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Question

Question: What is the oxidation state for the ion \(C{{r}_{2}}O_{7}^{2-}\) ....

What is the oxidation state for the ion Cr2O72C{{r}_{2}}O_{7}^{2-} .

Explanation

Solution

The oxidation state is nothing but the number of electrons gained or lost by an element during the formation of a chemical bond with other atoms. The oxidation state of an element will be positive or negative and it is going to depend on the electrons gained or lost by the element in a chemical reaction.

Complete answer:
- In the question it is asked to find the oxidation state of chromium in Cr2O72C{{r}_{2}}O_{7}^{2-} .
- We can use the following way to calculate the oxidation state of an element which is present in a chemical.
- We know that the oxidation of an oxygen in normal oxide is ‘-2’.
- In the given chemical there are seven oxygen atoms means the charge by the seven oxygen atoms
= (7) (-2) = -14.
- We can see that there are already two electrons which are present in the oxygen of the given chemical.
- Therefore the oxidation state of the chromium in Cr2O72C{{r}_{2}}O_{7}^{2-} is as follows.
2x + (-14) = -2
2x = 14-2
x = 6.
- We have taken ‘2x’ for two chromium atoms.
- Therefore the oxidation state of chromium in Cr2O72C{{r}_{2}}O_{7}^{2-} is ‘+6’.

Note:
We have to consider the charge of the chemical also while calculating the oxidation state of an element in the chemical. We have to count the number elements are present in the given chemical while calculating the oxidation state.