What type of bond is formed between sodium and iodine in sodium iodide (NaI)? A. ionic B. metallic C. polar covalent D. nonpolar covalent
The Correct Answer and Explanation is:
The correct answer is A. ionic.
In sodium iodide (NaI), the bond between sodium (Na) and iodine (I) is an ionic bond. This is because of the significant difference in electronegativity between sodium and iodine, which causes the transfer of electrons from sodium to iodine.
Here’s a breakdown of the process:
- Electronegativity Difference: Sodium is a metal with a low electronegativity (0.93 on the Pauling scale), while iodine is a nonmetal with a much higher electronegativity (2.66 on the Pauling scale). This difference in electronegativity is key to the ionic bonding process.
- Electron Transfer: Sodium (Na) has one electron in its outermost shell, which it can easily lose to achieve a stable electron configuration (similar to the noble gas neon). When sodium loses this electron, it forms a Na⁺ ion, becoming positively charged.
- Formation of Iodide Ion: Iodine (I) has seven electrons in its outer shell and needs one more electron to complete its octet. When sodium loses its electron, iodine readily gains that electron, becoming I⁻, a negatively charged ion.
- Electrostatic Attraction: The oppositely charged ions, Na⁺ and I⁻, experience a strong electrostatic force of attraction, which is the characteristic feature of an ionic bond. This attraction holds the two ions together, forming the ionic compound sodium iodide (NaI).
In summary, the bond in NaI is ionic because it involves the transfer of an electron from sodium to iodine, creating oppositely charged ions that are attracted to each other by electrostatic forces.
