Question

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

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

Answer 1

Answer: (A)

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 $\left(C{N}^{-}\right)$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.