Download presentation
Presentation is loading. Please wait.
1
4.05 Atomic Structure and Electronic Configuration
Arranging the electrons in an atom Learning Objectives/Terminology- Aufbau Hund’s Rule Pauli Exclusion Principle Diamagnetism/Paramagnetism Ne-Va-S-P Dr. F. O. Garces Chemistry 100 Miramar College
2
Electronic Configuration
How are the electrons of an atom arranged in the atom? What are shells and orbitals arrange outside the nuclei of an atom? How is the e- arrangement liken to that of a Hotel room (Hotel del Orbital) ? What is the importance of the valence electrons and how do these influence chemistry?
3
Prelude to eConfiguration
Placing electrons in orbitals to complete the electron configuration.
4
Hotel del Orbitals @ Planet Claire
f 1 2 3 4 Like the hotel del Orbital, the shells of an atom are staggered and are filled based on their relative energies Filling Order (Aufbau Principle)
5
Hotel del Orbitals and a pictorial view of the shell arrangements
The filling order can be memorized by the following scheme 4 3 2 1
6
Shells and Orbitals
7
Shells and Orbitals 4d 5 5s 4p 4s 3d 4 3p 3 3s 1 2p 2 2 2s 3 4 5 1 1s
8
Relative Energies for Shells and Orbitals
p d f Relative Energies of the orbitals
9
The First 10 elements: H = 1s1 He = 1s2 Li = 1s2 2s1 Be = 1s2 2s2
Orbital Diagrams and Electron Configurations H = 1s1 He = 1s2 Li = 1s2 2s1 Be = 1s2 2s2 B = 1s2 2s2 2p1 C = 1s2 2s2 2p2 N = 1s2 2s2 2p3 O = 1s2 2s2 2p4 F = 1s2 2s2 2p5 Ne = 1s2 2s2 2p6 2s 2p
10
Electron Configuration Hydrogen and Helium
H = 1s1 1s 2s 2p He He = 1s2 1s 2s 2p No two electrons can have the same address (quantum number) Pauli Exclusion Principle
11
Relative Energies for Shells and Orbitals
p d f Relative Energies of the orbitals
12
Electron Configuration: Lithium to Beryllium
+ + Li = 1s 2s 2p Li = 1s2 2s1 Be + = 1s 2s 2p Be = 1s2 2s2 Building out from previous e-Config Aufbau Process
13
Relative Energies for Shells and Orbitals
p d f Relative Energies of the orbitals
14
Electron Configuration: Boron to Nitrogen
+ = C B B = 1s2 2s2 2p1 N = 1s2 2s2 2p3 1s 2s 2p C = 1s2 2s2 2p2 When an electron fills degenerate orbitals, it does so in an unoccupied orbital with the same spin as that of an occupied orbital Hund’s Rule e-Config with unpaired electrons Paramagnetic
15
Relative Energies for Shells and Orbitals
p d f Relative Energies of the orbitals
16
Electron Config: Oxygen to Neon
+ = 1s 2s 2p O = 1s2 2s2 2p4 + + = F 1s 2s 2p F = 1s2 2s2 2p5 Ne + = 1s 2s 2p Ne = 1s2 2s2 2p6 e-Config with all orbitals paired with e- Diamagnetic
17
Relative Energies for Shells and Orbitals
p d f Relative Energies of the orbitals
18
Electron Configuration using the Periodic Table
When using the Periodic table to determine the electron configuration of an atom, it is important to understand the layout of the periodic table. The row of a periodic table can be used to determine which energy level, 1, 2, 3... the valence electrons are located. The column of the periodic table helps determine how many valence electron an atom possesses. Elements in the 1A column (family) have one valence electron, in the IIA column, elements have two valence electron, in the IIIA family, elements have three valence electrons, and so on.
19
Electron Configuration using the Periodic Table
The elements designated in block is where the very last electrons are found in the s-orbitals. Therefore the elements in portion of the periodic table is referred to as the s-block elements. The elements designated in block is where the very last electrons are found in the p-orbitals. The elements found in this portion of the periodic table is referred to as the p-block elements. The same can be said about the block and -block with the elements found in this portion of the periodic table referred to as the d-block (transition metals) and f-block (man-made) elements. Sublevel Blocks to Write Electron Configurations
20
Electron Configuration: ne-va-s-p
ne - number of electrons; the total number of electrons va - valence electrons; the number of valence electrons s - shell of valence electrons p - previous noble gas e- config. for Sulfur ? H 1s1 Li 2s1 Na 3s1 K 4s1 Rb 5s1 Cs 6s1 Fr 7s1 Be 2s2 Mg 3s2 Ca 4s2 Sr 5s2 Ba 6s2 Ra 7s2 Sc 3d1 Ti 3d2 V 3d3 Cr 4s13d5 Mn 3d5 Fe 3d6 Co 3d7 Ni 3d8 Zn 3d10 Cu 4s13d10 B 2p1 C 2p2 N 2p3 O 2p4 F 2p5 Ne 2p6 He 1s2 Al 3p1 Ga 4p1 In 5p1 Tl 6p1 Si 3p2 Ge 4p2 Sn 5p2 Pb 6p2 P 3p3 As 4p3 Sb 5p3 Bi 6p3 S 3p4 Se 4p4 Te 5p4 Po 6p4 Cl 3p5 4p5 I 5p5 At 6p5 Ar 3p6 Kr 4p6 Xe 5p6 Rn 6p6 Y 4d1 La 5d1 Ac 6d1 Cd 4d10 Hg 5d10 Ag 5s14d10 Au 6s15d10 Zr 4d2 Hf 5d2 Db 6d2 Nb 4d3 Ta 5d3 Jl 6d3 Mo 5s14d5 W 6s15d5 Rf 7s16d5 Tc 4d5 Re 5d5 Bh 6d5 Ru 4d6 Os 5d6 Hn 6d6 Rh 4d7 Ir 5d7 Mt 6d7 4d8 5d8
21
Electron Configuration: Sulfur
ne - number of electrons; the total number of electrons va - valence electrons; the number of valence electrons s - shell of valence electrons p - previous noble gas s - shell of valence electrons = 3 Va - valence electrons; the number of valence electrons = 6 H 1s1 Li 2s1 Na 3s1 K 4s1 Rb 5s1 Cs 6s1 Fr 7s1 Be 2s2 Mg 3s2 Ca 4s2 Sr 5s2 Ba 6s2 Ra 7s2 Sc 3d1 Ti 3d2 V 3d3 Cr 4s13d5 Mn 3d5 Fe 3d6 Co 3d7 Ni 3d8 Zn 3d10 Cu 4s13d10 B 2p1 C 2p2 N 2p3 O 2p4 F 2p5 Ne 2p6 He 1s2 Al 3p1 Ga 4p1 In 5p1 Tl 6p1 Si 3p2 Ge 4p2 Sn 5p2 Pb 6p2 P 3p3 As 4p3 Sb 5p3 Bi 6p3 S 3p4 Se 4p4 Te 5p4 Po 6p4 Cl 3p5 4p5 I 5p5 At 6p5 Ar 3p6 Kr 4p6 Xe 5p6 Rn 6p6 Y 4d1 La 5d1 Ac 6d1 Cd 4d10 Hg 5d10 Ag 5s14d10 Au 6s15d10 Zr 4d2 Hf 5d2 Db 6d2 Nb 4d3 Ta 5d3 Jl 6d3 Mo 5s14d5 W 6s15d5 Rf 7s16d5 Tc 4d5 Re 5d5 Bh 6d5 Ru 4d6 Os 5d6 Hn 6d6 Rh 4d7 Ir 5d7 Mt 6d7 4d8 5d8 p - previous noble gas = Ne (10 e-) ne - number of electrons; the total number of electrons; this equals the number of protons or atomic number = 16 e- config for Sulfur S = [Ne]3s23p4
22
Electron arrangement for the Sulfur atom
p d f The 16 electrons for sulfur occupy the shells & orbitals of sulfur from the lowest energy to the highest. e- config for Sulfur S = [Ne] 3s2 3p4 16 total electrons Relative Energies of the orbitals and the filling order. S
23
Electron arrangement for the Scandium atom
p d f The 16 electrons for sulfur occupy the shells & orbitals of sulfur from the lowest energy to the highest. e- config for Scandium Sc = [Ar]4s23d1 21 total electrons Relative Energies of the orbitals and the filling order. Sc
24
Electron arrangement for the Scandium atom
p d f The 16 electrons for sulfur occupy the shells & orbitals of sulfur from the lowest energy to the highest. e- config for Tin Sn = [Kr]5s24d105p2 Relative Energies of the orbitals and the filling order. 50 total electrons Sn
25
The First 20 elements: H = 1s1 He = 1s2 Li = 1s2 2s1 Be = 1s2 2s2
B = 1s2 2s2 2p1 C = 1s2 2s2 2p2 N = 1s2 2s2 2p3 O = 1s2 2s2 2p4 F = 1s2 2s2 2p5 Ne = 1s2 2s22p6 Na = 1s2 2s22p63s1 Mg = 1s2 2s22p63s2 Al = 1s2 2s22p63s23p1 Si = 1s2 2s22p63s23p2 P = 1s2 2s22p63 s23p1 S = 1s2 2s2 2p63s23p4 Cl = 1s2 2s2 2p63s2 3p5 Ar = 1s2 2s2 2p6 3s2 3p6 1s 2p 3p 3s
26
SOLUTION SAMPLE PROBLEM 3.8
Orbital Diagrams and Electron Configurations For each of the following elements, write the stated type of electron notation: a. orbital diagram for silicon b. electron configuration for phosphorus c. abbreviated electron configuration for chlorine SOLUTION a. Silicon in Period 3 has atomic number 14, which tells us that it has 14 electrons. To write the orbital diagram, we draw boxes for the orbitals up to 3p. Add 14 electrons, starting with the 1s orbital. Show paired electrons in the same orbital with opposite spins, and place the last 2 electrons in different 3p orbitals.
27
SOLUTION CONTINUED STUDY CHECK SAMPLE PROBLEM 3.8
Orbital Diagrams and Electron Configurations For each of the following elements, write the stated type of electron notation: a. orbital diagram for silicon b. electron configuration for phosphorus c. abbreviated electron configuration for chlorine SOLUTION CONTINUED b. The electron configuration gives the electrons that fill the sublevels in order of increasing energy. Phosphorus is in Group 5A (15) in Period 3. In Periods 1 and 2, a total of 10 electrons fill sublevels: 1s2, 2s2, and 2p6. In Period 3, 2 electrons go into 3s2. The 3 remaining electrons (total of 15) are placed in the 3p sublevel. c. In chlorine, the previous noble gas is neon. For the abbreviated configuration, write [Ne] for 1s22s22p6 followed by the electrons in the 3s and 3p sublevels. STUDY CHECK Write the complete and abbreviated electron configurations for sulfur.
28
SOLUTION STUDY CHECK SAMPLE PROBLEM 3.9
Using Sublevel Blocks to Write Electron Configurations Use the sublevel blocks on the periodic table to write the electron configuration for bromine. SOLUTION STEP 1 Bromine is in the p block and in Period 4. STEP 2 Beginning with 1s2, go across the periodic table writing each filled sublevel block as follows: STEP 3 Five electrons in the 4p sublevel for Br (4p5) completes the electron configuration for Br: STUDY CHECK Write the electron configuration for tin.
29
Assignment: Determine the electron configuration.
1. Write out the electron configurations for the following atoms and ions. Determine the number of unpaired electrons in the ground state. 13Al 12Mg+2 50Sn 15P-3 34Se-1 32Ge+2 2. Write the electron box diagram for the following elements. Which have identical electron configuration (isoelectronic) ? S-2 Cl Ar Ca+2 3. What does the term: Aufbau, Pauli Exclusion and Hund’s Rule mean? 4. How does atomic radius change as one goes left to right along the periodic table? Explain*. 5. How does the ionization change as one goes down the periodic table? Explain*. * Explain, means explain why.
Similar presentations
