Question

part 3b 9 marks choose 3 of 4 (lesson 3b) draw the following covalent bonds a) pcl3 b) so2 c) cbr4 d) seo3

Explanation:

Step1: Determine valence electrons

For $PCl_3$: Phosphorus (P) has 5 valence electrons and chlorine (Cl) has 7 valence electrons. Total valence electrons = 5+(3×7)=26.
For $SO_2$: Sulfur (S) has 6 valence electrons and oxygen (O) has 6 valence electrons. Total valence electrons = 6+(2×6)=18.
For $CBr_4$: Carbon (C) has 4 valence electrons and bromine (Br) has 7 valence electrons. Total valence electrons = 4+(4×7)=32.
For $SeO_3$: Selenium (Se) has 6 valence electrons and oxygen (O) has 6 valence electrons. Total valence electrons = 6+(3×6)=24.

Step2: Draw Lewis structure for $PCl_3$

Place P in the center. Connect P to 3 Cl atoms with single - bonds. Each Cl has 3 lone - pairs. P has 1 lone - pair.

Step3: Draw Lewis structure for $SO_2$

Place S in the center. Connect S to 2 O atoms. One S - O is a single bond and the other is a double bond. S has 1 lone - pair, and the O atoms have appropriate lone - pairs to complete their octets.

Step4: Draw Lewis structure for $CBr_4$

Place C in the center. Connect C to 4 Br atoms with single - bonds. Each Br has 3 lone - pairs.

Answer:

Draw the Lewis structures as described above for three of the four compounds ($PCl_3$, $SO_2$, $CBr_4$).