A) \[-4,\,\,-2,\,\,+2,\,\,0,\,\,+4\]
B) \[+2,\,\,4,\,\,0,\,\,-2,\,\,-4\]
C) \[+4,\,\,+0,\,\,-2,\,\,-2,\,\,+4\]
D) \[0,\,\,2,\,\,-2,\,\,4,\,\,4\]
Correct Answer: A
Solution :
Let the oxidation number of C is x. \[\underset{\begin{smallmatrix} x+4\times (+1)=0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-4 \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,\] \[\underset{\begin{smallmatrix} x+3(+1)-1=0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-2 \end{smallmatrix}}{\mathop{C{{H}_{3}}Cl}}\,\] \[\underset{\begin{smallmatrix} x+1+3\times (-1)=0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=+2 \end{smallmatrix}}{\mathop{CHC{{l}_{3}}}}\,\] \[\underset{\begin{smallmatrix} x+2\times (+1)+2\times (-1)=0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0 \end{smallmatrix}}{\mathop{C{{H}_{2}}C{{l}_{2}}}}\,\] \[\underset{\begin{smallmatrix} x+4\times (-1)=0 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=+4 \end{smallmatrix}}{\mathop{CC{{l}_{4}}}}\,\] Hence, the oxidation number of\[C\]in\[C{{H}_{4}}\], \[C{{H}_{4}},\,\,C{{H}_{3}}Cl,\,\,CHC{{l}_{3}},\,\,C{{H}_{2}}C{{l}_{2}}\]and\[CC{{l}_{4}}\]is respectively \[-4,\,\,-2,\,\,+2,\,\,0\]and\[+4\].
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