Potassium iodide (KI) exhibits predominantly ionic bonding. The
and
ions have electron structures that are identical to which two inert gases?
The Correct Answer and Explanation is:
Potassium iodide (KI) exhibits predominantly ionic bonding, where potassium (K) donates an electron to become a positively charged ion (K⁺), and iodine (I) accepts that electron to become a negatively charged ion (I⁻). These ions each have electron structures that resemble the electron configurations of the nearest inert gases.
- Potassium ion (K⁺): When potassium loses one electron to become K⁺, it achieves an electron configuration similar to argon (Ar), which has an atomic number of 18. Potassium (K) originally has 19 electrons, but after losing one electron, K⁺ has 18 electrons, matching the electron configuration of argon, which is [Ne]3s²3p⁶.
- Iodide ion (I⁻): When iodine gains an electron to form I⁻, its electron configuration resembles that of xenon (Xe), which has an atomic number of 54. Iodine (I) has 53 electrons, and by gaining one additional electron, I⁻ has 54 electrons, mirroring the electron configuration of xenon, which is [Kr]4d¹⁰5s²5p⁶.
The reason KI forms ionic bonds is due to the significant difference in electronegativity between potassium and iodine. Potassium, being a metal with low electronegativity, readily loses its outer electron, while iodine, a non-metal with higher electronegativity, readily accepts the electron. This results in the formation of oppositely charged ions, K⁺ and I⁻, which are held together by strong electrostatic forces of attraction, characteristic of ionic bonds.
The tendency of these ions to achieve electron configurations resembling those of inert gases is a driving force behind the formation of ionic compounds like KI. This process allows both ions to reach a stable, low-energy state akin to that of noble gases, which are chemically inert due to their full outer electron shells.
