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Q.
The oxidation number of $C$ in $CH _{4}, CH _{3} Cl , CH _{2} Cl _{2}, CHCl _{3}$ and $CCl _{4}$ is respectively
AMUAMU 2002
Solution:
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 $x .$
$\Rightarrow C H_{4}$
$x+4(+1)=0$
$x=-4$
$\Rightarrow C H_{3} C l$
$x+3(+1)+(-1)=0$
$x+3-1=0$
$x=-2$
$\Rightarrow C H_{2} C l_{2}$
$x+2(+1)+2(-1)=0$
$x=0$
$\Rightarrow C H C l_{3}$
$x+(+1)+3(-1)=0$
$x+1-3=0$
$ x=2$
$\Rightarrow C C l_{4}$
$x+4(-1)=0$
$x=+4$
Therefore, the O.N. of $C$ in $C H_{4}, C H_{3} C l, C H_{2} C l_{2}, C H C l_{3}$ and $C C l_{4}$ are $-4,-2,0,+2,+4$ respectively.