Questions: Your instructor added question feedback to this assessment Assignment Content Question 1 Consider the following Lewis structure: H-O=N=O How many valence electrons are present in the entire structure: 24 How many bonding pairs of electrons are in the entire structure: 5 How many lone pairs of electrons are in the entire structure: 9 Feedback Valence electrons are the electrons that each atom is either sharing with other atoms or holding as lone pairs. The number of valence electrons in the structure can be found by counting the numbers of dots and lines in the structure. Each line represents a pair of shared electrons and each dot represents one lone electron. A bonding pair of electrons is represented as a single line. A lone pair (nonbonding pair) of electrons is represented as a pair of dots.

Solution Steps

To determine the number of valence electrons in the entire structure, we need to count the valence electrons for each atom in the molecule. The structure given is H-O=N=O.

• Hydrogen (H) has 1 valence electron.
• Oxygen (O) has 6 valence electrons.
• Nitrogen (N) has 5 valence electrons.

Sum the valence electrons for all atoms in the molecule: \[ 1 \text{ (H)} + 6 \text{ (O)} + 5 \text{ (N)} + 6 \text{ (O)} = 18 \text{ valence electrons} \]

Each bonding pair of electrons is represented by a single line in the Lewis structure. The given structure H-O=N=O has:

• 1 single bond between H and O
• 1 double bond between O and N
• 1 double bond between N and O

This results in: \[ 1 \text{ (H-O)} + 2 \text{ (O=N)} + 2 \text{ (N=O)} = 5 \text{ bonding pairs} \]

Lone pairs are represented as pairs of dots. In the given structure:

• The first oxygen (O) has 2 lone pairs.
• The nitrogen (N) has 1 lone pair.
• The second oxygen (O) has 2 lone pairs.

This results in: \[ 2 \text{ (O)} + 1 \text{ (N)} + 2 \text{ (O)} = 5 \text{ lone pairs} \]

Final Answer