Complete this table. name empirical formula barium iodide rubidium fluoride
The Correct Answer and Explanation is:
Of course. Here is the completed table with the correct empirical formulas, followed by a detailed explanation.
| name | empirical formula |
| barium iodide | BaI₂ |
| rubidium fluoride | RbF |
Explanation:
An empirical formula represents the simplest whole-number ratio of atoms or ions within a chemical compound. For ionic compounds, which are typically formed between a metal and a nonmetal, the chemical formula is the empirical formula. The key to determining this formula is to balance the electrical charges of the ions to form a neutral compound.
1. Barium Iodide (BaI₂)
To find the formula for barium iodide, the constituent ions and their charges must be identified.
- Barium (Ba) is an alkaline earth metal, located in Group 2 of the periodic table. Elements in this group consistently lose two valence electrons to form a stable cation with a +2 charge. Thus, the barium ion is Ba²⁺.
- Iodine (I) is a halogen, found in Group 17. Halogens tend to gain one electron to achieve a stable electron configuration, forming an anion with a -1 charge. The iodide ion is I⁻.
For the compound to be neutral, the sum of the positive and negative charges must be zero. One barium ion (Ba²⁺) has a charge of +2. To balance this, two iodide ions (I⁻) are needed, as each has a charge of -1. The calculation is: (1 × +2) + (2 × -1) = 0.
This results in a ratio of one barium ion to two iodide ions. The cation (Ba) is written first, and the number of ions is indicated by a subscript. Therefore, the empirical formula for barium iodide is BaI₂.
2. Rubidium Fluoride (RbF)
A similar process is followed for rubidium fluoride.
- Rubidium (Rb) is an alkali metal from Group 1. These metals lose one valence electron to form a cation with a +1 charge. The rubidium ion is Rb⁺.
- Fluorine (F) is the most electronegative halogen in Group 17. It gains one electron, forming the fluoride anion with a -1 charge, F⁻.
In this case, the +1 charge of the rubidium ion perfectly balances the -1 charge of the fluoride ion. The ratio is one-to-one. The calculation is: (1 × +1) + (1 × -1) = 0.
Because the simplest whole-number ratio is 1:1, only one of each ion is needed for a neutral compound. When the subscript is 1, it is omitted from the formula. The empirical formula for rubidium fluoride is RbF.
