Q. \( \mathrm{NH_3} \) lewis dot.

Q: How many valence electrons are in \mathrm{NH_3} and how are they used?

Nitrogen contributes 5 and each H contributes 1, giving 5 + 3 = 8 valence electrons. Six go into three N–H bonds. The remaining two form one lone pair on N.

Q: How do lone pairs affect bond angles in \mathrm{NH_3} ?

The lone pair compresses bond angles from the tetrahedral 109.5^\circ. So \mathrm{NH_3} has a smaller H–N–H bond angle of about 107^\circ.

Answer

NH\(_3\) has nitrogen in the center and three hydrogens around it.

N has \(5\) valence electrons. Each H has \(1\) valence electron, so total valence electrons \(=5+3\times 1=8\).

Form three N–H single bonds (use \(6\) electrons total). The remaining \(2\) electrons are a lone pair on nitrogen.

Lewis dot structure:
N in the center with one lone pair and three single bonds to H.

\[
\begin{array}{c}
\ \ \ \ H \\
\ \ \ \ | \\
\ H – N : – H \\
\ \ \ \ | \\
\ \ \ \ H
\end{array}
\]

Final result: Nitrogen has \(1\) lone pair and forms \(3\) single bonds with three H atoms.

Detailed Explanation

Goal: Draw the Lewis dot structure (Lewis dot diagram) for the molecule \( \mathrm{NH_3} \) (ammonia).

Step 1: Count total valence electrons.

We need valence electrons only.

For each element:

\( \mathrm{N} \) is in Group 15, so it has \(5\) valence electrons.
\( \mathrm{H} \) is in Group 1, so each \( \mathrm{H} \) has \(1\) valence electron. There are \(3\) hydrogens, so \(3 \times 1 = 3\).

Total valence electrons:

\[
5 + 3 = 8
\]

So \( \mathrm{NH_3} \) has \(8\) total valence electrons to place in the Lewis structure.

Step 2: Choose the central atom.

Nitrogen is the least electronegative and has enough bonding capacity, so it is the central atom. Hydrogen atoms go around it.

Skeleton shape: one nitrogen in the center with three hydrogens attached.

Step 3: Place bonding pairs first.

An \( \mathrm{N-H} \) bond is a single bond, which is made of a bonding pair of electrons (two electrons shared between N and H).

We need three \( \mathrm{N-H} \) single bonds to attach three hydrogens.

Number of bonding pairs present after making three single bonds:

Three bonds means \(3\) bonding pairs.
Each bond uses \(2\) electrons.

Electrons used so far:

\[
3 \times 2 = 6
\]

Remaining electrons:

\[
8 – 6 = 2
\]

Step 4: Place the remaining electrons as lone pair(s) on the central atom.

The remaining \(2\) electrons must form one lone pair. Since all three hydrogens already have bonds (so they do not get lone pairs in a typical Lewis structure for \( \mathrm{NH_3} \)), the lone pair goes on nitrogen.

Therefore, nitrogen has :

\(3\) bonding pairs (from the three \( \mathrm{N-H} \) bonds)
\(1\) lone pair

Step 5: Check electron count and typical valence rule.

Hydrogen: each \( \mathrm{H} \) has a duet (it is satisfied by the single bond), so it has \(2\) electrons around it.

Nitrogen: has \(3\) bonds (6 bonding electrons around it) plus \(1\) lone pair (2 nonbonding electrons), for a total of \(8\) electrons in its valence shell. This matches the total valence electron accounting.

Final Answer: Lewis dot structure for \( \mathrm{NH_3} \).

Nitrogen in the center with three single bonds to three hydrogens, and one lone pair on nitrogen.

Lewis dot diagram (text form):

      ..
H  -  N  -  H
      |
      H

Clear depiction in words:

Three single \( \mathrm{N-H} \) bonds
One lone pair on \( \mathrm{N} \)

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General Chemistry FAQs

What is the Lewis dot structure of \( \mathrm{NH_3} \)?

Place N in the center with three single bonds to H. Nitrogen has one lone pair. Valence electrons total: \(5 + 3(1) = 8\). So you draw 3 N–H bonds and 2 nonbonding electrons on N.

How many valence electrons are in \( \mathrm{NH_3} \) and how are they used?

Nitrogen contributes 5 and each H contributes 1, giving \(5 + 3 = 8\) valence electrons. Six go into three N–H bonds. The remaining two form one lone pair on N.

Which atom has the lone pair in \( \mathrm{NH_3} \) and how many electrons are in it?

The lone pair is on the nitrogen atom. Nitrogen keeps 2 nonbonding electrons, shown as one pair of dots around N.

What is the molecular geometry and electron-domain geometry of \( \mathrm{NH_3} \) from the Lewis structure?

With 4 electron domains (3 bonds + 1 lone pair), the electron-domain geometry is tetrahedral. The molecular shape is trigonal pyramidal because one domain is a lone pair.

Does \( \mathrm{NH_3} \) follow the octet rule on nitrogen?

Yes. Nitrogen forms 3 single bonds (uses 6 bonding electrons) plus 1 lone pair (2 electrons), totaling 8 electrons around N.

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

Using the Lewis structure with one lone pair on N: nitrogen formal charge is 0. Each hydrogen formal charge is 0 because each H has one bond and no lone pairs.

How do lone pairs affect bond angles in \( \mathrm{NH_3} \)?

The lone pair compresses bond angles from the tetrahedral \(109.5^\circ\). So \( \mathrm{NH_3} \) has a smaller H–N–H bond angle of about \(107^\circ\).

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