Question

Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine. $NaCl{O}_4,NaCl{O}_3,NaClO,KCl{O}_2,C{l}_2{O}_7,Cl{O}_3,C{l}_{2}O,NaCl,C{l}_2,Cl{O}_2$

Which oxidation state is not present in any of the above compounds?

Answer 1

$NaCl{O}_4$

Let oxidation number of $Cl$ be X. Considering $+1$ for sodium and $-2$ for oxygen,
$+1+X+(-2)x4=0$
$1+X-8=0$
$X-7=0$
$X=+7$
Therefore, the oxidation number of $Cl$ is $+7$

$NaCl{O}_3$
Let the oxidation number of $Cl$ be X. Considering $+1$ for sodium and $-2$ for oxygen,
$+1+X+(-2)x3=0$
$1+X-6=0$
$X-5=0$
$X=+5$
Therefore, the oxidation number of $Cl$ is $+5$

$NaClO$

Let oxidation number of $Cl$ be X. Considering $+1$ for sodium and $-2$ for oxygen,
$+1+X+(-2)=0$
$X=+1$
Therefore, the oxidation number of $Cl$ is $+1$

$KCl{O}_2$

Let oxidation number of $Cl$ be X. Considering $+1$ for potassium and $-2$ for oxygen,
$+1+X+(-2)2=0$
$1+X-4=0$
$X=+3$
Therefore, the oxidation number of $Cl$ is $+3$

$C{l}_2{O}_7$

Let oxidation number of $Cl$ be X. Considering $-2$ for oxygen,
$2X+(-2)\times 7=0$
$2X-14=0$
$X=+7$
Therefore, the oxidation number of $Cl$ is $+7$

$C{l}_{2}O$

Let theoxidation number of $Cl$ be X. Considering $-2$ for oxygen,
$2X+(-2)=0$
$2X-2=0$
$X=+2/2$
$X=+1$
Therefore, the oxidation number of $Cl$ is $+1$

$NaCl$
Let the oxidation number of $Cl$ be X. Considering $+1$ for sodium,
$+1+X=0$
$X=-1$
Therefore, the oxidation number of $Cl$ is $-1$

$C{l}_2$

Let the oxidation number of $Cl$ be X.
$C{l}_2$
$2X=0$
$X=0$
Therefore, the oxidation number of $Cl$ is $0$

$Cl{O}_2$

Let the oxidation number of $Cl$ be X.
$X+(-2)\times 2=0$
$X-4=0$
$X=+4$
Therefore, the oxidation number of $Cl$ is $+4$

Compounds in increasing order of oxidation number of chlorine are as follow;

$NaCl<Cl<C{l}_{z}O=NaClO<KCl{O}_z<Cl{O}_z<NaCl{O}_3<Cl{O}_3<C{l}_z{O}_7=NaCl{O}_4$
$(-2)$ oxidation state is not found in any of these compounds.