When asked to draw the Lewis structure for nitrogen monoxide (NO), how many total electrons should be counted?

To determine the total number of electrons to be counted for drawing the Lewis structure of nitrogen monoxide (NO), one must first consider the contributions from both nitrogen and oxygen atoms based on their respective valence electrons.

Nitrogen is in group 15 of the periodic table and has five valence electrons. Oxygen is in group 16 and has six valence electrons. Therefore, the total number of valence electrons from both atoms can be calculated as follows:

• Nitrogen: 5 valence electrons
• Oxygen: 6 valence electrons

Adding these yields a total of 11 valence electrons:

5 (from nitrogen) + 6 (from oxygen) = 11 valence electrons

This total accounts for the number of electrons that will be used to form bonds and lone pairs in the Lewis structure, which is critical for accurately depicting the molecular structure of nitrogen monoxide. Drawing the Lewis structure will involve creating a bond between the nitrogen and oxygen atoms and distributing any remaining electrons as lone pairs while ensuring that both atoms fulfill the octet rule as best as possible.

Thus, the correct count of total electrons to be counted for the Lewis structure of nitrogen monoxide is indeed 11.