Question

6.61 using electronegativity values in figure 6.11, for each pair of bonds, indicate the more polar bond and use an arrow to show the direction of polarity in each bond:
(a) c-o and c-n
(b) p-br and p-cl
(c) b-o and b-s
(d) b-f and b-i
6.62

Explanation:

Step1: Determine electronegativity values

Assume electronegativity values: \(C = 2.5\), \(O=3.5\), \(N = 3.0\), \(P = 2.1\), \(Br=2.8\), \(Cl = 3.0\), \(B=2.0\), \(S = 2.5\), \(F = 4.0\), \(I = 2.5\)

Step2: Calculate electronegativity differences

• For \(C - O\): \(\Delta\chi=3.5 - 2.5=1.0\)
• For \(C - N\): \(\Delta\chi=3.0 - 2.5 = 0.5\)
• For \(P - Br\): \(\Delta\chi=2.8 - 2.1=0.7\)
• For \(P - Cl\): \(\Delta\chi=3.0 - 2.1 = 0.9\)
• For \(B - O\): \(\Delta\chi=3.5 - 2.0=1.5\)
• For \(B - S\): \(\Delta\chi=2.5 - 2.0 = 0.5\)
• For \(B - F\): \(\Delta\chi=4.0 - 2.0=2.0\)
• For \(B - I\): \(\Delta\chi=2.5 - 2.0 = 0.5\)

Step3: Determine polarity direction

The more electronegative atom has the negative end of the bond dipole.

• \(C - O\): \(\overset{\delta+}{C}-\overset{\delta -}{O}\)
• \(C - N\): \(\overset{\delta+}{C}-\overset{\delta -}{N}\)
• \(P - Br\): \(\overset{\delta+}{P}-\overset{\delta -}{Br}\)
• \(P - Cl\): \(\overset{\delta+}{P}-\overset{\delta -}{Cl}\)
• \(B - O\): \(\overset{\delta+}{B}-\overset{\delta -}{O}\)
• \(B - S\): \(\overset{\delta+}{B}-\overset{\delta -}{S}\)
• \(B - F\): \(\overset{\delta+}{B}-\overset{\delta -}{F}\)
• \(B - I\): \(\overset{\delta+}{B}-\overset{\delta -}{I}\)

Answer:

(a) \(C - O\) is more polar. Polarity: \(\overset{\delta+}{C}-\overset{\delta -}{O}\), \(\overset{\delta+}{C}-\overset{\delta -}{N}\)
(b) \(P - Cl\) is more polar. Polarity: \(\overset{\delta+}{P}-\overset{\delta -}{Br}\), \(\overset{\delta+}{P}-\overset{\delta -}{Cl}\)
(c) \(B - O\) is more polar. Polarity: \(\overset{\delta+}{B}-\overset{\delta -}{O}\), \(\overset{\delta+}{B}-\overset{\delta -}{S}\)
(d) \(B - F\) is more polar. Polarity: \(\overset{\delta+}{B}-\overset{\delta -}{F}\), \(\overset{\delta+}{B}-\overset{\delta -}{I}\)