Q. Formal charge of \( \mathrm{NH_3} \).

Q: What is the formal charge calculation method for \mathrm{NH_3} ?

Use \text{FC} = \text{valence} - (\text{nonbonding} + \tfrac{1}{2}\text{bonding}) . For N in \mathrm{NH_3} , valence =5, nonbonding =2 electrons, bonding =3 single bonds =6 electrons, so \text{FC}=5-(2+\tfrac{6}{2})=0 .

Q: What is the formal charge on nitrogen in \mathrm{NH_3} ?

\mathrm{NH_3} has neutral total charge. Nitrogen’s formal charge is 0. Each N–H bond is a single bond, and N has one lone pair (2 nonbonding electrons), giving \text{FC}_N=0 .

Q: What are the nonbonding electrons on nitrogen in \mathrm{NH_3} , and how do they affect formal charge?

Nitrogen has one lone pair, so nonbonding electrons on N =2. In \text{FC} = 5-(2+\tfrac{6}{2}) , these 2 electrons reduce formal charge by 2, leading to net \text{FC}_N=0 .

Answer

For ammonia \( \text{NH}_3 \), nitrogen has \(5\) valence electrons and forms \(3\) single bonds with three hydrogens. Nitrogen has one lone pair.

Formal charge formula: \( \text{FC} = \text{valence} – \text{bonding electrons}/2 – \text{nonbonding electrons} \).

For nitrogen: valence \(=5\). Bonding electrons \(=6\) (three N–H single bonds), so \(6/2=3\). Nonbonding electrons \(=2\) (one lone pair).

\[ \text{FC(N)} = 5 – 3 – 2 = 0 \]

For each hydrogen, \( \text{FC} = 0 \) as well.

Final result: The formal charge of nitrogen in \( \text{NH}_3 \) is \(0\) (and each hydrogen also has formal charge \(0\)).

Detailed Explanation

We want the formal charge on \( \mathrm{NH_3} \) (ammonia). In \( \mathrm{NH_3} \), the nitrogen atom is the central atom bonded to three hydrogens.

Step 1: Write the formal charge formula.

For any atom \(X\), the formal charge is

\[
\text{Formal charge on } X
= \text{Valence electrons of } X
– \left( \text{Nonbonding electrons on } X
+ \frac{1}{2}\left(\text{Bonding electrons involving } X\right)\right)
\]

Step 2: Determine the valence electrons for nitrogen.

Nitrogen is in group \(15\), so it has \(5\) valence electrons.

\[
\text{Valence electrons of N} = 5
\]

Step 3: Count nonbonding electrons on nitrogen in \( \mathrm{NH_3} \).

Ammonia has one lone pair on nitrogen, so nitrogen has \(2\) nonbonding electrons.

\[
\text{Nonbonding electrons on N} = 2
\]

Step 4: Count bonding electrons involving nitrogen.

Nitrogen forms three single bonds to three hydrogens. Each single bond contains \(2\) bonding electrons. So total bonding electrons involving N are

\[
3 \text{ bonds} \times 2 \text{ electrons per bond} = 6
\]

Step 5: Apply the formal charge formula to nitrogen.

\[
\text{Formal charge on N}
= 5 – \left(2 + \frac{1}{2}(6)\right)
\]

\[
= 5 – \left(2 + 3\right)
\]

\[
= 5 – 5
= 0
\]

Step 6: Check the hydrogens (optional but confirms the structure).

Each hydrogen has \(1\) valence electron and participates in one single bond (no lone pairs on H). So each hydrogen has formal charge \(0\).

Final Answer:

The formal charge of the nitrogen in ammonia \( \mathrm{NH_3} \) is \(0\). (All atoms in the neutral \( \mathrm{NH_3} \) molecule have formal charge \(0\).)

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General Chemistry FAQs

What is the formal charge calculation method for \( \mathrm{NH_3} \)?

Use \( \text{FC} = \text{valence} - (\text{nonbonding} + \tfrac{1}{2}\text{bonding}) \). For N in \( \mathrm{NH_3} \), valence \(=5\), nonbonding \(=2\) electrons, bonding \(=3\) single bonds \(=6\) electrons, so \( \text{FC}=5-(2+\tfrac{6}{2})=0 \).

What is the formal charge on nitrogen in \( \mathrm{NH_3} \)?

\( \mathrm{NH_3} \) has neutral total charge. Nitrogen’s formal charge is \(0\). Each N–H bond is a single bond, and N has one lone pair (2 nonbonding electrons), giving \( \text{FC}_N=0 \).

What is the formal charge on each hydrogen in \( \mathrm{NH_3} \)?

Each hydrogen has one bonding pair and no lone pairs. For H: valence \(=1\), nonbonding \(=0\), bonding \(=2\) (one single bond), so \( \text{FC}_H=1-(0+\tfrac{2}{2})=0 \).

Does \( \mathrm{NH_3} \) have any formal charge on any atom?

No. In the usual Lewis structure, all atoms have formal charge \(0\): \( \text{FC}_N=0 \) and \( \text{FC}_{H1}= \text{FC}_{H2}= \text{FC}_{H3}=0 \).

If I draw a different Lewis structure for ammonia, can formal charges change?

Yes, but that structure may violate typical octet rules. For \( \mathrm{NH_3} \), the valid structure has three N–H single bonds and one lone pair on N, yielding formal charges \(0\) on all atoms.

What are the nonbonding electrons on nitrogen in \( \mathrm{NH_3} \), and how do they affect formal charge?

Nitrogen has one lone pair, so nonbonding electrons on N \(=2\). In \( \text{FC} = 5-(2+\tfrac{6}{2}) \), these 2 electrons reduce formal charge by 2, leading to net \( \text{FC}_N=0 \).

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