Question

1. choose the best answer.

which structure represents a correctly drawn lewis structure for ammonia (nh₃)?
h—n—h
|
h
h—:n:—h
|
h
n—h—h—h
h—:n:—h
|
h

Explanation:

To determine the correct Lewis structure for \( \text{NH}_3 \) (ammonia), we analyze the valence electrons:

• Nitrogen (\( \text{N} \)) has 5 valence electrons.
• Hydrogen (\( \text{H} \)) has 1 valence electron, and there are 3 H atoms, so \( 3 \times 1 = 3 \) valence electrons from H.
• Total valence electrons: \( 5 + 3 = 8 \).

In a Lewis structure, we form single bonds between \( \text{N} \) and each \( \text{H} \) (3 single bonds, using \( 3 \times 2 = 6 \) electrons). The remaining electrons ( \( 8 - 6 = 2 \) electrons, or 1 lone pair) should be on the \( \text{N} \) atom.

• The first option: No lone pairs on \( \text{N} \), incorrect (missing lone pair).
• The second option: \( \text{N} \) has 1 lone pair (2 electrons) and 3 single bonds to H. This uses \( 3 \times 2 + 2 = 8 \) electrons, which matches the total valence electrons.
• The third option: Incorrect bonding (linear arrangement, not the correct structure for \( \text{NH}_3 \)).
• The fourth option: Too many lone pairs (5 electrons, which is incorrect as \( \text{N} \) should have 1 lone pair (2 electrons) after bonding).

Answer:

The second option: \( \boldsymbol{\text{H}-\stackrel{\boldsymbol{\cdot\cdot}}{\text{N}}-\text{H}} \) (with the lone pair on N and three single bonds to H)