Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 20

Q. 20.14

Using average bond energies from Table A.9 of the appendix, calculate the difference in electronegativity between H and F.

Table A.9            Average Bond Energies
Bond Bond Energy/kJ mol^{−1}
Br−Br 193
C−C 343
C=C 615
C≡C 812
C−Cl 326
C−F 490
C−H 416
C−N 290
C≡N 891
C−O 351
C=O 724
C=O (in CO_{2}) 799
C≡O 1046
Cl−Cl 244
F−F 158
H−Br 366
H−Cl 432
H−H 436
H−F 568
H−I 298
H−N 391
H−S 367
N−N 160
N≡N 946
N−O 176
O−H 464
O−O 144
O=O (in O_{2}) 498
From F. T.Wall, Chemical Thermodynamics, 3rd ed.,W. H. Freeman,San Francisco, 1974, p. 63.

Step-by-Step

Verified Solution

We omit the units on the factors:

\left|X_{H}-X_{Li} \right|=(0.102)[568 − (1/2)(436 + 158)]^{1/2}=1.7            (Table value _ 1.9)