Step 1: Understanding the Question:
The question asks to compare the count of bonding electrons (those shared between atoms) and non-bonding electrons (lone pairs) in various molecules and find the one that fits a specific ratio of 3:2.
Step 2: Detailed Explanation:
Let's analyze Nitrogen (\(N_2\)):
Nitrogen has 5 valence electrons. In \(N_2\), two nitrogen atoms form a triple bond (\(N \equiv N\)) to achieve an octet.
Number of bonding electrons = \(3 \text{ bonds} \times 2 = 6\).
Each nitrogen atom also has one lone pair. Total non-bonding electrons = \(2 \times 2 = 4\).
Ratio (Bonding : Non-bonding) = \(6 : 4 = 3 : 2\). This matches the requirement.
Let's analyze Oxygen (\(O_2\)):
Oxygen has 6 valence electrons. Two oxygen atoms form a double bond (\(O = O\)).
Bonding electrons = \(2 \times 2 = 4\).
Each oxygen has 2 lone pairs. Total non-bonding = \(2 \times 4 = 8\).
Ratio = \(4 : 8 = 1 : 2\).
Let's analyze Hydrogen Chloride (\(HCl\)):
Formed by a single bond (\(H - Cl\)).
Bonding electrons = 2.
Chlorine has 3 lone pairs (6 electrons). Hydrogen has zero.
Total non-bonding = 6.
Ratio = \(2 : 6 = 1 : 3\).
Let's analyze Fluorine (\(F_2\)):
Formed by a single bond (\(F - F\)).
Bonding electrons = 2.
Each fluorine has 3 lone pairs. Total non-bonding = 12.
Ratio = \(2 : 12 = 1 : 6\).
Step 3: Final Answer:
In the Nitrogen molecule (\(N_2\)), the ratio of bonding electrons to non-bonding electrons is exactly 3:2.