1 electron filling order figure 8.5. 2 electron configurations and the periodic table figure 8.7
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Electron Electron Filling Filling OrderOrder
Figure 8.5Figure 8.5
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Electron Configurations Electron Configurations and the Periodic Tableand the Periodic Table
Figure 8.7
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Relationship of Electron Relationship of Electron Configuration and Region Configuration and Region
of the Periodic Tableof the Periodic Table
• Gray = s blockGray = s block
• Orange = p blockOrange = p block
• Green = d blockGreen = d block
• Violet = f blockViolet = f block
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PhosphorusPhosphorusPhosphorusPhosphorus
All Group 5A All Group 5A elements have elements have [core ] ns[core ] ns2 2 npnp3 3
configurations configurations where n is the where n is the period number.period number.
Group 5AGroup 5A
Atomic number = 15Atomic number = 15
1s1s2 2 2s2s2 2 2p2p6 6 3s3s2 2 3p3p33
[Ne] 3s[Ne] 3s2 2 3p3p33
1s
2s
3s3p
2p
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LithiumLithiumLithiumLithium
Group 1A
Atomic number = 3
1s22s1 ---> 3 total electrons
1s
2s
3s3p
2p
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Electron PropertiesElectron Properties
Diamagnetic: NOT attracted to a magnetic fieldParamagnetic: substance is attracted to a magnetic field. Substance has unpaired electrons.
Diamagnetic: NOT attracted to a magnetic fieldParamagnetic: substance is attracted to a magnetic field. Substance has unpaired electrons.
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8
NeonNeonNeonNeon
Group 8A
Atomic number = 10
1s2 2s2 2p6 --->
Diamagnetic
1s
2s
3s3p
2p
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BerylliumBerylliumBerylliumBeryllium
Group 2A
Atomic number = 4
1s22s2
Diamagnetic
1s
2s
3s3p
2p
10
BoronBoronBoronBoron
Group 3A
Atomic number = 5
1s2 2s2 2p1
Paramagnetic
1s
2s
3s3p
2p
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CarbonCarbonCarbonCarbon
Group 4A
Atomic number = 6
1s2 2s2 2p2
Paramagnetic1s
2s
3s3p
2p
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FluorineFluorineFluorineFluorine
Group 7A
Atomic number = 9
1s2 2s2 2p5 --->
Paramagnetic
1s
2s
3s3p
2p
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Ion ConfigurationsIon ConfigurationsIon ConfigurationsIon Configurations
How do we know the configurations of ions?
Determine the magnetic properties of ions.
Ions with UNPAIRED ELECTRONS are
PARAMAGNETIC.
Without unpaired electrons DIAMAGNETIC.
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transition metal ionstransition metal ionstransition metal ionstransition metal ions
Fe [Ar] 4s2 3d6 loses 2 electrons ---> Fe2+ [Ar] 4s0 3d6
• loses 3 electrons ---> Fe3+ [Ar] 4s0 3d5
4s 3d 3d4s
Fe Fe2+
4s 3d 3d4s
Fe Fe2+
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Transition MetalsTransition Metals
How do they fill? How can we determine?
CopperCopperIronIron
ChromiumChromium
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Ion Configurations MnIon Configurations MnIon Configurations MnIon Configurations Mn
Mn [Ar] 4s2 3d5 ---> Mn5+ [Ar] 4s03d2
loses 5 electrons ---> Mn5+ [Ar] 4s2 3d0
4s 3d 3d4s
Fe Fe2+D P
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PERIODIPERIODIC C
TRENDSTRENDS
PERIODIPERIODIC C
TRENDSTRENDS
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Effective Nuclear Charge, Effective Nuclear Charge, Z*Z*
Effective Nuclear Charge, Effective Nuclear Charge, Z*Z*
• Z* is the nuclear charge experienced by Z* is the nuclear charge experienced by the outermost electrons.the outermost electrons. See p. 295 and Screen 8.6.See p. 295 and Screen 8.6.
• Explains why E(2s) < E(2p)Explains why E(2s) < E(2p)
• Z* increases across a period owing to Z* increases across a period owing to incomplete shielding by inner electrons.incomplete shielding by inner electrons.
• Estimate Z* by --> [ Estimate Z* by --> [ Z - (no. inner electrons) Z - (no. inner electrons) ]]
• Charge felt by 2s e- in Li Charge felt by 2s e- in Li Z* = 3 - 2 = 1 Z* = 3 - 2 = 1
• Be Be Z* = 4 - 2 = 2Z* = 4 - 2 = 2
• B B Z* = 5 - 2 = 3Z* = 5 - 2 = 3 and so on!and so on!
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Effective Nuclear Charge, Effective Nuclear Charge, Z*Z*
Effective Nuclear Charge, Effective Nuclear Charge, Z*Z*
• Z* is the nuclear charge experienced by Z* is the nuclear charge experienced by the outermost electrons.the outermost electrons. See p. 295 and Screen 8.6.See p. 295 and Screen 8.6.
• Explains why E(2s) < E(2p)Explains why E(2s) < E(2p)
• Z* increases across a period owing to Z* increases across a period owing to incomplete shielding by inner electrons.incomplete shielding by inner electrons.
• Estimate Z* by --> [ Estimate Z* by --> [ Z - (no. inner electrons) Z - (no. inner electrons) ]]
• Charge felt by 2s e- in Li Charge felt by 2s e- in Li Z* = 3 - 2 = 1 Z* = 3 - 2 = 1
• Be Be Z* = 4 - 2 = 2Z* = 4 - 2 = 2
• B B Z* = 5 - 2 = 3Z* = 5 - 2 = 3 and so on!and so on!
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EffectiveEffective Nuclear Charge, Z* Nuclear Charge, Z*
• Atom Z* Experienced by Electrons in Valence Orbitals
• Li +1.28
• Be -------
• B +2.58
• C +3.22
• N +3.85
• O +4.49
• F +5.13
Increase in Increase in Z* across a Z* across a periodperiod
21General Periodic General Periodic TrendsTrends
• Atomic and ionic sizeAtomic and ionic size
• Ionization energyIonization energy
• Electron affinityElectron affinity
Higher effective nuclear chargeElectrons held more tightly
Larger orbitals.Electrons held lesstightly.