What is the oxidation numbers of (a) N in #NH_4^+#?

2 Answers
anor277
Mar 12, 2018

Well, as usual, the oxidation number of #H# is #+I# as is typical....

Explanation:

And the sum of the individual oxidation numbers is equal to the charge on the ion...

And so #N_"oxidation number"+4xxI^+=+1#

#N_"oxidation number"=-III#

For a few more examples .... see [this older answer.](https://socratic.org/questions/assign-oxidation-numbers#570916)

Nam D.
Mar 12, 2018

#-3#

Explanation:

We have the ammonium ion, #NH_4^+#.

As you can see from the formula, it has a #+1# charge.

Since nitrogen is more electronegative than hydrogen, hydrogen will occupy a #+1# charge. There are four hydrogen atoms in this ion, so the total charge of the hydrogens is #+1*4=+4#.

Let #x# be the oxidation number of #N# in #NH_4^+#.

We got:

#x+(+4)=+1#

Treating them as normal numbers, we get

#x+4=1#

#x=1-4#

#=-3#

So, nitrogen would have a #-3# charge, which is also its usual charge.

Related questions
See all questions in Oxidation Numbers
Impact of this question
49205 views around the world
You can reuse this answer
Creative Commons License