JEE MAIN - Chemistry (2002 - No. 44)
Oxidation number of Cl in CaOCl2 (bleaching powder) is :
zero, since it contains Cl2
-1, since it contains Cl-
+1, since it contains ClO-
+1 and -1 since it contains ClO- and Cl-
Explanation
$$CaOC{l_2} - $$or it can also be written as $$Ca\mathop {\left( {OCl} \right)}\limits_{{x_1}} \mathop {Cl}\limits_{{x_2}} $$
hence oxidation no of $$Cl$$ in $$OC{l^ - }$$ is
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 + {x_2} = - 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = 2 - 1 = + 1 \cr} $$
now oxidation no. of another $$Cl$$ is $$-1$$ as it is present as $$C{l^ - }.$$
hence oxidation no of $$Cl$$ in $$OC{l^ - }$$ is
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2 + {x_2} = - 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} = 2 - 1 = + 1 \cr} $$
now oxidation no. of another $$Cl$$ is $$-1$$ as it is present as $$C{l^ - }.$$
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