Question #f6f9a

1 Answer
Jun 16, 2017

Answer:

In #"NaCl": -1#
In #"KClO"_2:+3#
In #"Cl"_2: 0#
In #"ClO"_4^-#

Explanation:

Recall: The following rules for assigning oxidation numbers
enter image source here
Credit: Tyler DeWitt

Keeping those rules In mind, let's look at each compound:

#"NaCl"#

#"Na"# is located in the 1A group so it has an oxidation number of #+1#. That also means that #"Cl"# has an oxidation of #-1# because the sum of a neutral (meaning the compound has no charge) compound is equal to 0 The sum of the is zero since #1-1=0#

With oxidation numbers Sodium Chloride #("NaCl")# is:

#"Na"^(+1)"Cl"^(-1)#

#---------------------#

Now looking at #"KClO"_2#

#"K"# (Potassium) lies in the 1A group so it has an oxidation number of #+1#

O (Oxygen) has an oxidation of #-2# with the exception of peroxide but there is no peroxide in our compound. However, if we look at the compound we can see that we have two oxygen atoms and so we have to multiply the oxidation number by the number of atoms we have: In this case, #-2*2=-4#

Thus far we have:

#"K"^(+1)"Cl"^(?)"O"_2^(-4)#

Remember: the sum of of O.N's in a neutral compound is #0#

Therefore #"Cl"# has an O.N of #+3# because #1+3-4=0#

#---------------------#

For #"Cl"_2#, you should notice that this is in fact not a compound but rather a diatomic molecule which is a molecule that consists of two atoms from the same element like (#"H"_2#). Given that there is only one element here and referring back to our rules, an element by itself has an O.N of #0#

#---------------------#

Lastly: #"ClO"_4^-)#

This one is interesting because this is known as a polyatomic ion. #"ClO"_4^-)# or perchlorate ion has an overall charge of #-1#. This will be important to know:

"O" has an O.N of #-8# because #-2*4=-8#

To determine the O.N of chlorine we must keep in mind the sum of O.N's is NOT ZERO. In this case the sum it is equal to the ion charge #(-1)#

So: Chlorine would have an O.N of #+7# because #7-8=-1#

I hope this helped, but if you're still stuck or if you need a better explanation of this concept, check out the video below (I bet he can explain it way better than me!):