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

2 Answers
Mar 12, 2018

Answer:

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.

Mar 12, 2018

Answer:

#-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.