Question

The oxidation number of C in ${CH}_4,{CH}_{3}Cl,{CH}_{2}C{l}_2,CHC{l}_3$ 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 $-4,-2,0,+2,+4$

Electronegativity order in carbon, hydrogen, and chlorine is-

$Cl>C>H$

Hence, oxidation state of hydrogen and chlorine in all the given compounds will be +1 and -1 respectively.

In $C{H}_4$-

Let the oxidation state of carbon in $C{H}_4$ be x.

$\therefore x+(4\times (+1))=0$

$\Rightarrow x=-4$

In $C{H}_{3}Cl$-

Let the oxidation state of carbon in $C{H}_{3}Cl$ be x.

$\therefore x+(3\times (+1))+(-1)=0$

$\Rightarrow x+3-1=0$

$\Rightarrow x=-2$

In $C{H}_2{Cl}_2$-

Let the oxidation state of carbon in $C{H}_3{Cl}_2$ be x.

$\therefore x+(2\times (+1))+(2\times (-1))=0$

$\Rightarrow x+2-2=0$

$\Rightarrow x=0$

In $CH{Cl}_3$-

Let the oxidation state of carbon in $CH{Cl}_3$ be x.

$\therefore x+1+(3\times (-1))=0$

$\Rightarrow x+1-3=0$

$\Rightarrow x=+2$

In $C{Cl}_4$-

Let the oxidation state of carbon in $C{Cl}_4$ be x.

$\therefore x+(4\times (-1))=0$

$\Rightarrow x=+4$

Hence, the correct option is $C$.