Question

For each pair of bonds, indicate the more polar bond, and use an arrow to show the direction of polarity in each bond.Use electronegativitya. C-O and C-Nb. P-Br and P-Clc. B-O and B-Sd. B-F and B-I

Answer 1

Step 1

Data needed: electronegativities from all atoms involved here

(Please, look them up in the periodic table)

C 2.55

O 3.44

N 3.04

Br 2.96

P 2.19

Cl 3.16

B 2.04

S 2.58

F 3.98

I 2.66

Step 2

Polarity is compared using the next formula:

ΔE = subtraction

e.g. for C-O ΔE = E oxygen - E carbon = 3.44 - 2.55 = 0.89 (most polar)

For C-N: ΔE = 3.04 - 2.55 = 0.49

So, for a. C-O <=== C-N

b. P-Br and P-Cl

ΔE (P-Br) = 2.96 - 2.19 = 0.77

ΔE (P-Cl) = 3.16 - 2.19 = 0.97

P-Br ===> P-Cl

c. B-O and B-S

ΔE ( B-O) = 3.44 - 2.04 = 1.4

ΔE (B-S) = 2.58 - 2.04 = 0.54

B-O <=== B-S

d. B-F and B-I​

ΔE ( B-F) = 3.98 - 2.04 = 1.94

ΔE (B-I​) = 2.66 - 2.04 = 0.62

B-F <=== B-I​

Answer:

a. C-O <=== C-N

b. P-Br ===> P-Cl

c. B-O <=== B-S

d. B-F <=== B-I​