Atomic structure Greeks  Rutherford GCSE ATOM: Electrons in shells Ionisation Energy: Experimental evidence A level Atom and the existence of orbitals Orbitals: How they are filled Orbitals: Shapes, energies, distance from nucleus Ionisation energy: Fitting the theory to practice.

Greeks to Bohr 2000 years of history in a blinking of an eye Democritus (400BC) – Paper folding Dalton (1803) – States the obvious as he looks at carbon and sulphur Dobereiner (1810) - Triads Newlands (1866) – Musician who liked science Mendeleev (1869) – Played cards and came up with a table Thompson (1900) – Got the mass:charge ratio right but then pulled out a plum Rutherford (1911) – Can paper stop a bullet? Bohr (1913)– Looked at the universe and thought he solved the structure of the atom Chadwick (1932) – Found the invisible man Today – Quarks and other strange particles.

GCSE ATOM: Electrons in shells Nucleus Space 99.9% 1 st Shell 3 rd Shell 4 th Shell 2 nd Shell

Ionisation Energy The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms / ions 1 st IE: The energy required to remove 1 mole of electrons from 1 mole of gaseous atoms X (g)  X + (g) + e - 2 nd IE: The energy required to remove 1 mole of electrons from 1 mole of gaseous singly charged positive ions X + (g)  X +2 (g) + e -

Ionisation energy No of Protons INCREASES IE, as it is harder to remove an electron from an atom with a highly positive Nucleus No of shells DECREASES IE, as it is easier to remove an electron from an atom the further it is from the nucleus Shielding (the reduction of the effective nuclear charge by the inner shells, so that the outer shell electrons experience less of an attraction to the nucleus) DECREASES IE, as it is easier to remove an electron from an atom in which the effective nuclear charge has been reduced. IMPORTANT: SHIELDING IS THE MOST IMPORTANT FACTOR

Ionisation energy- expected variation DOWN GROUP DECREASES- A new shell is added, hence the level of shielding increases. This reduces the effective nuclear charge making it easier to remove an electron. THIS IS WHAT WE FIND IN OUR PLOT. ACROSS A PERIOD INCREASES- electrons are added to the same shell, hence the level of shielding stays the same Protons are added to the nucleus so the attraction between the outer shell and the nucleus increases. WE BASICALLY FIND THIS BUT THERE A FEW ANNOMALIES.

REJECTION OF GCSE ATOM Note in IE graph Na is less than Ar (This is repeated Li/Ne, H/He) Expected from GCSE ideas? Na is less than Ne OR Na is less than Li (This is repeated Li/He, K/Na, Cl/F) Expected from GCSE ideas? 1 st IE Be being greater than 1 st IE B (This is repeated with Mg / Al and Ca / Ga) 1 st IE N being greater than 1 st IE O (This is repeated with P / S and As / Se) The lack of change of 1 st IE between Sc  Cu

A level Atom and the existence of orbitals Within a shell there must be another environment for electrons to exist in. Each shell made up of orbitals. Each shell or orbital has a SPECIFIC ENERGY VALUE ASSOCIATED WITH IT Hold 2 electrons maximum 1 st shell – 1 types 2 nd shell – 2 typessp 3 rd shell – 3 typess pd 4 th shell – 4 typesspd (f)

Shapes and energies Shape Energies The nearer the nucleus an orbital is the lower its energy. Orbitals of lowest energy (nearest the nucleus filled first)

Filling the orbitals (Auf Bau Building principle) Orbitals nearest to the nucleus first. Hence H  Be H = 1s 1 He = 1s 2 (FULL) Li = 1s 2 2s 1 Be = 1s 2 2s 2 (FULL)

Filling the orbitals Be  B Be = 1s 2 2s 2 (FULL) B = 1s 2 2s 2 2p 1 USE TO EXPLAIN 1 st IE GRAPH ANNOMALY Be/B

Filling the p orbitals Pauli exclusion principle P orbitals Must be three of them All identical except they are perpendicular to each other Distance Huge distance between 1 st and 2 nd shell Small distance between 2s and 2p orbitals 2p 2s 1s

Filling the orbitals B  N B = 1s 2 2s 2 2p 1 C = 1s 2 2s 2 2p 1 2p 1 (written 1s 2 2s 2 2p 2 ) N= 1s 2 2s 2 2p 1 2p 1 2p 1 (written 1s 2 2s 2 2p 3 ) Each electron placed in different p orbital to minimise electron electron repulsion

Filling the orbitals N  O N= 1s 2 2s 2 2p 1 2p 1 2p 1 (written 1s 2 2s 2 2p 3 ) O = 1s 2 2s 2 2p 2 2p 1 2p 1 (written 1s 2 2s 2 2p 4 ) New electron in O placed in a p orbital which already has an electron in it. USE TO EXPLAIN 1 st IE GRAPH ANNOMALY N/O

Filling the orbitals O  Na O= 1s 2 2s 2 2p 2 2p 1 2p 1 (written 1s 2 2s 2 2p 4 ) F= 1s 2 2s 2 2p 5 Ne= 1s 2 2s 2 2p 6 At Neon the 2 nd shell is full, so new electon placed in the s orbital in 3 rd shell Na= 1s 2 2s 2 2p 6 3s 1 or [Ne]3s 1

Filling the orbitals Pattern repeated from Na  Ar K same chemistry as Na, Li Electron must be placed in same orbital type 4s is nearest to nucleus than 3d K = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Ca = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2

3d nearest to nucleus once an electron is placed in it. USE TO EXPLAIN 1 st IE GRAPH ANNOMALY Sc  Cu Sc = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 Ti = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 2 ( 3d 1 3d 1 3d 0 3d 0 3d 0 ) V = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 ( 3d 1 3d 1 3d 1 3d 0 3d 0 ) Cr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 ( 3d 1 3d 1 3d 1 3d 1 3d 1 ) (All half-filled) Mn = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2 ( 3d 1 3d 1 3d 1 3d 1 3d 1 ) Fe = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 ( 3d 2 3d 1 3d 1 3d 1 3d 1 ) Co = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 7 4s 2 ( 3d 2 3d 2 3d 1 3d 1 3d 1 ) Ni = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2 ( 3d 2 3d 2 3d 2 3d 1 3d 1 ) Cu= 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1 ( 3d 2 3d 2 3d 2 3d 2 3d 2 ) (3 rd SHELL full) Zn = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 ( 3d 2 3d 2 3d 2 3d 2 3d 2 )

Filling the orbitals Zn  Kr (pattern as before Mg  Ar) Ga = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 1 Ge = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 2 As = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 3 Se = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 4 Br = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 Kr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6

Electrons in orbitals They spin Same spin if in different orbitals of the same energy (called degenerate orbitals) Eg 1 electron in each of the 3*p orbitals Opposite spins if in same orbital Reduces repulsion

Filling orbitals part 2 Represent electrons as arrows showing spin by direction Represent orbitals as a box Place the electrons in the boxes Set up boxes to represent each orbital Fill them 1s 2s 2p 1s 2s 2p

Distance from nucleus and energy 1s 2s 2p 3s 3p 3d 4s 4p Distance from nucleus or Energy Order up to Ca 1s 2s 2p 3s 3p 4s 4p 3d Order from Sc  Kr

Removal of electrons Electrons are removed from the orbital of highest energy (furthest from nucleus) The number of electrons removed corresponds to the charge on the ion In bonding enough electrons are removed to leave the atom with a FULL OUTER SHELL

Addition of electrons Electrons are added to the orbital of highest energy In bonding enough electrons are added to leave the atom with a FULL OUTER SHELL The number of electrons added corresponds to the charge on the ion

Electronic configuration of ions Na +1 1s 2 2s 2 2p 6 F -1 1s 2 2s 2 2p 6 O -2 1s 2 2s 2 2p 6 Ca +2 1s 2 2s 2 2p 6 3s 2 3p 6 Br -1 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6

Fe +3 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Mn +2 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Ga +3 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 As +2 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 1 S -1 1s 2 2s 2 2p 6 3s 2 3p 5 Fe +2 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 or 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 Ge -2 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 4 Electronic configuration of ions

Removal of electrons Removing electrons from Li atoms 1 st electron: Shielded by 1 shell 2 nd electron:No shielding and ion is +vely charged 3 rd electron:No shielding and ion is more +vely charged

Removing electrons from Al atoms 1 st electron: Shielded by 2 shell 2 nd electron:Also shielded by 2 shells and ion is +vely charged 3 rd electron: Also shielded by 2 shells and ion is more +vely charged 4 th electron:Only shielded by 1 shell 5 th electron: Also shielded by 1 shell and ion is more +vely charged 6 th electron:Also shielded by 1 shell and ion is more +vely charged 7 th electron: Also shielded by 1 shell and ion is more +vely charged

Successive Ionisation energy curves Successive Ionisation energies always increasing Ion gets more positively charged (e no longer = p) Electron removed from a shell nearer the nucleus Shielding is dramatically reduced. Ionisation energy increases dramatically The group of the element can be determined by looking at the graph of successive IE. Group = IE before huge increase The number of electrons in each shell can also be determined (and hence the old fashioned electronic configuration) The graph must be read backwards.

Examples