Problem 7 - Chemistry Matter and Change

Question

Determine the oxidation number of nitrogen in each of these molecules or ions. $$ \text { a. } \mathrm{NH}_{3} \quad \text { b. KCN } \quad \text { c. } \mathrm{N}_{2} \mathrm{H}_{4} $$

Step-by-Step Solution

Answer

In \(\text{NH}_3\) it's \(-3\), in \(\text{KCN}\) it's \(-3\), and in \(\text{N}_2\text{H}_4\) it's \(-2\).

Step 1: Determine Oxidation Number in NH3

In the molecule \( \text{NH}_3 \), nitrogen is bonded to hydrogen. The oxidation number for hydrogen is usually \( +1 \). Let's call the oxidation number of nitrogen \( x \). Since the molecule is neutral, the sum of oxidation numbers must equal zero. Setting up the equation: \[ x + 3(+1) = 0 \] Simplifying, \[ x + 3 = 0 \] thus, \[ x = -3 \]. Therefore, the oxidation number of nitrogen in \( \text{NH}_3 \) is \( -3 \).

Step 2: Determine Oxidation Number in KCN

In the compound \( \text{KCN} \), recognize that it consists of \( \text{K}^+ \), \( \text{C} \), and \( \text{N} \). The potassium ion \( \text{K}^+ \) has an oxidation number of \( +1 \). The cyanide ion \( \text{CN}^- \) has an overall charge of \( -1 \). Let \( y \) be the oxidation number of carbon and \( z \) be that of nitrogen. The equation for \( \text{CN}^- \) is \[ y + z = -1 \]. Since carbon in cyanide typically has the oxidation number \( +2 \), substitute into the equation: \[ +2 + z = -1 \] Solving this gives \[ z = -3 \]. Hence, the oxidation number of nitrogen in \( \text{KCN} \) is \( -3 \).

Step 3: Determine Oxidation Number in N2H4

In the molecule \( \text{N}_2\text{H}_4 \), find the oxidation number of nitrogen. Let \( a \) be the oxidation number of nitrogen. Hydrogen has an oxidation number of \( +1 \). The sum of oxidation numbers in a neutral molecule is zero. The equation is: \[ 2a + 4(+1) = 0 \] Simplifying gives: \[ 2a + 4 = 0 \] thus, \[ 2a = -4 \] and \[ a = -2 \]. Therefore, the oxidation number of nitrogen in \( \text{N}_2\text{H}_4 \) is \( -2 \).

Key Concepts

NH3 oxidation

Understanding the oxidation number of nitrogen in ammonia, or \( \text{NH}_3 \), is a common chemistry exercise that reveals how electrons are shared or transferred between elements. First, let's identify each element in \( \text{NH}_3 \). In this molecule, nitrogen (N) is bonded to three hydrogen atoms (H). Hydrogen typically has an oxidation number of \( +1 \). Since the molecule is neutral, the total sum of oxidation numbers must be zero.

Given the equation: \( x + 3(+1) = 0 \), where \( x \) represents the oxidation number of nitrogen. This simplifies to \( x + 3 = 0 \). Solving for \( x \) gives \( x = -3 \). Thus, the oxidation number of nitrogen in ammonia is \(-3\). The negative oxidation number indicates that nitrogen is gaining electrons (or donating fewer) compared to its elemental form, which is a fundamental aspect of its chemical behavior in this compound.

KCN oxidation

The compound potassium cyanide, or \( \text{KCN} \), is a fascinating case study in oxidation numbers. It is composed of potassium \( (\text{K}^+) \), carbon \( (\text{C}) \), and nitrogen \( (\text{N}) \). Potassium, a metal in group 1, typically has an oxidation number of \( +1 \).

In this molecule, the most complicated part is the cyanide ion \( (\text{CN}^-) \), which holds a charge of \( -1 \). We need to determine individual oxidation numbers for carbon and nitrogen within this ion. Generally, carbon in cyanide tends to have an oxidation number of \( +2 \).

• If carbon is \( +2 \), then the equation \( +2 + z = -1 \) applies, where \( z \) is the oxidation number of nitrogen.
• Solve for \( z \): \( z = -3 \).

Thus, in \( \text{KCN} \), nitrogen has an oxidation number of \(-3\). The oxidation state helps us understand nitrogen's role in forming the stable and often dangerous cyanide group.

N2H4 oxidation

Hydrazine, expressed chemically as \( \text{N}_2\text{H}_4 \), presents an intriguing structure with two nitrogen atoms. Here, hydrogen once again has an oxidation number of \( +1 \). Because the molecule is neutral, the sum of all oxidation numbers should equal zero.

To determine nitrogen's oxidation number, designate \( a \) as the oxidation number for a single nitrogen atom. The equation for the compound is \[ 2a + 4(+1) = 0 \]. Simplifying yields \( 2a + 4 = 0 \). From here:

• Subtract 4 from both sides: \( 2a = -4 \).
• Then divide by 2: \( a = -2 \).

Thus, each nitrogen atom in \( \text{N}_2\text{H}_4 \) has an oxidation number of \(-2\). This oxidation state indicates that nitrogen in hydrazine has accepted electrons, which aligns with its chemical properties such as being a powerful reducing agent that participates in combustion and other oxidative reactions.