13 (Aluminum, Al). Atomic radius = 0.130 nm.
• Element M: Atomic number 17 (Chlorine, Cl). Atomic radius = 0.099 nm.

Answer: Element M has a smaller atomic radius than element E. This is because both elements are in the same period, but element M has a higher nuclear charge (more protons) than element E. The increased nuclear attraction pulls the outer electrons closer to the nucleus, resulting in a smaller atomic radius.

• Element E: Aluminum (Al), forms $\mathrm{Al^{3+}}$ ions.
• Element M: Chlorine (Cl), forms $\mathrm{Cl^-}$ ions.

Answer: $\boxed{\mathrm{AlCl_3}}$

• Oxide of D: $\mathrm{MgO}$ (Magnesium oxide). Boiling point of oxide = 3075 °C.
• Oxide of L: $\mathrm{SO_2}$ (Sulfur dioxide). Boiling point of oxide = -72 °C.

Answer: The oxide of D ($\mathrm{MgO}$) has a very high boiling point because it is an ionic compound with a *