What is the oxidation state of each element in K2Cr2O7?

K2Cr2O7, also known as potassium dichromate, is a common chemical compound used in various industrial applications. It is an orange-red crystalline solid that is highly soluble in water. The oxidation state of each element in this compound is crucial to understanding its chemical properties and reactions. In this article, we will explore the oxidation states of potassium (K), chromium (Cr), and oxygen (O) in K2Cr2O7.

Potassium (K) is an alkali metal that belongs to Group 1 of the periodic table. It has a single valence electron, which it readily loses to form a +1 oxidation state. In K2Cr2O7, there are two potassium atoms, so the total contribution from potassium is +2.

Chromium (Cr) is a transition metal that can exhibit multiple oxidation states. In K2Cr2O7, it has an oxidation state of +6. This is because oxygen (O) typically has an oxidation state of -2 in most compounds. Since there are seven oxygen atoms in K2Cr2O7, the total contribution from oxygen is -14. The sum of the oxidation states of all elements in a compound must equal zero, so the oxidation state of chromium can be calculated as follows:

2(K) + 2(Cr) + 7(O) = 0
2(+1) + 2(x) + 7(-2) = 0
2 + 2x – 14 = 0
2x = 12
x = +6

Therefore, the oxidation state of chromium in K2Cr2O7 is +6.

Oxygen (O) is a nonmetal that usually has an oxidation state of -2 in most compounds. In K2Cr2O7, there are seven oxygen atoms, so the total contribution from oxygen is -14.

In summary, the oxidation states of each element in K2Cr2O7 are as follows:
– Potassium (K): +1
– Chromium (Cr): +6
– Oxygen (O): -2

Understanding the oxidation states of these elements is essential for predicting the chemical reactions and properties of potassium dichromate.

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