Atomic Structure Composition of an atom Atoms are made up of 3 fundamental subatomic particles: Relative mass Relative electric charge Position in atom Protons (p) Neutrons (n) Electrons (e - ) nucleus nucleus 01 around the / 1840 nucleus

Effect of electric fields on beams of p, n & e - In an electric field: -+-+ p n e-e- The e - is deflected more than p as it has a small mass.

Fleming’s Left-Hand Motor Rule ThuMb : Force, Motion Fore finger : Magnetic Field Centre finger : Current Current flow in the same direction as proton

Effect of magnetic fields on beams of p, n & e In a magnetic field: p n e-e- NB: Electrons are deflected to a greater extent than protons because electrons have a much smaller mass. The magnetic field line is moving perpendicular into the screen

Atomic number & Atomic mass E A Z where A: mass no. (total no. of p & n) Z: atomic no. (no. of p) N: no. of neutrons ( = A - Z ) NB: 1. Since the atomic no. (no. of p) = no. of electrons, therefore, there is no net charge on an atom 2. Loss of electrons give rise to cations (positively charged ions, no. of p > e - ) while the gain of electrons give rise to anions (negatively charged ions, no. of p < e - )

Try these : AlAl p n e - S p n e Na p n e

Isotopes Isotopes are atoms of the same element with same atomic no. but different mass no. (or different no. of neutrons) For examples: 1 H, 2 H, 3 H are isotopes of hydrogen 35 C l, 37 C l are isotopes of chlorine Try this: Which of the following 3 atoms are isotopes of the same element? A B C mass no no. of neutrons atomic no

Isotopes n Isotopes of an element have similar chemical properties This is because they have the same no. of electrons and the chemical properties depend upon the transfer and redistribution of electrons. Isotopes of an element have different physical properties. Isotopes of an element have different physical properties. This is because they have different no. of neutrons and hence different masses. eg. 37 C l has a higher density than 35 C l.

Electronic Energy levels or shells Electrons move around the nucleus of an atom in definite energy levels or shells which are identified by numbers called principal quantum numbers (p.q.n.), n. K 19 : electrons are arranged in 4 shells ( 2, 8, 8, 1) energy level / shell no. n = 4 n = 3 n = 2 n = 1 no. of electrons ____ Highest energy level (most easily removed) Lowest energy level farthest from nucleus nearest to nucleus Increasing energy

Energy sublevels or subshells Energy levels are split into sublevels / subshells. ( labelled s, p, d, f ) Energy level n = 1 n = 2 n = 3 n = 4 no. of subshells types of subshells s s, p s, p, d s, p, d, f labelled as 1s 2s, 2p 3s, 3p, 3d 4s, 4p, 4d, 4f

Electrons & Orbitals Electrons are in constant motion. Within each subshells, electrons can be found in certain regions known as orbitals. An orbital is a region or volume of space within which there is a high probability of finding an electron. Each orbital can accommodate two electrons. type of subshell no. of orbitals s p d f max. no. of electrons

Relative energies of energy levels & sublevels 4f 4d 4p 3d 4s 3p 3s 2p 2s 1s n = 4 n = 3 n = 2 n = 1 increasing energy NB: 3d higher energy than 4s

Try this: Shell no. Maximum no. of electrons n = 1 n = 2 n = 3 n = = = = 32 General Formula = 2n 2

Shapes of Orbitals Shape: spherical size of s orbitals: 4s > 3s > 2s > 1s  1s z y x  y zx 2s

 x z y pxpx Shape : dumb-bell different axes of symmetry mutually perpendicular to one another orbitals(p x, p y, p z ) with the same p.q.n. have the same energy i.e. they are degenerate.  x z y pypy z  x y pzpz

Electronic Configuration 1. Pauli Exclusion Principle: An atomic orbital can be occupied by only two electrons and their spins are always opposite.  Electrons are arranged in orbitals according to a set of rules.

Electronic Configuration 2. Aufbau (or building up) Principle: Electrons fill the orbitals of lowest energy first and then a higher energy. 6s... 5s5p... 4s4p4d4f 3s3p3d 2s2p 1s Increasing energy

Electronic Configuration 3. Hund’s Rule: When electrons are added successively to a set of orbital of the same energy level, they occupy them singly first before any pairing occurs.  2p

Ionisation Energies Information about the arrangement of electrons has been obtained by bombarding the atoms with streams of fast- moving electrons and so removing electrons from the atom. This method allows us to measure approximately the energy required to remove an electron, called the ionisation energy. First ionisation energy of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous atoms. M (g) M + (g) + e  H = + value kJ mol -1

Examples: Na (g) Na + (g) + e  H = kJ mol -1 Mg (g) Mg + (g) + e  H = kJ mol -1 Second ionisation energy of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous ions. M + (g) M 2+ (g) + e  H = + value kJ mol -1

Example: Na + (g) Na 2+ (g) + e  H = kJ mol -1 Mg + (g) Mg 2+ (g) + e  H = kJ mol -1 NB: 1st IE < 2nd IE < 3rd IE < <

Factors affecting the first IE of elements Atomic radius (distance of outermost electron from the nucleus) As atomic radius increases, outer electrons experience less nuclear attraction, hence 1st IE decreases. Size of nuclear charge (no. of p in nucleus) As size of nuclear charge increases, outer electrons experience greater nuclear attraction, hence 1st IE increases.

Screening or shielding effect by inner-shells electrons Outer electrons are shielded from nuclear attraction by repelling effect of inner-shells electrons. As no. of inner shells increases, the shielding effect by inner-shells electron increases, and hence 1st IE decreases. NB: Electrons in the same shell exert a negligible shielding effect on each other.

Try this: Compare the 1st IE of 5 B & 6 C. C has a greater nuclear charge than B (because C has a greater no. of proton)  atomic radius of C is smaller than that of B.  outer electron of C would experience greater nuclear attraction. Both B & C have  the same shielding effect by inner- shell electrons (because they have the same no. of inner principal quantum shell) As a result, 1st IE of C > 1st IE of B. 5 B : 1s 2 2s 2 2p 1 6 C : 1s 2 2s 2 2p 2

Trend of 1st IE across a period atomic number 1st IE / kJmol -1 Li Be B C N O F Ne Na Mg Period 2 Period 3 ( 2,3,3 pattern)

Across any period, there is a general increase in the 1st IE due to  the increase in the nuclear charge &  the decrease in atomic radius across the period. (the shielding effect remains almost the same from one element to the other because electrons are added successively to the same shell)

However, the 1st IE of B is lower than that of Be. 5 B : 1s 2 2s 2 2p 1 4 Be : 1s 2 2s 2 2p electron of B is of a higher energy than the 2s electron of Be. Hence, a lower amount of energy is required to remove the 2p electron from B than to remove the 2s electron from Be.

the 1st IE of O is lower than that of N. 8 O : 1s 2 2s 2 2p 4 7 N : 1s 2 2s 2 2p 3 The mutual repulsion of the paired electrons in a 2p orbital of O makes the removal of one electron from that orbital easier compared to a 2p electron of N which does not experience such repulsion.

Down a group, the 1st IE decreases. This is due to  the increase in atomic radius  the increase in shielding effect by inner-shell electrons. These two factors outweigh the effect of the increasing nuclear charge.

Try this: Explain why the 1. the 1st IE of Al < 1st IE of Mg 2. the 1st IE of S < 1st IE of P atomic number 1st IE / kJmol -1 Na Mg Al Si P S Cl Ar

Relevance of successive IEs to electronic configuration of atom o no. of electrons removed Log 10 IE A plot of successive IEs of Mg ( 1s 2 2s 2 2p 6 3s 2 )

no. of electrons removed Log 10 IE n = 1 Closest to the nucleus n = 3 Furthest away from the nucleus n = 2 A plot of successive IEs of Mg: 1s 2 2s 2 2p 6 3s 2 1s 2 2p 6 2s 2 3s 2 drastic “jump”  change of p.q. shell. Gradual increase  electrons in the same p.q. shell

Try this: o no. of electrons removed Log 10 IE

o no. of electrons removed Log 10 IE A plot of successive log 10 IEs of Element X against the no. of electrons removed. n = 1 n = 2 n = 3 Electrons in the outermost shell Drastic “jump”

n = 3 n = 2 n = 1 It has 3 principal quantum shells and its electronic configuration is 1s 2 2s 2 2p 6 3s 2 3p 2 Element X belongs to Group IV because it has 4 valence electrons. Valence electrons

Sketch a graph of the first seven successive log 10 IEs of A l against the no. of electrons removed. Log 10 IE no. of electrons removed Drastic “jump” 13 A l : 1s 2 2s 2 2p 6 3s 2 3p 1 3p 1 3s 2 2p 4

no. of electrons removed IE / kJ mol -1 A plot of successive IEs of A l against the no. of electrons removed. 13 A l : 1s 2 2s 2 2p 6 3s 2 3p 1 Drastic “jump” 1s 2 2s 2 2p 6 3p 1 Moderate “jump”  presence of subshells 3s 2

The first six successive IEs of an element X are as follows: In which group does the element belong to? kJmol   Element X belongs to group III. There is a drastic increase in the 4th IE  the 4th electron to be removed is in an inner and more stable principal quantum shell that is closer to the nucleus.  there are 3 valence electrons.

The successive IEs of an element Y are as follows: In which group does element Y belong?  Ans: Element Y belongs to Group II.

N92P1Q1 (a) Write down the electronic configuration of the selenium atom. (b) How would you expect the first ionisation energy of selenium to compare with that of (i) sulphur (ii) bromine Give your reasoning.

(a) 34 Se: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 4 Gp VI (b) 16 S: 1s 2 2s 2 2p 6 3s 2 3p 4 Gp VI The 4p electron of Se to be removed is ii) further away from the nucleus compared with the 3p electron to be removed from S. i) more shielded by inner-shells’ electrons and Therefore, the 1st IE of Se is expected to be lower than that of S.

(a) 34 Se: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 4 Gp VI (b) 35 Br: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 Gp VII Se has a ii) larger atomic radius than Br i) smaller nuclear charge and Therefore, the 1st IE of Se is expected to be lower than that of Br. same period