Which of the following best describes the Lewis dot structure for the iodide ion in nickel(III) iodide (NiI3)?

• A. Each I atom has 6 valence electrons, represented by 2 lone pairs and 2 bonding pairs with Ni.
• B. Each I atom has 8 valence electrons, represented by 4 lone pairs and no bonding pairs.
• C. Each I atom has 8 valence electrons, represented by 3 lone pairs and 1 bonding pair with Ni.
• D. Each I atom has 7 valence electrons, represented by 3 lone pairs and 1 bonding pair with Ni.

Answer: (C)

Verified step by step guidance

1. Recall that the iodide ion (I⁻) is derived from iodine, which has 7 valence electrons in its neutral state because it is in group 17 of the periodic table.
2. Since the iodide ion carries a negative charge (I⁻), it has gained one extra electron, increasing its total valence electrons to 8.
3. In the Lewis dot structure, these 8 valence electrons are arranged as 4 pairs: 3 lone pairs (6 electrons) and 1 bonding pair (2 electrons) shared with the nickel atom.
4. The bonding pair represents the single covalent bond between the iodide ion and the nickel (Ni) atom, while the 3 lone pairs remain localized on the iodide ion.
5. Therefore, the correct Lewis structure for each iodide ion in NiI₃ shows 8 valence electrons: 3 lone pairs and 1 bonding pair with Ni.