CHEM 130 Lecture Notes - Lecture 11: Lattice Energy, Electronegativity, Transition Metal

36 views4 pages
School
Department
Course

Document Summary

You can do this for more complicated formations too. Same process, you just have more steps associated. Na is larger than li so it going to have a less negative lattice energy. The distance between na and f is going to be larger so less stabilization- less negative. The ionization will be a little less positive because easier to pull an electron out of. The charge on mg is 2+, charge one o is. 2-, would would expect 4x the charge but mg is larger than li and o is larger than f so increase in q1 and q2 is balanced in radius, ends up being a little less than 4x as large. Clicker: the transition metal iron (fe) can exist as either a +2 or +3 ion. Percent ionic character of a bond: we can calculate this by: (the measured dipole moment of x-y) / (calculated dipole moment of x+ y-) x 100%

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers

Related textbook solutions

Related Documents

Related Questions