How did they get 3? Do you always have to divide?
The oxidation state of Cr in the given oxide is +6. The electronic configuration of Cr can be written as [Ar] 3d5 4s1. In the +6 oxidation state, Cr loses its six electrons (5 electrons from 3d sub-shell and 1 electron from 4s orbital). On the other hand, the electronic configuration of oxygen atom is 1s2 2s2 2p4. Since it is a normal oxide, the general oxidation state of oxygen is taken to be -2. Let the number of oxygen atoms involved in the formula of the oxide be x. The oxidation state of one oxygen atom is -2. Hence, the oxidation state of x atoms will be -2x. The sum of oxidation states of the atoms is equal to zero in order to maintain the neutrality. This can be shown as 6 – 2x = 0. The value of x is obtained as 3. Hence, there are three oxygen atoms in the given oxide.
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