Question

Identify the following hydrides as ionic, metallic, or molecular: (a) \(\mathrm{BaH}_{2}\), (b) \(\mathrm{H}_{2} \mathrm{Te}\), (c) \(\mathrm{TiH}_{1.7}\).

Short answer

(a) \(\mathrm{BaH}_{2}\) is an ionic hydride, as it consists of a metal (Ba) and a non-metal (H) that form ionic bonds. (b) \(\mathrm{H}_{2} \mathrm{Te}\) is a molecular hydride, as it consists of two non-metals (H and Te) that form covalent/molecular bonds. (c) \(\mathrm{TiH}_{1.7}\) is a metallic hydride, as it involves a transition metal (Ti) and a non-metal (H) that form non-stoichiometric metallic bonds.

Step-by-step solution

(a) Identify type of hydride for \(\mathrm{BaH}_{2}\)

For BaH2, we will proceed by analyzing the participating elements. Barium (Ba) is an alkaline earth metal belonging to Group 2 in the periodic table, and Hydrogen (H) is a non-metal in Group 1. Generally, when metals form bonds with non-metals, they transfer electrons and form ionic compounds. Hence, we can conclude that BaH2 is an ionic hydride.

(b) Identify type of hydride for \(\mathrm{H}_{2} \mathrm{Te}\)

H2Te consists of Hydrogen (H) and Tellurium (Te). Tellurium (Te) is a non-metal in Group 16 of the periodic table. As we have two non-metals here, they typically form covalent/molecular bonds. Therefore, H2Te is a molecular hydride.

(c) Identify type of hydride for \(\mathrm{TiH}_{1.7}\)

TiH1.7 involves Titanium (Ti) and Hydrogen (H). Titanium (Ti) is a transition metal from the d-block, and combines with the non-metal Hydrogen (H). Non-stoichiometric hydrides, such as TiH1.7, usually involve transition metals and are characterized by metallic bonding. Hence, TiH1.7 is a metallic hydride.