QUESTION IMAGE
Question
arrange the highlighted bonds in the table below in decreasing order of polarity. that is, pick 1 for the most polar bond, pick 2 for the next most polar bond, and so on.
bond polarity
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To determine the polarity of bonds, we use the concept of electronegativity difference between the bonded atoms. The greater the electronegativity difference, the more polar the bond.
Step 1: Identify the bonds
- First bond: \( \text{C-F} \) (carbon - fluorine)
- Second bond: \( \text{C-O} \) (carbon - oxygen)
- Third bond: \( \text{C-C} \) (carbon - carbon)
Step 2: Recall electronegativity values
- Electronegativity of \( \text{C} \): ~2.5
- Electronegativity of \( \text{F} \): ~4.0
- Electronegativity of \( \text{O} \): ~3.5
- Electronegativity of \( \text{C} \): ~2.5
Step 3: Calculate electronegativity differences
- For \( \text{C-F} \): \( |4.0 - 2.5| = 1.5 \)
- For \( \text{C-O} \): \( |3.5 - 2.5| = 1.0 \)
- For \( \text{C-C} \): \( |2.5 - 2.5| = 0.0 \)
Step 4: Rank by polarity
The greater the electronegativity difference, the more polar the bond. So the order from most polar to least polar is: \( \text{C-F} \) (1), \( \text{C-O} \) (2), \( \text{C-C} \) (3)
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- For the \( \text{C-F} \) bond (first row): 1
- For the \( \text{C-O} \) bond (second row): 2
- For the \( \text{C-C} \) bond (third row): 3