C2HClF2 (2-Chloro-1,1-difluoroethylene) oxidation states

Oxidation State Calculator

Oxidation states of C₂HClF₂ (2-Chloro-1,1-difluoroethylene):

Oxidation States

Oxidation states of C₂HClF₂

Element	Symbol	Charge	Count
Carbon	C	0	1
Carbon	C	+2	1
Chlorine	Cl	-1	1
Fluorine	F	-1	2
Hydrogen	H	+1	1

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Compound properties

Calculating oxidation states

To calculate oxidation states of atoms in a compound, enter its chemical formula and click 'Calculate'. In chemical formula you may use:

Any chemical element. Capitalize the first letter in chemical symbol and use lower case for the remaining letters: Ca, Fe, Mg, Mn, S, O, H, C, N, Na, K, Cl, Al.
Functional groups: D, T, Ph, Me, Et, Bu, AcAc, For, Tos, Bz, TMS, tBu, Bzl, Bn, Dmg
parenthesis () or brackets [].
Common compound names.

Examples of compounds with oxidation states: H2O, CO2, CH4, NH3, HCl, H2SO4, KMnO4, NaCl, CaCl2, FeCl3, Al2O3, methane, ammonia, water, carbon dioxide.

Oxidation state calculator shows the oxidation states of atoms in chemical compounds through visual diagrams.

What is an oxidation state?

An oxidation state (also called oxidation number) is the hypothetical charge of an atom if all of its bonds to other atoms are fully ionic. It describes the degree of oxidation (electron loss) of an atom in a chemical compound. Oxidation states can be positive, negative, or zero, and they help predict chemical formulas and reactions. The sum of oxidation states in a neutral compound must equal zero.

How are oxidation states calculated?

This calculator determines oxidation states by analyzing the molecular structure and electronegativity of atoms. Here's how the process works:

Step 1: Molecular structure analysis

The calculator first analyzes the chemical bonds in the molecule to understand how atoms are connected. Each bond represents a pair of electrons shared between atoms.

Step 2: Electronegativity-based electron assignment

For each bond, electrons are assigned to the more electronegative atom:

If atom A is more electronegative than atom B, all bonding electrons go to atom A
If both atoms have equal electronegativity, electrons are shared equally
Lone pair electrons always belong to the atom they're on

Step 3: Oxidation state calculation

The oxidation state is calculated as:
Oxidation state = Valence electrons - Assigned electrons

Where:

Valence electrons = number of electrons in the outermost shell of the neutral atom
Assigned electrons = electrons 'owned' by the atom based on electronegativity

Example: Water (H₂O)

Let's see how this works for water:

Oxygen (electronegativity 3.44) is more electronegative than hydrogen (2.20)
In each O-H bond, both electrons are assigned to oxygen
Oxygen also has 2 lone pairs (4 electrons) that belong to it
Total electrons assigned to oxygen: 4 (from bonds) + 4 (lone pairs) = 8
Oxygen's oxidation state: 6 (valence) - 8 (assigned) = -2
Each hydrogen gets 0 electrons from bonds
Each hydrogen's oxidation state: 1 (valence) - 0 (assigned) = +1

Traditional rules for determining oxidation states

The following rules provide a quick way to determine oxidation states without detailed calculations:

Rule 1: Pure elements

The oxidation state of any pure element is 0.
Examples: Na, Cl₂, O₂, S₈ all have oxidation states of 0.

Rule 2: Monatomic ions

The oxidation state of a monatomic ion equals its charge.
Examples: Na⁺ has oxidation state +1, Cl⁻ has oxidation state -1.

Rule 3: Oxygen

Oxygen typically has an oxidation state of -2 in compounds.
Exception: In peroxides (H₂O₂), oxygen has an oxidation state of -1.

Rule 4: Hydrogen

Hydrogen typically has an oxidation state of +1 in compounds.
Exception: In metal hydrides (NaH), hydrogen has an oxidation state of -1.

Rule 5: Fluorine

Fluorine always has an oxidation state of -1 in compounds.

Rule 6: Neutral compounds

The sum of oxidation states in a neutral molecule must equal zero.
Example: In H₂O, H has +1 and O has -2, so (2 × +1) + (-2) = 0.

Rule 7: Polyatomic ions

The sum of oxidation states in a polyatomic ion equals the ion's charge.
Example: In SO₄²⁻, S has +6 and O has -2, so (+6) + (4 × -2) = -2.

Example calculation: H₂SO₄

Let's determine the oxidation state of sulfur in sulfuric acid (H₂SO₄):

H has oxidation state +1 (rule 4)
O has oxidation state -2 (rule 3)
Let S have oxidation state x
Sum must equal 0: (2 × +1) + x + (4 × -2) = 0
Solving: 2 + x - 8 = 0, so x = +6
Therefore, sulfur has oxidation state +6 in H₂SO₄

Lesson on oxidation states

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