Question

In which of the following, oxidation state of nitrogen is lowest?

• A. NH$_2$OH
• B. NH$_4$Cl
• C. N$_2$H$_4$
• D. HNO$_2$

Hint:

To find the oxidation state of an element in a compound, use the standard rules: O is -2 (except in peroxides), H is +1 (except in metal hydrides), halogens are -1 (except when bonded to a more electronegative element). The sum of oxidation states must equal the overall charge of the molecule or ion.

Answer 1

The Correct Option is B

We need to calculate the oxidation state of Nitrogen (N) in each compound. Let the oxidation state of N be $x$.
We use the standard rules for assigning oxidation states: Oxygen is usually -2, Hydrogen is usually +1, and Chlorine (as a halide ion) is -1. The sum of oxidation states in a neutral compound is 0.
(A) NH$_2$OH (Hydroxylamine): The compound can be seen as NH$_2$ and OH. $x + 2(+1) + (-2) + (+1) = 0$. $x + 3 - 2 = 0$. $x + 1 = 0 \implies x = -1$.
(B) NH$_4$Cl (Ammonium chloride): This is an ionic compound composed of the ammonium ion (NH$_4^+$) and the chloride ion (Cl$^-$). We calculate the oxidation state of N in NH$_4^+$. $x + 4(+1) = +1$. $x+4=1 \implies x = -3$.
(C) N$_2$H$_4$ (Hydrazine): $2x + 4(+1) = 0$. $2x + 4 = 0$. $2x = -4 \implies x = -2$.
(D) HNO$_2$ (Nitrous acid): $(+1) + x + 2(-2) = 0$. $1 + x - 4 = 0$. $x - 3 = 0 \implies x = +3$.
Comparing the oxidation states of Nitrogen in each compound: (A) -1 (B) -3 (C) -2 (D) +3
The lowest (most negative) oxidation state is -3, which is found in NH$_4$Cl.