Assign oxidation numbers..?

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1 Answer
Mar 11, 2018

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

Explanation:

#HCl#...we gots #stackrel(+I)H-stackrel(-I)Cl#

#Cl_2# is a neutral diatomic molecule, we gots #stackrel(0)Cl-stackrel(0)Cl#

And #"sodium hypochlorite"#, we gots #""^(-)stackrel(-II)O-stackrel(+I)Cl#...and #stackrel(+I)Na^+#

...for #"chloride"# we simply have #stackrel(-I)Cl#, i.e. the charge on the atomic ion is the oxidation number....

And for #"chloric acid"#, we gots #stackrel(+I)H-O-stackrel(+V)Cl(=O)_2#...the oxygens are all #stackrel(-II)O#...and its conjugate base, #ClO_3^(-)#, RETAINS these oxidation numbers....

#"Potassium chlorite"#, we gots #K^+#, and #""^(+III)ClO_2^(-)#

And finally with #"perchloric acid"# we gots #Cl(VII+)#...

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 #HClO_4#..the weighted sum of the elemental oxidation numbers is ZERO....i.e. #+I_H -4xxII_(O)+VII_(Cl)=0# as required.

Happy?