Which two bonds are most similar in polarity? Which two bonds are most similar in polarity? O-F and Cl-F B-F and Cl-F I-Br and Si-Cl Al-Cl and I-Br C-Cl and Be-Cl
The Correct Answer and Explanation is:
Correct Answer: O–F and Cl–F
Explanation
Polarity in chemical bonds is determined by the difference in electronegativity between the two atoms involved. The greater the difference, the more polar the bond. To determine which two bonds are most similar in polarity, we need to compare the electronegativity differences for each bond.
Here are the relevant electronegativities (based on the Pauling scale):
- Oxygen (O): 3.44
- Fluorine (F): 3.98
- Chlorine (Cl): 3.16
- Boron (B): 2.04
- Iodine (I): 2.66
- Bromine (Br): 2.96
- Silicon (Si): 1.90
- Aluminum (Al): 1.61
- Carbon (C): 2.55
- Beryllium (Be): 1.57
Now, calculate the electronegativity differences:
- O–F: |3.44 − 3.98| = 0.54
- Cl–F: |3.16 − 3.98| = 0.82
- B–F: |2.04 − 3.98| = 1.94
- I–Br: |2.66 − 2.96| = 0.30
- Si–Cl: |1.90 − 3.16| = 1.26
- Al–Cl: |1.61 − 3.16| = 1.55
- C–Cl: |2.55 − 3.16| = 0.61
- Be–Cl: |1.57 − 3.16| = 1.59
Now compare the electronegativity differences:
- O–F and Cl–F both have small and similar electronegativity differences (0.54 and 0.82 respectively). This means their polarities are relatively close compared to the other pairs listed.
Other pairs show large differences in bond polarity or are composed of atoms with quite different electronegativities, making them dissimilar.
Thus, O–F and Cl–F are the two bonds most similar in polarity, as both involve fluorine bonded to highly electronegative atoms and have close electronegativity differences.
