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Atomic Structure, Bonding andPeriodicity
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Atomic Structure
Fundamental Particles Mass Number and Isotopes
Mass Spectrometer Electron Arrangement Trends Ionisation Energies
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Atoms have a small central nucleus made up of protons and neutronsaround which there are electrons.
AtomicParticle
RelativeMass
RelativeCharge
neutron 1 0
proton 1 +1
electron negligible -1
Fundamental Particles
Electron
Neutron
Proton
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Mass Spectrometer
Thesample isput in viaan injector
Anelectrongunionisesthe atoms
Chargedplatesacceleratepositiveions
A curved magnetdeflects thepositive ions. Thelighter ions aredeflected more
The ions are detectedto produce a massspectrum
Ionisation acceleration deflection detection
The mass spectrum gives the relativemasses of each isotope and theabundance of each isotope.
The relative atomic mass is theweighted average mass of an atom of anelement compared with 1/12 of the massof 12 C.We can calculate this from massspectra
Example: Copper 69% of copper atoms have arelative mass of 63.
31% of copper atoms have arelative mass of 65.The weighted average is calculatedas follows:(0.69 x 63) + (0.31 x 65) =63.62
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Arrangement of Electrons Energy Levels or Shells
The simplest model of electrons has them orbiting in shellsaround the nucleus. Each successive shell is further from thenucleus and has a greater energy.
Sub Shells and Orbitals This model can be further refined by the concept of sub shells
and orbitals. Sub shells are known by letters s, p, d, and f. The s sub shell can
contain 2 electrons, p 6, d 10 and f 14. Electrons occupy negative charge clouds called orbitals, each
orbital can hold only 2 electrons. Each type of shell has adifferent type of orbital.
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How we write electron configurations Electrons fill the lowest energy level first this means it is generally easy to predict how the
electrons will fill the orbitals (it gets more complicated with the transition metals).
Arrangement of Electrons
Element Electronconfiguration
H 1s
He 1s 2
Li 1s 22s
Be 1s 22s 2
B 1s22s
22p
1
C 1s 22s 22p 2
N 1s 22s 22p 3
O 1s 22s 22p 4
F 1s 22s 22p 5
Element Electron configuration
Ne 1s 22s 22p 6Na 1s 22s 22p 63s 1
Mg 1s 22s 22p 63s 2
Al 1s 22s 22p 63s 23p 3
Si 1s 22s 22p 63s 23p 3
P 1s 22s 22p 63s 23p 3
S 1s 22s 22p 63s 23p 4
Cl 1s 22s 22p 63s 23p 5
Ar 1s 22s 22p 63s 23p 6
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Trends in Ionisation EnergiesThe first ionisation energies of group 2
electrons
0200
400600800
1000
Be Mg Ca Sr BaGroup 2 elements
E n e r g y
( K j / m o
l )
Going down a group in the periodic tablethere are more filled energy levels between thenucleus and the outer most electronsthese shield the outer electrons from theattractive force of the positive nucleus.
as the radius of the atom increases, the distancebetween the nucleus and the outer electronincreases and therefore the force of attractionbetween the nucleus and outer most electrons isreduced.These factors mean that less energy is neededto remove the first electron from an atom at thebottom of the group compared to one at the topof the group.
First ionisation energies of period 3elements
0
500
1000
1500
2000
Na Mg Al Si P S Cl Ar Period 3 elements
E n e r g y
( K j / m o
l )
Going across a period of elements:there are more protons in each nucleus so thenuclear charge in each element increases,this increases the attractive force acting onthe outer most electrons.there is no significant increase in shielding aseach successive electron enters the sameenergy level as the one before.Overall more energy is needed to remove thefirst electron from its outermost shell.
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Amount of Substance
Amount of Substance Calculations
Balancing Equations Reacting Quantities
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Calculations Amount of substance, n
n = massMr
Solution calculations The concentration of
solutions are measured inmol dm 3 c= concentration in dm 3 V = volume in cm 3
n = Vc1000
The Ideal Gas equation Providing that the
pressure and temperatureof gases are the same,equal volumes of twogases can be assumed tohave the same number of moles. pV = nRT
p is the pressure in Pa
V is the volume in m3
T is the temperature inKelvin
R= 8.31 JK -1 mol -1 Pr
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Balancing Equations
Chemical reactions involve the rearrangement of atoms not the making or destroying of atoms.
It is necessary to make sure that you have the
same amount of atoms on both sides of theequation.
State symbols can also be added to show thephysical condition of the reactants and products
(s) solid, (l) liquid, (g) gas, (aq) aqueous
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Reacting Quantities The numbers in a balanced equation give the ratio of the amount of
each substance in the reaction. We can use this information tocalculate quantities of reactants or products.
50g of CaCO 3 are heated how much CaO will be formed First write a balanced equation:
CaCO 3(s) CaO (s) + CO 2(g) Then calculate the Mr of the substances we are interested in:
CaCO3
40 + 12 + (3 x 16) = 100 CaO 40 + 16 =56
Calculate the number of moles of CaCO 3 used. n=mass/Mr 50/100 = 0.5 mol
From the equation we can see that one mole of CaCO 3 producesone mole of CaO. Therefore 0.5 mol of CaCO 3 produces 0.5 mol of
CaO. Finally calculate the mass of 0.5 mol of CaO: n=mass/Mr, mass=Mr x n, 0.5 x 56 = 28g
Therefore 50g of CaCO 3 produces 28g CaO
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Bonding
The nature of bonds Bond polarity and the polarisation of ions
Intermolecular forces Hydrogen Bonding
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Hydrogen Bonding Hydrogen bonds are a special case of
permanent dipole-permanent dipolebonding.
It exists where an electronegative
element such as oxygen, chlorinefluorine or nitrogen is bonded tohydrogen.
Hydrogen bonding causes stronger intermolecular bonds than wouldotherwise be predicted this increasesthe boiling point of substances such aswater.O
HHd+ d+
d-
O
HHd+ d+
d-
OHH
d+ d+
d-
O
HHd+ d+
d-
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Periodicity Chemists classify elements according to their position in the periodic
table. Periodicity is the term used to describe the repeating pattern of
properties observed within the periodic table.
H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac
s- block d-block p-block
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Trends in Period Three of the Periodic TableProperty Trend from
left to right
Explanation
Atomic radius decreases because the nuclear chargeincreases
First ionisationenergy
increases because the nuclear chargeincreases
Electronegativity increases because the nuclear chargeincreases
Electricalconductivity
Increases untilthe non metals
because the metals have anincreased number of delocalisedelectrons
Boiling point andMelting point
Increases untilthe middle thendecreases
Because these properties dependon the forces between the particles.This depends on the structure of theelement which varies from metallicto giant covalent to simple
molecular.
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