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Atomic structure Topic 2

Isotopes - questions What is the relative atomic mass of chlorine if it has two isotopes with the following abundances: 35Cl at 75% and 37Cl at 25%? Boron exists in 2 isotopic forms, 10B and 11B. Use your periodic table to find the abundances of the two isotopes.

Answers 35.5amu 10B = 19.00% and 11B = 81.00%

THE MASS SPECTROMETER Boron Zirconium
Calculate the RAM of these elements from there mass spectra: Boron Zirconium 100 51.5 23 17.1 17.4 11.2 2.8

THE MASS SPECTROMETER CALCULATING RAM OF ELEMENTS
(10 x 23) + (100 x 11) 123 RAM = 10.8 Boron 100 23

THE MASS SPECTROMETER CALCULATING RAM OF ELEMENTS RAM = 91.3
(90 x 51.5) + (91 x 11.2) + (92 x 17.1) + (94 x 17.4) + (96 x 2.8) 100 RAM = 91.3 51.5 17.1 17.4 11.2 2.8

Spectrums Distinguish between a continuous spectrum and a line spectrum.

CONTINUOUS SPECTRUM A continuous spectrum is produced when white light is passed through a prism. It shows all colors in an unbroken sequence of frequencies, such as the spectrum of visible light.

LINE SPECTRUM A line spectrum is an emission spectrum that has sharp lines produced by specific frequencies of light. It is produced by excited atoms and ions as they fall back to a lower energy level. Different elements have different line spectra so they can be used to identify unknown elements.

Bohr model

s ORBITAL The smallest orbital is the “s” orbital. The “s” orbital:
Has only 1 shape (holds 2 e-) Is spherical in shape Is the lowest energy orbital

p ORBITALS The 2nd orbital shape is the “p” orbital shape.
There are 3 “p” shapes, each holding 2 electrons, for a total of 6 electrons in the “p” orbital. The “p” orbitals are: Dumbbell-shaped arranged at right angles with the nucleus at the center. Higher in energy than the “s”

d ORBITALS The 3rd orbital shape is the “d” orbital shape.
There are 5 “d” orbital shapes, for a total of 10 electrons in the “d” orbital. “d” orbitals are higher in energy than “p” orbitals.

f ORBITALS The last orbital shape is the “ f ” orbital shape.
“ f ” orbitals have irregular shapes due to quantum tunneling. There are 7 “ f ” shapes, for a total of 14 electrons. Electrons in f orbitals are very high in energy.

More orbital information
The main energy level or shell is given an integer number, n, and can hold a maximum of electrons, 2n2. Within an energy level, the “s” orbital has the lowest energy followed by “p”, then “d”, then “f” with the highest energy. Sub-levels contain a fixed number of orbitals, regions of space where there is a high probability of finding an electron.

ORBITALS IN ENERGY LEVELS
Principle Energy Level # Sub Levels (orbitals) # e- In each sublevel Total # Electrons in level n=1 1 1s 2 n=2 2s,2p 2,6 8 n=3 3 3s,3p,3d 2,6,10 18 n=4 4 4s,4p,4d,4f 2,6,10,14 32

ORBITAL ENERGIES The Aufbau diagram shows us the
energies of the various orbitals. The lowest energy is the 1s orbital. As you follow the arrows down, energy increases.

Quantum Mechanical Model
The scientists Heisenberg, de Broglie and Schrodinger developed the current model of the atom called the Quantum Mechanical Model. The electrons do not travel in precise orbits, but in wave functions called orbitals. HEISENBERG UNCERTAINTY PRINCIPLE: We are limited in just how precisely we can know both the position and momentum of a particle at a given time. The wave function or orbital has a 90% probability of finding the electron within it.

THREE MAIN RULES FOR e- CONFIG
Aufbau principle – electrons enter orbitals of lowest energy first Pauli exclusion principle – an orbital can only hold 2 electrons with opposite spins. Hund’s rule (Bus rule) – electrons enter orbitals singly until they have to pair up.

Write the electron configurations for the following
sodium________________________________________________ iron ________________________________________________ bromine ________________________________________________ barium ________________________________________________ neptunium ________________________________________________ Write the Noble Gas (abbreviated) electron configurations of the following elements: 6) cobalt________________________________________________ 7) silver________________________________________________ 8) tellurium________________________________________________ 9) radium________________________________________________ 10) lawrencium________________________________________________

Determine what elements are denoted by the following electron configurations:
1s22s22p63s23p4 ____________________ 1s22s22p63s23p64s23d104p65s1 ____________________ [Kr] 5s24d105p3 ____________________ [Xe] 6s24f145d6 ____________________ [Rn] 7s25f11 ____________________ Determine which of the following electron configurations are not valid: State which rule has been violated. 1s22s22p63s23p64s24d104p5 ____________________ 1s22s22p63s33d5 ____________________ [Ra] 7s25f8 ____________________ [Kr] 5s24d105p5 ____________________ [Xe] ____________________

Lastly

Higher level

IONISATION ENERGY WHAT IS IONISATION ENERGY?
“The energy required to remove 1 electron from each atom in 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions” Energy required to remove electrons Measured in kJmol-1 Given the abbreviation IE Values tell us a lot about the electronic configuration of elements

IONISATION ENERGY SUCCESSIVE IONISATION ENERGIES
We can measure the energies required to remove each electron in turn, from outer electrons to inner electrons Values increase with each successive ionisation event Why might this be? Successive Ionisation Energies of Na

To form a positive ion, need energy to overcome attraction from nucleus. As each electron is removed from an atom, the remaining ion becomes more positively charged. Removing the next electron away from an increasing positive charge is more difficult and the ionisation energy is even larger.

Na (g)  Na+ (g) + e st IE = kJmol-1 Na+ (g)  Na2+ (g) + e nd IE = kJmol-1 Na2+ (g)  Na3+ (g) + e rd IE = kJmol-1 Notice how successive ionisation energies are written Start with the end product of previous ionisation event MUST include the gaseous state!

From this graph we can see evidence for electron shells Na looks to have 1 electron that’s easy to remove  furthest from nucleus 8 nearer to the nucleus  bit harder to remove 2 very close to the nucleus  nearest to +ve charge and hardest to remove

IONISATION ENERGY FACTORS AFFECTING IONISATION ENERGY As electrons are negatively charged and protons in the nucleus are positively charged, there will be an attraction between them. The greater the pull of the nucleus, the harder it will be to pull an electron away from an atom. Nuclear attraction of an electron depends on: Atomic radius Nuclear Charge Electron shielding or screening

IONISATION ENERGY ATOMIC RADIUS:
Greater the atomic radius, the smaller the nuclear attraction experienced by the outer electrons NUCLEAR CHARGE: The greater the nuclear charge, the greater the attractive force on the outer electrons Lithium has a greater nuclear charge so you may expect it to have a higher ionisation energy. However it is affected to a greater extent by the atomic radius and the electron shielding Hydrogen Helium Lithium 519 kJ mol-1 1310 kJ mol-1 2370 kJ mol-1

IONISATION ENERGY ELECTRON SHIELDING
Inner shells of electrons repel the outer-shell electrons Known as ‘Electron shielding’ or ‘screening’ More inner shells  larger the screening and smaller the nuclear attraction of outer electrons Hydrogen Helium Lithium 519 kJ mol-1 1310 kJ mol-1 2370 kJ mol-1

IONISATION ENERGY TRENDS ACROSS PERIODS
Look at the graph 1) Describe the patterns you see 2) Can you suggest an explanation for them

IONISATION ENERGY TRENDS ACROSS PERIODS
There is a ‘general increase’ across a period before the value drops dramatically for the start of another period. This is because nuclear charge is increasing He Ne Ar Kr Xe

IONISATION ENERGY EXPLANATION
HYDROGEN EXPLANATION Despite having a nuclear charge of only 1+, Hydrogen has a relatively high 1st Ionisation Energy as its electron is closest to the nucleus and has no shielding. 1st IONISATION ENERGY / kJmol-1 1s ATOMIC NUMBER 1

HELIUM EXPLANATION Helium has a much higher value because of the extra proton in the nucleus. The additional charge provides a stronger attraction for the electrons making them harder to remove. 1st IONISATION ENERGY / kJmol-1 1s ATOMIC NUMBER 2

LITHIUM EXPLANATION There is a substantial drop in the value for Lithium. Despite the increased nuclear charge, there is electron shielding from the 1s orbital. The 2s electron is also further away from the nucleus. It is held less strongly and needs less energy for removal. 1st IONISATION ENERGY / kJmol-1 1s 1s 2s ATOMIC NUMBER 3

BERYLLIUM EXPLANATION The value for Beryllium is higher than for Lithium due to the increased nuclear charge. There is no extra shielding. 1st IONISATION ENERGY / kJmol-1 1s 1s 2s 1s 2s ATOMIC NUMBER 4

There is a DROP in the value for Boron. This is because the extra electron has gone into one of the 2p orbitals. The increased shielding makes the electron easier to remove It was evidence such as this that confirmed the existence of sub-shells. If there hadn’t been any sub-shell, the value would have been higher than that of Beryllium. 1s BORON 1st IONISATION ENERGY / kJmol-1 1s 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 5

CARBON EXPLANATION The value increases again for Carbon due to the increased nuclear charge. 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 6

NITROGEN EXPLANATION The value increases again for Nitrogen due to the increased nuclear charge. As before, the extra electron goes into the vacant 2p orbital. There are now three unpaired electrons. 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 7

There is a DROP in the value for Oxygen. The extra electron has paired up with one of the electrons already in one of the 2p orbitals. The repulsive force between the two paired-up electrons means that less energy is required to remove one of them. 1s OXYGEN 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s p 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 8

FLUORINE EXPLANATION The value increases again for Fluorine due to the increased nuclear charge. The 2p orbitals are almost full. 1s 2s p 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s p 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 9

NEON EXPLANATION The value increases again for Neon due to the increased nuclear charge. The 2p orbitals are now full so the next electron in will have to go into the higher energy 3s orbital. 1s 2s p 1s 2s p 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s p 1s 2s 1s 2s p 1s 2s ATOMIC NUMBER 10

There is a substantial drop in the value for Sodium. This is because the extra electron has gone into an orbital in the next energy level. This means there is an extra shielding effect of filled inner 1s, 2s and 2p energy levels. 1s SODIUM 1s 2s p 1s 2s p 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s p 1s 2s 1s 2s p 1s 2s 1s 2s p 3s ATOMIC NUMBER 11

The value for Magnesium is higher than for Sodium due to the increased nuclear charge. There is no extra shielding. The trend is similar to that at the start of the 2nd period. 1s MAGNESIUM 1s 2s p 1s 2s p 1s 2s p 1st IONISATION ENERGY / kJmol-1 1s 1s 2s p 1s 2s p 1s 2s 1s 2s p 3s 1s 2s p 1s 2s 1s 2s p 3s ATOMIC NUMBER 12

IONISATION ENERGY TREND IN PERIOD 3
Draw a graph for the first ionisation energy of elements in period 3 For each point draw the box and arrow electron configuration For each point explain why the first ionisation energy either increases or decreases Na Mg Al Si P S Cl Ar 496 738 578 789 1012 1000 1251 1521

IONISATION ENERGY TREND IN PERIOD 3 Drop from Mg  Al:
Mg: 1s2 2s2 2p6 3s2 Al: 1s2 2s2 2p6 3s2 3p1 Drop from P  S: P: 1s2 2s2 2p6 3s2 3p3 S: 1s2 2s2 2p6 3s2 3p4 The outer electron in Al has moved into the 3p orbital. It takes less energy to remove an electron from here than the 3s In P each of the 3p orbitals has 1 electron. In S one must have 2. The repulsion between these 2 paired electrons makes it easier to remove one

IONISATION ENERGY TREND DOWN GROUPS
General trend is a decrease as we go down a group The outer electron is in a level that gets further from the nucleus Nuclear charge increases however this is more than cancelled out by the electron shielding from lower energy shells

IONISATION ENERGY EXTENSION QUES: Which has the higher value, the 1st I.E. of sodium or the 2nd I.E. of magnesium? A: The 2nd I.E. of magnesium The 1st I.E. of sodium involves the following change Na(g) Na+(g) 1s2 2s2 2p6 3s s2 2s2 2p6 The 2nd I.E. of magnesium involves the same change in electron configuration… Mg+(g) Mg2+(g) 1s2 2s2 2p6 3s s2 2s2 2p6 However, magnesium has 12 protons in its nucleus, whereas sodium only has 11. The greater nuclear charge means that the electron being removed is held more strongly and more energy must be put in to remove it.

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