Write the Lewis structure for NH[latex]_{3}[/latex].

Since hydrogen is always terminal, put nitrogen in the central position. H [latex]\underset{H}{N} [/latex] H Total number of electrons for Lewis structure = [latex] \left( \begin{matrix} \text{number of} \\ \text{valence} \\ e^- \text{for }N \end{matrix} \right) [/latex] +3 [latex] \left( \begin{matrix} \text{number of } \\ \text{valence }\\ e^- \text{for } H \end{matrix}\right) [/latex] = 5 + 3(1) = 8 Bonding electrons are first. H:[latex]\underset{\overset{..}{H} }{N} [/latex]:H (6 of 8 electrons used) Lone pairs on terminal atoms are next, but none are needed on hydrogen.Lone pairs on central atom are last. [latex] \begin{matrix}H -\underset{|} {\ddot{N} } -H\\ H\end{matrix}[/latex] (8 of 8 electrons used) Since all of the atoms have octets (or duets for hydrogen), the Lewis structure for NH[latex]_{3}[/latex] is complete as shown in step 3.

Question 9.5: Write the Lewis structure for NH3.

Principles of Chemistry: A Molecular Approach

Write the Lewis structure for NH $_{3}$ .

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Since hydrogen is always terminal, put nitrogen in the central position.

H $\underset{H}{N}$ H

Total number of electrons for Lewis structure =

$\left( \begin{matrix} \text{number of} \\ \text{valence} \\ e^- \text{for }N \end{matrix} \right)$ +3 $\left( \begin{matrix} \text{number of } \\ \text{valence }\\ e^- \text{for } H \end{matrix}\right)$

= 5 + 3(1) = 8

Bonding electrons are first.

H: $\underset{\overset{..}{H} }{N}$ :H

(6 of 8 electrons used)
Lone pairs on terminal atoms are next, but none are needed on hydrogen.Lone pairs on central atom are last.