Question

part 2 (1 point) rank the bonds in the two molecules shown in order of increasing polarity. (electronegativity values are: c, 2.5; f, 4.0; h, 2.1; o, 3.5) covalent bond (4 items) (drag and drop into the appropriate area) relative polarity most polar 1 c - h 2 o - h 3 c - f 4 c - o least polar

Explanation:

Step1: Calculate electronegativity difference for C - H

The electronegativity of C is 2.5 and of H is 2.1. The difference $\Delta\chi_{C - H}=2.5 - 2.1=0.4$.

Step2: Calculate electronegativity difference for O - H

The electronegativity of O is 3.5 and of H is 2.1. The difference $\Delta\chi_{O - H}=3.5 - 2.1 = 1.4$.

Step3: Calculate electronegativity difference for C - F

The electronegativity of C is 2.5 and of F is 4.0. The difference $\Delta\chi_{C - F}=4.0 - 2.5=1.5$.

Step4: Calculate electronegativity difference for C - O

The electronegativity of C is 2.5 and of O is 3.5. The difference $\Delta\chi_{C - O}=3.5 - 2.5 = 1.0$.

Step5: Rank the bonds

Based on the electronegativity - difference values, the order of increasing polarity is: C - H (least polar), C - O, O - H, C - F (most polar).

Answer:

1. C - H
2. C - O
3. O - H
4. C - F