Question

How many unbonded $e^{-}$ are present on N in NO?

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B

3

For NO to achieve octet, it needs share only 2 electrons with Nitrogen (7N).

$\therefore :\stackrel{\cdot }{N}=\stackrel{\cdot \cdot }{O}:$

After this oxygen has 8$e^{-}$, but N is left with only 7 electrons.

$\therefore$ Oxygen shares one more of its electron to form an odd electron bond