A) zero and 4
B) zero and -4
C) 2 and 4
D) -2 and -4
Correct Answer: A
Solution :
Let the oxidation state of C in\[C{{H}_{2}}C{{l}_{2}}\]is\[x\]. \[x+2+2(-1)=0\] \[x+2-2=0\] \[x=0\] Let the oxidation state of C in\[CC{{l}_{4}}\]is\[x\] \[x+4(x-1)=0\] \[x-4=0\] \[x=+4\] Hence, the oxidation number of C-atom in \[C{{H}_{2}}C{{l}_{2}}\]and\[CC{{l}_{4}}\]are zero and +4 respectively.You need to login to perform this action.
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