# Assign oxidation numbers..?

Mar 11, 2018

An old story...simple solution....see this link....

#### Explanation:

$H C l$...we gots $\stackrel{+ I}{H} - \stackrel{- I}{C} l$

$C {l}_{2}$ is a neutral diatomic molecule, we gots $\stackrel{0}{C} l - \stackrel{0}{C} l$

And $\text{sodium hypochlorite}$, we gots ""^(-)stackrel(-II)O-stackrel(+I)Cl...and $\stackrel{+ I}{N} {a}^{+}$

...for $\text{chloride}$ we simply have $\stackrel{- I}{C} l$, i.e. the charge on the atomic ion is the oxidation number....

And for $\text{chloric acid}$, we gots $\stackrel{+ I}{H} - O - \stackrel{+ V}{C} l {\left(= O\right)}_{2}$...the oxygens are all $\stackrel{- I I}{O}$...and its conjugate base, $C l {O}_{3}^{-}$, RETAINS these oxidation numbers....

$\text{Potassium chlorite}$, we gots ${K}^{+}$, and ""^(+III)ClO_2^(-)

And finally with $\text{perchloric acid}$ we gots $C l \left(V I I +\right)$...

As always, the SUM of the oxidation numbers in a given compound or ion EQUALS the charge on the ion. For neutral molecules, such as $H C l {O}_{4}$..the weighted sum of the elemental oxidation numbers is ZERO....i.e. $+ {I}_{H} - 4 \times I {I}_{O} + V I {I}_{C l} = 0$ as required.

Happy?