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Periodic table. Electron configuration of carbon atoms and molecules. . John Summerscales. Fundamental particles in atom. Atomic number = number of protons for balanced charge (in atom) = number of electrons value is characteristic of a specific element
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Periodic table.Electron configuration of carbon atoms and molecules. John Summerscales
Fundamental particles in atom • Atomic number = number of protons • for balanced charge (in atom) = number of electrons • value is characteristic of a specific element • Atomic weight = number of (protons + neutrons) • can have partial values as isotopes have different numbers of neutrons and different proportions
The atom (simple model) If K=1, L=2, M=3, N=4, then at each level there are 2n2 electrons: n=1 gives 2 electrons n=2 gives 8 electrons n=3 gives 18 electrons n=4 gives 32 electrons K L M N
The atom (simple model) If n = shell number - 1, then each shell has 2(2n+1) additional electrons: n=0 gives 2 electrons (s-block) n=1 gives 6 electrons (p-block) n=2 gives 10 electrons (d-block) n=3 gives 14 electrons (f-block) nucleus (protons and neutrons) level 1 = 2 level 2 = 2+6 = 8 level 3 = 2+6+10 = 18 level 4 = 2+6+10+14 =32
The Periodic Table Periodic table of the Elements s-block p-block d-block f-block
Inert gases Fibres - glass: B, O, Al, Si aramid: H, C, N, O Resins - H, C, N, O Periodic table of the Elements This column has a full electron shell- the most stable configuration
Ionic bonding Periodic table of the Elements This column lose one electron to become X+ This column lose two electrons to become X2+ This column lose three electrons to become X3+
Ionic bonding Periodic table of the Elements This column gains one electron to become X- This column gains two electrons to become X2- This column gains three electrons to become X3-
Carbon (and Si, Ge, Sn, Pb) Periodic table of the Elements This column could become either X4+ or X4- ??
Carbon 4+ or 4- ?? • In practice, • six protons holding three electrons> strong force/electron (difficult to remove 4th e-) • six protons holding ten electrons> weak force/electron (difficult to retain 10th e-) • so, carbon shares electrons > covalent bonding • one bond ... is ... two shared electrons
Carbon - covalent bonding • carbon has 4 electrons in the outer shell needs four electrons to fill shell • can share with 2, 3 or 4 other atoms • 4 other atoms = 4 x single (σ) bonds • 3 other atoms = 3 x σ and 1 x double (π) bond • 2 other atoms = two σ and two π bonds - 1 x single and 1 x triple (2π) bonds • but ....
Carbon - electron orbitals • electron shells divide into electron orbitals • each has up to two electrons of opposite spin • electrons enter empty orbitals first • at level 2 of Periodic Table, maximum of: • 2 electrons in a spherical orbital • 2 electrons in a dumbbell orbit on x-axis • 2 electrons in a dumbbell orbit on y-axis • 2 electrons in a dumbbell orbit on z-axis
Electron orbitals 1s 2s2px2py Note: the orbitals are not drawn to scale. They are probabilities of finding an electron.The pz orbital is normal to the plane of this image.
Electron orbitals (px, py, pz) Animation
Electron configurations Periodic table of the 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 2s2 2p6 2p1 = px1 2p2 = px1 py1 2p3 = px1 py1 pz1 2p4 = px2 py1 pz1 2p5 = px2 py2 pz1 2p6 = px2 py2 pz2
Methane (CH4) • carbon bonded to four hydrogen atoms • if each H bonds to a different electron orbitalthe resulting molecule is asymmetric • symmetrical molecules have lowest energyand are thus the most stable form • so (2s + 2px +2py + 2pz) reorganise tofour hybrid sp3 orbitals (think s1p3 !!)oriented along each line from theapex to the centre of a tetrahedron
CH4 tetrahedron • Pyramid with a triangular base • carbon nucleus at centre • hydrogen at each apex • sp3 orbital on each line from apex to base centre
Electron orbitals (sp3 hybrid molecular orbital) y x forward behind plane forward
add a methylene group -CH2- • Methane CH4 generic CnH2n+2 • Ethane C2H6 • Propane C3H8 • Butane C4H10 • Pentane C5H12 • Hexane C6H14 • Heptane c7H16 • Octane C8H18 • ...paraffins ... polyethylene
... with one double bond: • Methene n/ageneric CnH2n • Ethene C2H4 a.k.a. ethylene • Propene C3H6 a.k.a. propylene • Butene C4H8 a.k.a. butylene • Pentene C5H10 • Hexene C6H12 • Heptene c7H14 • Octene C8H16 • etcetera ....
... with one triple bond: • Methyne n/ageneric CnH2n-2 • Ethyne C2H2 a.k.a. acetylene • Propyne C3H4 • Butyne C4H6 • Pentyne C5H8 • Hexyne C6H10 • Heptyne c7H12 • Octyne C8H14 • etcetera ....
sp2 hybrid orbital • 2s + 2px + 2py hybridise to 3 x sp2 orbitals • 2pz orbital remains and forms double bond < plan view (excl. pz) side view > pz
Double bond (C=C) π (1e-) σ (2e-) π (1e-) half of double (π) bond electrons above atom centres centres single (σ) bond on line of atom centres half of double (π) bond electrons below atom centres Triple bond (C=C) has π orbitalsabove, below, in front and behind the σ bond
Consider σ and πbonds as springs compression tension torsion
Hybrid orbitals - summary • sp3 bonds to 4 other atoms4σ (single) bonds bond angle = 109° 28’ (tetrahedral molecule) • sp2 bonds to 3 other atoms3σ and 1πbond bond angle = 120° (triangular molecule)1σand1πbond = the double bond (i.e. 1+1 = 2) • sp bonds to 2 other atoms2σ and 2πbonds bond angle = 180° (linear molecule)1σand2πbonds = the triple bond (i.e. 1+2 = 3)
Benzene (C6H6 - cyclohextriene) • ring of six carbon atomsignore H atoms to give C at each corner tri-ene is three double bonds symmetry results in hexagonal molecule symmetry gives lowest energy so stable molecule
Benzene (Kekulé resonance) • left molecule is same as right molecule but upside down • double bonds constantly switch positions • change is so fast that upper 3 electrons appear as a single ring lower 3 electrons appear as a single ring
Benzene ring • delocalised (conjugated) electrons C-C bond length 1.54 Å C:C bond in benzene 1.39 Å C=C bond length 1.33 Å
Chemical bond typeand chemical bond densityeach determine materialstiffness/strengthand chemical durability Conclusion: