Question

name
lewis structures worksheet
total the valence electrons, then use lewis structures to show the covalent bonding in the following:
cbr₄
pcl₃
ocl₂
n₂
c₂h₄

Explanation:

Step1: Determine valence electrons for each atom

• For \(OCl_{2}\):
• Oxygen (\(O\)) has \(6\) valence electrons (\(Group - 6A\)).
• Chlorine (\(Cl\)) has \(7\) valence electrons (\(Group - 7A\)).
• Total valence electrons \(=6+(2\times7)=6 + 14=20\).
• For \(N_{2}\):
• Nitrogen (\(N\)) has \(5\) valence electrons (\(Group - 5A\)).
• Total valence electrons \(=2\times5 = 10\).
• For \(C_{2}H_{4}\):
• Carbon (\(C\)) has \(4\) valence electrons (\(Group - 4A\)).
• Hydrogen (\(H\)) has \(1\) valence electron (\(Group - 1A\)).
• Total valence electrons \(=(2\times4)+(4\times1)=8 + 4=12\).

Step2: Draw Lewis structures

• For \(OCl_{2}\):
• Oxygen is the central atom. It forms single bonds with two chlorine atoms.
• After bonding, oxygen has two lone pairs and each chlorine has three lone pairs. Lewis structure: \(\ce{:\ddot{O} - Cl:}\) (with appropriate lone pairs on each atom).
• For \(N_{2}\):
• Nitrogen atoms form a triple bond. Each nitrogen has one lone pair. Lewis structure: \(\ce{:N#N:}\).
• For \(C_{2}H_{4}\):
• Carbon atoms form a double bond. Each carbon forms single bonds with two hydrogen atoms. Lewis structure: \(\ce{H_{2}C=CH_{2}}\) (with appropriate electron distribution).

Answer:

Formula	Total Valence \(e^{-}\)	Lewis Structure
\(N_{2}\)	\(10\)	\(\ce{:N#N:}\)
\(C_{2}H_{4}\)	\(12\)	\(\ce{H_{2}C=CH_{2}}\)