Question

The oxidation state of nitrogen varies from:
(A) -3 to +5
(B) 0 to +5
(C) -3 to 1
(D) +3 to +5

Answer 1

Hint: Write the electronic configuration of nitrogen. Determine how many electrons can be accepted and donated by a nitrogen atom. Think about compounds formed by nitrogen and find their oxidation states to get the range.

Complete step by step solution:
- Nitrogen is the $7^{th}$ element in the periodic table. Its electronic configuration is given by,
$${}^{7}N=1s^{2}2s^{2}2p^3$$
- Nitrogen has five electrons in its outermost orbital. It requires three more electrons to complete its valence shell and attain the noble gas configuration. That means, nitrogen has five valence electrons and has a valency of 3. So, nitrogen can either accept 3 electrons or donate 5 electrons.
- Therefore, the oxidation state of nitrogen ranges from -3 to +5.

So, the correct option is (A).

Additional information:
- Nitrogen forms a variety of compounds both organic and inorganic. Some compounds are nitric acid, ammonia, amines, nitro compounds, etc.
- Let’s find oxidation state of nitrogen in nitric acid, $HN{O}_3$
- Hydrogen is +1, oxygen is -2 and let nitrogen be ‘x’.
$\begin{array}{rl}& 1+x+(3\times -2)=0\\ & x=+5\end{array}$
- So, the highest oxidation state nitrogen has is +5.
-Let’s find oxidation state of nitrogen in ammonia, $N{H}_3$
$$\begin{array}{rl}& x+(3\times +1)=0\\ & x=-3\end{array}$$
- Therefore, the lowest oxidation state nitrogen has is -3.
- The organic compounds are derivatives of inorganic compounds only like amines are derivatives of ammonia. So, they all will have nitrogen oxidation states in between -3 to +5.

Note: Remember that though nitrogen is electronegative in nature, it still can donate its electrons to form stable compounds. Don’t get confused with its electronegativity character and oxidation state. Always write electronic configuration and try writing oxidation states of some nitrogen-containing compounds to avoid confusion.