Ionization Energy ↑, Atomic Size 11 Metallic Character • For atoms, ionization energy (IE) is the energy required for the complete removal of 1 mol of e 's from 1 mol of gaseous atoms: Atom(g) - - Ion+(g) + e AE = IE, > 0 Increases → higher zeff valence orbitals are at lower E harder to lose e from higher IE→ ↑ lower n, valence orbitals are at er E er to lose e from er IE ↑ IONIZATION ENERGY • Metals tend to lose e 's in their chemical changes. Nonmetals tend to gain e 's in their chemical changes. + lower zeff, Valence orbitals are at huaherE = easier to lose e from =hiaher MC+ ļ higher n, valence orbitals are at er E er to lose e from er MC I Increases METALLIC BEHAVIOR • e s' are easier to lose and harder to add farther from the nucleus! → Recall the valence orbitals trends in size & E to figure the others.
Ionization Energy ↑, Atomic Size 11 Metallic Character • For atoms, ionization energy (IE) is the energy required for the complete removal of 1 mol of e 's from 1 mol of gaseous atoms: Atom(g) - - Ion+(g) + e AE = IE, > 0 Increases → higher zeff valence orbitals are at lower E harder to lose e from higher IE→ ↑ lower n, valence orbitals are at er E er to lose e from er IE ↑ IONIZATION ENERGY • Metals tend to lose e 's in their chemical changes. Nonmetals tend to gain e 's in their chemical changes. + lower zeff, Valence orbitals are at huaherE = easier to lose e from =hiaher MC+ ļ higher n, valence orbitals are at er E er to lose e from er MC I Increases METALLIC BEHAVIOR • e s' are easier to lose and harder to add farther from the nucleus! → Recall the valence orbitals trends in size & E to figure the others.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Ionization Energy 11 Atomic Size 1↑ Metallic Character
• For atoms, ionization energy (IE) is the energy required for the
complete removal of 1 mol of e 's from 1 mol of gaseous atoms:
Atom(g) Ion*(g) + e
AE = IE, > 0
→ higher zeff valence orbitals are at louer E
- horder to lose e from =>
Increases
higher IE>
↑ lower n, valence orbitals are at er E
er to lose e from er IE ↑
IONIZATION ENERGY
• Metals tend to lose e 's in their chemical changes.
Nonmetals tend to gain e 's in their chemical changes.
e lower Zeffr
= easier to lose e from higher MC+
| higher n, valence orbitals are at
er to lose e from er MC I
valence orbitals are at uaherE
Increases
er E
METALLIC BEHAVIOR
• e s' are easier to lose and harder to add farther from the nucleus!
→ Recall the valence orbitals trends in size & E to figure the others.
Increase
eases
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