A) -4, -2, 0, +2, +4
B) +2, 4, 0, -2, -4
C) 4, 2, 0, -2, -4
D) 0, 2, -2, 4, 4
Correct Answer: A
Solution :
Key Idea The O.N. of hydrogen is +1 when combined with non-metals and is -1 when combined with active metals called metal hydrides. The O.N. of monoatomic ion \[(C{{N}^{-}})\] is same as the charge on it, so the O.N. of chlorine is -1. Let the oxidation number of C is x. \[\Rightarrow \] \[C{{H}_{4}}\] \[x+4\,(+1)=0\] \[x=-4\] \[\Rightarrow \] \[C{{H}_{3}}Cl\] \[x+3\,(+1)+(-1)=0\] \[x+3-1=0\] \[x=-2\] \[\Rightarrow \] \[C{{H}_{2}}C{{l}_{2}}\] \[x+2(+1)+2(-1)=0\] \[x=0\] \[\Rightarrow \] \[CHC{{l}_{3}}\] \[x+(+1)+3(-1)=0\] \[x+1-3=0\] \[x=2\] \[\Rightarrow \] \[CC{{l}_{4}}\] \[x+4(-1)=0\] \[x=+4\] Therefore, the O.N. of C in\[C{{H}_{4}},C{{H}_{3}}Cl,C{{H}_{2}}C{{l}_{2}},CHC{{l}_{3}}\] and \[CC{{l}_{4}}\] are -4, - 2, 0, + 2, + 4 respectively.
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