Calculate the oxidation number (O.N.) of the atom in each of the compounds.

(d) Nitrogen in ammonium ion [tex]\(\left( NH_4^{+} \right)\)[/tex]



Answer :

To calculate the oxidation number (O.N) of nitrogen in the ammonium ion ([tex]\( \text{NH}_4^+ \)[/tex]), follow these steps:

1. Determine the known oxidation numbers:
- The oxidation number of hydrogen (H) is generally [tex]\( +1 \)[/tex].

2. Set up the equation:
- Let [tex]\( x \)[/tex] be the oxidation number of nitrogen (N).
- The ammonium ion ([tex]\( \text{NH}_4^+ \)[/tex]) has one nitrogen atom and four hydrogen atoms.
- The algebraic sum of the oxidation numbers of all atoms in the ion should equal the charge on the ion.

3. Write the equation:
- The oxidation number of nitrogen [tex]\( x \)[/tex] plus the sum of the oxidation numbers of the four hydrogen atoms (each with an oxidation number of [tex]\( +1 \)[/tex]) equals the net charge of the ion ([tex]\( +1 \)[/tex]).
- This can be written as:
[tex]\[ x + (4 \times +1) = +1 \][/tex]

4. Simplify to solve for [tex]\( x \)[/tex]:
[tex]\[ x + 4 = +1 \][/tex]
- Subtract 4 from both sides of the equation:
[tex]\[ x = +1 - 4 \][/tex]
- Simplify further:
[tex]\[ x = -3 \][/tex]

5. Conclusion:
- The oxidation number of nitrogen (N) in the ammonium ion ([tex]\( \text{NH}_4^+ \)[/tex]) is [tex]\( -3 \)[/tex].