Question

The oxidation numbers of C in $C{H}_4,C{H}_{3}Cl,C{H}_{2}C{l}_2,CHC{l}_3$ and $CC{l}_4$ are respectively.

• A. $+4,+2,0,-2,-4$
• B. $+2,+4,0,-4,-2$
• C. $-4,-2,0,+2,+4$
• D. $-2,-4,0,+4,+2$

Answer 1

The correct option is C $-4,-2,0,+2,+4$

Let oxidation state of carbon is $x$.

In $C{H}_4$, $x+4(+1)=0$

$x=-4$

In $C{H}_{3}Cl$, $x+3(+1)+(-1)=0$

$x=-2$

In $C{Cl}_4$, $x+4(-1)=0$

$x=+4$

In $CH{Cl}_3$, $x+1+3(-1)=0$

$x=+2$

In $C{H}_2{Cl}_2$, $x+2(+1)+2(-1)=0$

$x=+0$

So option C) is correct answer.