In which pair of species is the oxidation num | vedclass

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(B) To find the oxidation number of $Cl$ in each species:$1$. In $ClO^-$,$x + (-2) = -1 \implies x = +1$.$2$. In $HClO_3$,$1 + x + 3(-2) = 0 \implies

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$ICl, NaCl$

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(B) To find the oxidation number of $Cl$ in each species:$1$. In $ClO^-$,$x + (-2) = -1 \implies x = +1$.$2$. In $HClO_3$,$1 + x + 3(-2) = 0 \implies x = +5$.$3$. In $ICl$,$I$ is more electropositive than $Cl$,so $Cl$ has an oxidation state of $-1$.$4$. In $NaCl$,$Na$ is $+1$,so $Cl$ has an oxidation state of $-1$.$5$. In $ClF_3$,$F$ is $-1$,so $x + 3(-1) = 0 \implies x = +3$.Comparing the pairs,in the pair $ICl$ and $NaCl$,the oxidation number of $Cl$ is $-1$ in both.

Compound	Oxidation number
$1$. $C_6H_{12}O_6$ (Carbon)	$a$. $+1$
$2$. $CCl_4$ (Carbon)	$b$. $0$
$3$. $NH_4Cl$ (Nitrogen)	$c$. $+4$
$4$. $Ba(H_2PO_2)_2$ (Phosphorus)	$d$. $-3$

In which pair of species is the oxidation number of chlorine the same?

Similar Questions

Oxidation number of $S$ in $Na_2SO_4$ is

Match the following compounds with the oxidation number of the central atom (or specific atom in question):

Compound Oxidation number

$1$. $C_6H_{12}O_6$ (Carbon) $a$. $+1$

$2$. $CCl_4$ (Carbon) $b$. $0$

$3$. $NH_4Cl$ (Nitrogen) $c$. $+4$

$4$. $Ba(H_2PO_2)_2$ (Phosphorus) $d$. $-3$

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The oxidation state of $Mn$ in $K_2MnO_4$ is:

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