Q. \[ \mathrm{NH}_2^+ \] Lewis structure.

Q: What does NH2 typically refer to in Lewis structure problems?

It can mean the aminyl radical \mathrm{NH_2\cdot} , the amide anion \mathrm{NH_2^-} , or the amine substituent \mathrm{-NH_2} . Context tells you which species to draw.

Q: How many valence electrons for \mathrm{NH_2^-} and how do you place them?

Total valence electrons: 5 + 2(1) + 1 = 8. Place N central, make two N-H single bonds, then place the remaining four electrons as two lone pairs on N. That gives complete octet and single negative formal charge.

Q: How do you draw the Lewis structure for the aminyl radical \mathrm{NH_2\cdot} ?

Total valence electrons: 5 + 2(1) = 7. Draw N with two N-H single bonds, one lone pair, and one unpaired electron on N. Formal charge on N is zero, and the octet is incomplete by one electron.

Q: What are the formal charges on N in \mathrm{NH_2^-} and \mathrm{NH_2\cdot} ?

Use FC = valence - nonbonding - 1/2 bonding. For \mathrm{NH_2^-} , FC = 5 - 4 - \tfrac{1}{2}(4) = -1. For \mathrm{NH_2\cdot} , FC = 5 - 3 - \tfrac{1}{2}(4) = 0.

Q: How many lone pairs and unpaired electrons are on N in each species?

In \mathrm{NH_2^-} there are two lone pairs and no unpaired electrons. In \mathrm{NH_2\cdot} there is one lone pair and one unpaired electron on N.

Q: What is the hybridization and molecular geometry of NH2 species?

Nitrogen in both \mathrm{NH_2^-} and \mathrm{NH_2\cdot} is approximately sp^3 hybridized, giving trigonal pyramidal geometry. Bond angles are slightly less than the tetrahedral value, near 107 degrees for amine like species.

Q: Does each NH2 species satisfy the octet rule?

\mathrm{NH_2^-} satisfies the octet, with eight electrons around N. \mathrm{NH_2\cdot} does not fully satisfy the octet, having seven electrons around N because of the unpaired electron.

Answer

Valence electrons: N 5 plus 2 from H equals 7 minus 1 for the positive charge gives 6. Two N H bonds use 4 electrons leaving 2 as a lone pair on N. Formal charge on N is +1. Lewis structure:

\[
\left[
\begin{array}{c}
\cdot\cdot \\
\mathrm{H\!-\!N\!-\!H} \\
\end{array}
\right]^+
\]

Final result: the ion \( \mathrm{NH_2^+} \) has nitrogen with one lone pair and two single bonds, and the positive charge resides on nitrogen.

Detailed Explanation

Goal. Draw the Lewis structure of the species NH2+ .

Step 1. Count total valence electrons. Nitrogen has 5 valence electrons. Each hydrogen has 1 valence electron, so two hydrogens contribute 2 electrons. The overall positive charge means one electron is removed. Thus the total valence electron count is

\[
\text{Total valence electrons} = 5_{\mathrm{N}} + 2\times 1_{\mathrm{H}} – 1_{\text{(positive charge)}} = 6 \text{ electrons}.
\]

What to do. Write that number down and keep track of electrons that will be used for bonds and lone pairs.

Step 2. Draw a reasonable skeleton. Nitrogen is the central atom and bonds to the two hydrogens. Place single bonds between N and each H. Each N–H single bond uses 2 electrons, so two bonds use 4 electrons in total. Show the skeletal connectivity as

\[
\mathrm{H}-\mathrm{N}-\mathrm{H}.
\]

What to do. Subtract the electrons used for bonding from the total to find how many remain for lone pairs.

Step 3. Place remaining electrons as lone pairs. From the total 6 electrons, 4 are used in the two N–H bonds, leaving 2 electrons. Place those 2 electrons as a lone pair on nitrogen. So after placing the lone pair the electrons are distributed as: two N–H bonding pairs and one lone pair on N.

Step 4. Assign formal charges to check the charge distribution. Compute the formal charge on nitrogen using the usual formula: formal charge = valence electrons of atom − (nonbonding electrons + one half of bonding electrons). For nitrogen here:

\[
\text{Formal charge on N} = 5 – \bigl(2 + \tfrac{1}{2}\times 4\bigr) = 5 – (2 + 2) = +1.
\]

Each hydrogen has formal charge 0. The sum of formal charges is +1, which matches the given overall charge. What this tells you is that the positive charge resides on the nitrogen atom.

Step 5. Write the final Lewis structure. Show the lone pair as two dots on nitrogen and indicate the positive charge on nitrogen. One conventional way to draw it is

\[
:\mathrm{N}^{\oplus}: \quad \text{with two single bonds to H, written } \mathrm{H}-\mathrm{N}^{\oplus}-\mathrm{H}.
\]

Summary. NH2+ has 6 total valence electrons. The Lewis structure has two N–H single bonds, one lone pair on nitrogen, and a formal positive charge on nitrogen. The nitrogen is the atom carrying the +1 formal charge.

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Chemistry FAQs

What does NH2 typically refer to in Lewis structure problems?

It can mean the aminyl radical \( \mathrm{NH_2\cdot} \), the amide anion \( \mathrm{NH_2^-} \), or the amine substituent \( \mathrm{-NH_2} \). Context tells you which species to draw.

How many valence electrons for \( \mathrm{NH_2^-} \) and how do you place them?

Total valence electrons: \(5 + 2(1) + 1 = 8\). Place N central, make two N-H single bonds, then place the remaining four electrons as two lone pairs on N. That gives complete octet and single negative formal charge.

How do you draw the Lewis structure for the aminyl radical \( \mathrm{NH_2\cdot} \)?

Total valence electrons: \(5 + 2(1) = 7\). Draw N with two N-H single bonds, one lone pair, and one unpaired electron on N. Formal charge on N is zero, and the octet is incomplete by one electron.

What are the formal charges on N in \( \mathrm{NH_2^-} \) and \( \mathrm{NH_2\cdot} \)?

Use FC = valence - nonbonding - 1/2 bonding. For \( \mathrm{NH_2^-} \), FC = \(5 - 4 - \tfrac{1}{2}(4) = -1\). For \( \mathrm{NH_2\cdot} \), FC = \(5 - 3 - \tfrac{1}{2}(4) = 0\).

How many lone pairs and unpaired electrons are on N in each species?

In \( \mathrm{NH_2^-} \) there are two lone pairs and no unpaired electrons. In \( \mathrm{NH_2\cdot} \) there is one lone pair and one unpaired electron on N.

What is the hybridization and molecular geometry of NH2 species?

Nitrogen in both \( \mathrm{NH_2^-} \) and \( \mathrm{NH_2\cdot} \) is approximately \( sp^3 \) hybridized, giving trigonal pyramidal geometry. Bond angles are slightly less than the tetrahedral value, near 107 degrees for amine like species.

Does each NH2 species satisfy the octet rule?

\( \mathrm{NH_2^-} \) satisfies the octet, with eight electrons around N. \( \mathrm{NH_2\cdot} \) does not fully satisfy the octet, having seven electrons around N because of the unpaired electron.

Where is the extrelectron located in \( \mathrm{NH_2^-} \) and what effect does it have on reactivity?

The extrelectron resides in nitrogen lone pair orbital, increasing basicity and nucleophilicity. \( \mathrm{NH_2^-} \) is strong base and nucleophile compared with neutral amines or the aminyl radical.

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