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Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
NaClO_{4}, NaClO_{3}, NaClO, KClO_{2}, Cl_{2}O_{7}, ClO_{3}, Cl_{2}O, NaCl, Cl_{2}, ClO_{2}.
Which oxidation state is not present in any of the above compounds?

Answers (1)

For all the chlorine atoms, we will assume the oxidation state to be x, and for atoms Na, K = +1, O = -2. Now, we know that, the sum of oxidation numbers of all the atoms in a compound is the same as the charge on that compound for an ion or zero.


NaClO_{4}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +7

NaClO_{3}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +5

In the Cl_{2}O_{7} structure there are two atoms of chlorine with varying types of O ? O, Cl ? Cl. The structure is explained as follows: -

Cl_{2}O_{7}\rightarrow 2\times x+7\times -2 = 0\rightarrow x = +7

ClO_{3}\rightarrow x+2\times -2 = 0\rightarrow x = +6

Cl_{2}O\rightarrow 2 \times x -2 = 0\rightarrow x = +1

NaCl\rightarrow +1+ x = 0\rightarrow x = -1

Cl_{2}\rightarrow As chlorine is in its naturalthe oxidation number is zero.

ClO_{2}\rightarrow x +2\times -2= 0\rightarrow x = +4

Therefore the ascending order is,

NaCl (-1), Cl_{2}(0), Cl_{2}O(+1), KClO_{2}(+3), ClO_{2}(+4), NaClO_{3}(+5), ClO_{3}(+6), Cl_{2}O_{7}=NaClO_{4}(+7).

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