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Question: Bleaching powder \((CaOC{{l}_{2}})\) has two types of chlorine atoms with oxidation number of +1 and...

Bleaching powder (CaOCl2)(CaOC{{l}_{2}}) has two types of chlorine atoms with oxidation number of +1 and -1. It can be represented as:
A. Ca2+(ClO2−)Cl−C{{a}^{2+}}(Cl{{O}_{2}}^{-})C{{l}^{-}}
B. Ca2+(ClO3−)Cl−C{{a}^{2+}}(Cl{{O}_{3}}^{-})C{{l}^{-}}
C. Ca2+(ClO+)Cl−C{{a}^{2+}}(Cl{{O}^{+}})C{{l}^{-}}
D. Ca2+(ClO−)Cl−C{{a}^{2+}}(Cl{{O}^{-}})C{{l}^{-}}

Explanation

Solution

Check the oxidation states of all the molecules and whether they add up to the corresponding charge given in each option. The overall charge of the molecule should be 0 and the overall atoms present in the options and the molecule should also be equal. Calculate your answer based on this.

Complete answer:
According to the formation and molecular structure of CaOCl2CaOC{{l}_{2}} molecule or Bleaching powder, we see that 1 calcium atom, 1 oxygen atom and 2 chlorine atoms are present. Thus, we can automatically rule out options ‘A’ and ‘B’ as our answers since they contain 2 and 3 oxygen atoms respectively.
We know the charge of CaCa here is +2 and one of the chlorine atoms has the charge of -1. From this we can say that the ClOClO ion should have the charge of -1 for the molecule to be neutral. We can see this case in option ‘D’.
To verify that this is in fact the correct answer we will verify using the oxidation states. For the ClOClO ion to have a -1 charge; we know that oxygen has an oxidation state of -2, to calculate the oxidation state of ClCl we will equate the individual oxidation states with the charge on the ion. Thus, we will get:

& \text{oxidation state of }O\text{ + oxidation state of }Cl=-1 \\\ & \text{oxidation state of }Cl=-1+2 \\\ & \text{oxidation state of }Cl=+1 \\\ \end{aligned}$$ Here, we can also see that the oxidation states of the chlorine involved in $ClO$ as well the solitary chlorine both have different oxidation states +1 and -1 respectively as is required in the question. **Hence, the correct answer is ‘D. $C{{a}^{2+}}(Cl{{O}^{-}})C{{l}^{-}}$’** **Note:** Remember that the oxidation states of the $Cl$ atoms independently should be different, not the charges on the ions they are involved in. As is seen in option ‘C’ the ions involving $Cl$ have different charges, but they will lead the overall charge of the molecule to be +2.