To determine which of the given bonds is the most polar, we need to understand the concept of electronegativity. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The greater the difference in electronegativity between the two atoms in a bond, the more polar the bond will be. Let's proceed step-by-step to find out which bond is the most polar.
1. List the elements and their electronegativity values:
- Oxygen (O): 3.5
- Sulfur (S): 2.5
- Phosphorus (P): 2.1
- Silicon (Si): 1.8
- Arsenic (As): 2.0
- Bromine (Br): 2.8
- Iodine (I): 2.5
2. Identify the bonds and their electronegativity differences:
- As-I bond:
- Electronegativity of As (Arsenic) = 2.0
- Electronegativity of I (Iodine) = 2.5
- Difference = |2.0 - 2.5| = 0.5
- S-O bond:
- Electronegativity of S (Sulfur) = 2.5
- Electronegativity of O (Oxygen) = 3.5
- Difference = |2.5 - 3.5| = 1.0
- Si-Br bond:
- Electronegativity of Si (Silicon) = 1.8
- Electronegativity of Br (Bromine) = 2.8
- Difference = |1.8 - 2.8| = 1.0
- O-P bond:
- Electronegativity of O (Oxygen) = 3.5
- Electronegativity of P (Phosphorus) = 2.1
- Difference = |3.5 - 2.1| = 1.4
3. Calculate the differences and compare:
- As-I bond difference: 0.5
- S-O bond difference: 1.0
- Si-Br bond difference: 1.0
- O-P bond difference: 1.4
4. Determine the bond with the greatest electronegativity difference:
- The electronegativity differences are: 0.5, 1.0, 1.0, and 1.4.
- The bond with the greatest difference is as follows:
[tex]\[
\text{O-P bond difference: } 1.4
\][/tex]
The bond [tex]\( \text{O-P} \)[/tex] has the highest electronegativity difference of [tex]\(1.4\)[/tex], making it the most polar bond among the given set of bonds.
