Q. \( \mathrm{NH_3} \) oxidation number.

Q: What is the oxidation number of nitrogen in \mathrm{NH_3} ?

Nitrogen has oxidation number -3 in \mathrm{NH_3} because hydrogen is +1 each, and the sum must equal the neutral molecule’s total 0.

Q: What oxidation number does hydrogen have in \mathrm{NH_3} ?

In \mathrm{NH_3} , each hydrogen has oxidation number +1 since hydrogen is typically +1 when bonded to nonmetals like nitrogen.

Q: How do I calculate \mathrm{NH_3} oxidation numbers quickly?

Use \sum \text{oxidation numbers} = 0 for neutral compounds. Set hydrogen as +1 (three H atoms), then solve for nitrogen: x + 3(+1)=0, so x=-3.

Q: What is the oxidation number of nitrogen in \mathrm{NH_4^+} ?

In \mathrm{NH_4^+} , hydrogen is still +1 each. With total charge +1: x + 4(+1)=+1, so x=-3.

Q: What is the oxidation number of nitrogen in \mathrm{N_2H_4} (hydrazine)?

Let nitrogen be x. Hydrogen is +1 each: 2x + 4(+1)=0. Thus 2x+4=0, so x=-2 per nitrogen atom.

Q: What oxidation number does \mathrm{NH_3} indicate for nitrogen’s oxidation state trend?

Nitrogen in \mathrm{NH_3} is at a low oxidation state -3. It increases toward positive values as \mathrm{NH_3} is oxidized.

Answer

To find the oxidation number of nitrogen in ammonia \( \text{NH}_3 \), use the rule that the sum of oxidation numbers in a neutral compound is zero.

Hydrogen is usually \(+1\), and there are three hydrogens:

\[
3(+1) + x = 0
\]

where \(x\) is the oxidation number of nitrogen.

So:

\[
x = -3
\]

Final result: The oxidation number of nitrogen in \( \text{NH}_3 \) is \( -3 \).

Detailed Explanation

To find the oxidation number of nitrogen in ammonia, NH\(3\), we use the standard rules for oxidation numbers.

Step 1: Identify known oxidation numbers.

In most high-school oxidation-number problems, hydrogen has an oxidation number of \(+1\) when it is bonded to a nonmetal (and nitrogen in ammonia behaves like a nonmetal). So:

\[ \text{Oxidation number of H in NH}_3 \;=\; +1 \]

Step 2: Use the “sum of oxidation numbers” rule.

Because ammonia, NH\(3\), is a neutral compound, the total sum of oxidation numbers of all atoms equals \(0\).

\[ \text{Sum of oxidation numbers in NH}_3 \;=\; 0 \]

Step 3: Set up the equation.

Let the oxidation number of nitrogen be \(x\). There are three hydrogen atoms, each with oxidation number \(+1\). So the sum is:

\[ x + 3(+1) \;=\; 0 \]

Step 4: Solve for \(x\).

\[ x + 3 \;=\; 0 \]

\[ x \;=\; -3 \]

Final Answer:

The oxidation number of nitrogen in ammonia, NH\(3\), is \(-3\).

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General Chemistry FAQs

What is the oxidation number of nitrogen in \( \mathrm{NH_3} \)?

Nitrogen has oxidation number \(-3\) in \( \mathrm{NH_3} \) because hydrogen is \(+1\) each, and the sum must equal the neutral molecule’s total \(0\).

What oxidation number does hydrogen have in \( \mathrm{NH_3} \)?

In \( \mathrm{NH_3} \), each hydrogen has oxidation number \(+1\) since hydrogen is typically \(+1\) when bonded to nonmetals like nitrogen.

How do I calculate \( \mathrm{NH_3} \) oxidation numbers quickly?

Use \(\sum \text{oxidation numbers} = 0\) for neutral compounds. Set hydrogen as \(+1\) (three H atoms), then solve for nitrogen: \(x + 3(+1)=0\), so \(x=-3\).

Why is nitrogen \(-3\) in \( \mathrm{NH_3} \) but can be positive in other compounds?

Oxidation number depends on bonding and electronegativity. Nitrogen can shift to positive values when bonded to more electronegative atoms (like oxygen), e.g., in \( \mathrm{HNO_3} \).

What is the oxidation number of nitrogen in \( \mathrm{NH_4^+} \)?

In \( \mathrm{NH_4^+} \), hydrogen is still \(+1\) each. With total charge \(+1\): \(x + 4(+1)=+1\), so \(x=-3\).

What is the oxidation number of nitrogen in \( \mathrm{N_2H_4} \) (hydrazine)?

Let nitrogen be \(x\). Hydrogen is \(+1\) each: \(2x + 4(+1)=0\). Thus \(2x+4=0\), so \(x=-2\) per nitrogen atom.

What oxidation number does \( \mathrm{NH_3} \) indicate for nitrogen’s oxidation state trend?

Nitrogen in \( \mathrm{NH_3} \) is at a low oxidation state \(-3\). It increases toward positive values as \( \mathrm{NH_3} \) is oxidized.

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