Consider the six second- and third-row elements in groups 4A–6A of the periodic table:


Possible structures for the binary fluorides of each of these elements in its highest oxidation state are shown below.


(a) Identify the nonfluorine atom in each case, and write the molecular formula of each fluoride.

Verified step by step guidance

1. Identify the elements in groups 4A–6A of the periodic table from the second and third rows.
2. List the elements: Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb) for group 4A; Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi) for group 5A; Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po) for group 6A.
3. Determine the highest oxidation state for each element, which is equal to the group number.
4. Write the molecular formula for the binary fluorides by combining each element with fluorine (F) in its highest oxidation state.
5. For example, for Carbon in group 4A, the highest oxidation state is +4, so the formula is CF4. Repeat this for each element.

Key Concepts

Periodic Table Groups

The periodic table is organized into groups (columns) that share similar chemical properties. Groups 4A to 6A include elements such as carbon, silicon, germanium, tin, lead, and their corresponding nonmetals. Understanding the position of these elements helps predict their behavior and the types of compounds they form, including binary fluorides.

Oxidation States

Oxidation states indicate the degree of oxidation of an atom in a compound, reflecting the number of electrons lost or gained. In the context of binary fluorides, the highest oxidation state of an element determines how many fluorine atoms can bond with it. This is crucial for writing the correct molecular formulas for the fluorides of the specified elements.

Binary Compounds

Binary compounds consist of two different elements. In this case, the binary fluorides are formed by combining the nonfluorine atom from groups 4A to 6A with fluorine. Identifying the nonfluorine atom and understanding how it interacts with fluorine is essential for determining the correct molecular formulas for these compounds.