Question

This holds a sample of Barium iodide, $Ba{I_2}$, together:
A.Hydrogen bonding
B.Ionic bonding
C.Metallic bonding
D.Nonpolar covalent bonding
E.Polar covalent bonding

Answer 1

To answer this question, you must recall the type of bonding present in barium iodide. Barium is alkaline earth metals present in the group 2 of the modern periodic table and can lose electrons and iodine is a halogen in the group 17 and can gain electrons.

Complete step by step solution:
- We know higher the difference in electronegativities of two atoms, higher is the chances of formation of an ionic bond. Barium is a metal and iodine is a non-metal. Barium loses two electrons to gain stable noble gas like electronic configuration forming a bivalent cation and iodine accepts one electron for the same forming a monovalent anion.
- Two iodide ions are required to satisfy the valency and neutralize the charge of the barium cation and thus barium iodide is formed by the formation of ionic bonds between a barium cation and two iodide ions.
- Ionic bond is the chemical bond formed due to electrostatic attraction between cations and anions. In barium iodide ionic bonding is the force of attraction that holds the sample together.

Thus, the correct option is B.

Note: For ionic bonding to take place, there are two essential factors. The first is that the metal participating in the bond formation must have a low ionization potential. The second is that the non- metal should have a high electron affinity. Ionization potential is the energy required to lose an electron from the valence shell of an isolated atom in the gaseous state. Electron affinity is the energy released in addition of an electron to an isolated atom in the gaseous state.