Oxidation state of F is always either ______ or _____ .

Answer

To determine the oxidation states of fluorine, we can follow these steps:

### Step 1: Identify the Group of Fluorine
Fluorine is located in Group 17 of the periodic table. This group is known as the halogens.

**Hint:** Remember that elements in the same group often have similar chemical properties.

### Step 2: Understand the Electronic Configuration
Fluorine has an electronic configuration of 2,7, which means it has 7 electrons in its outermost shell.

**Hint:** The number of valence electrons can help predict how an element will react and what charge it may carry.

### Step 3: Determine the Tendency to Gain Electrons
Due to having 7 valence electrons, fluorine has a strong tendency to gain 1 electron to achieve a stable octet configuration. This makes it highly electronegative.

**Hint:** Electronegativity is a key factor in determining how an element will interact with others.

### Step 4: Analyze the Oxidation States
1. **When bonded to another fluorine atom:** In a diatomic molecule (F2), both atoms share electrons equally, resulting in an oxidation state of 0.

2. **When bonded to an electropositive element:** Fluorine will gain an electron from the electropositive element, resulting in an oxidation state of -1.

**Hint:** Consider the nature of the bond (covalent vs ionic) to understand the resulting oxidation state.

### Step 5: Conclude the Possible Oxidation States
Based on the above analysis, the oxidation states of fluorine can be summarized as:
- 0 (when bonded to itself)
- -1 (when bonded to more electropositive elements)

**Final Answer:** The oxidation state of F is always either **0** or **-1**.