A) - 1 and + 5
B) + 1 and + 4
C) + 5 and +3
D) - 1 and + 1
Correct Answer: D
Solution :
When chlorine reacts with dilute and cold \[\text{NaOH,}\]sodium chloride and sodium hypochloride are formed. \[2NaOH(cold)+C{{l}_{2}}\to NaCl+\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\text{soldium} \\ \text{hypochloride} \end{smallmatrix}}{\mathop{NaClO}}\,\,+{{H}_{2}}O\]Let oxidation state of Cl in NaCI is\[x.\] \[\text{NaCl}\] \[+\,1+x=0\] \[x=-1\] Let oxidation state of Cl in \[\text{NaClO}\]is \[x.\] \[\text{NaClO}\] \[\text{+}\,1+x-2=0\] \[x-1=0\] \[x=+1\] \[\therefore \]Oxidation states of chlorine changes from 0 to\[-1\] and \[+\,1.\]
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