Topic 3: Periodicity
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Effective nuclear charge (Z*/Z eff) is the … The equation is Z eff =
Nuclear charge felt by an electron in the atom Feel charge of nucleus but often shielded by other electrons Z (atomic number) - S (shielding constant)
Trend in Z eff across periodic table
Increase Electrons added to same shell at roughly same distance from nucleus Don’t shield each other perfectly
Order of orbitals in ability to shield (high to low)
s, p, d, f
Trend of Z eff down periodic table
Increase Increased number of protons Atomic orbitals less effective at shielding
atomic radius =
half the distance between the centres of neighbouring atoms in a solid or, for non-metals, in a homonuclear molecule (e.g. O2, P4, S8)
Reason for jump in atomic radius between halogen and noble gas
Large increase Noble gas only have Van Der Waals Inert and monatomic
Trend in atomic radii across the periodic table
Decrease as Z eff increases Outer electrons feeling more effect of nuclear charge due to Z eff, pulling in electrons tightly and decreasing atomic radii
Trend in atomic radii down a group
Increase This shows that the valence shell (n) which the outer electrons occupy is more important than Z eff here
Ionic radius
derived from the distance between the centres of adjacent ions in a crystal, as determined using X-ray diffraction studies.
Trend in ionic radius in period 2 and 3
2 halves (as you move from positively to negatively charged elements Decrease for increasing atomic number in each half (Li, Be, B) (C, N, O, F) Large increase from B to C e.g. For Li+ (Na+), Be2+ (Mg2+) & B3+ (Al3+) The nuclear charge increases; at the same time the positive charge carried by the element increases [from +1 to +3] Remaining electrons feel increasing Z eff + drawn closer to nucleus Ions decrease in size For C4- (Si4-), N3- (P3-), O2- (S2-), & F- (Cl-) The nuclear charge increases; at the same time the negative charge carried by the element decreases [from -4 to -1] The effect is that the electrons feel an increasing Z eff, and are drawn closer to the nucleus Ions decrease in size Same for period 3
Ionisation energy
the energy required to remove one mole of electrons from one mole of gaseous species in their ground (or standard) state
Trend in ionisation energy across a period
General increase Decrease Be (2s2) to B (2s2 2p1) and P (2s2 2p3) to S (2s2 2p4) (period 3 follows same pattern) B readily looses p electron, have to break into Be full 2s sub shell which requires more energy S has a paired electron in 2p sub shell, paired electrons repel each other so are one more easily, P has unpaired sp electrons
Trend in ionisation energy down periodic table Group … is the exception? … to … has the same ionisation energy as … … to … has an … ionisation energy as …
Decrease due to increasing quantum number n 13 Al to Ga: increase due to increased Z eff due to less effective shielding from 3d orbitals In to Tl: increase due to increased Z eff due to less effective shielding by 4f orbitals
Electronegativity Pauling’s values range from …
The power [ability] of anatom in a molecule to attract electrons to itself. 0 → 4
Trends in electronegativity
χ increases across a group due to increasing Z off χ decreases down a group due to increasing principle quantum number, n
Electron affinity Ea is … for exothermic Ea is … for endothermic
the energy released when 1 mole of electrons is added to 1 mole of gaseous atoms in their ground (or standard) state Positive Negative (Opposite to usual)
Electron gain enthalpy (ΔegHo)
the change in standard molar enthalpy when a gaseous atom gains an electron
Equation for relationship between electron affinity and electron gain enthalpy
ΔegHo = –Ea –(5/2)RT
Trends in electron affinity across a period
Increases due to increasing Z eff and decreasing atomic radii
Electron affinity for Be/Mg (from Li/Na)
Decrease Be 2s2 Mg 3s2 s sub shell full so added electron needs to occupy empty np shell
Electron affinity for noble gases (from halogens)
Decrease have a full shell so the added electron needs to enter the (n+1) shell
Electron affinity for N/P (from C/Si)
Decrease N: 2s2 2p3 configuration P: 3s2 3p3 configuration Adding electron must pair unpaired p electron which causes repulsion of electrons and reduces stability
Why does SiH4 react spontaneously in air to form SiO2 and water but CH4 is stable given thermodynamic data states they both should
Kinetics O from O2 would need to attack central C but C is well shielded Si is larger so has more room to attack so is sterically hindered Si has ability too expand its octet as its empty d-orbitals (3d) closer in energy to p-orbitals (3p) than for C as its 3d orbitals a lot higher in energy than 2p
Account for the trends in values of the atomic radii of the group 14 elements
General trend of increasing atomic radii as a group is descended is due to increasing values of n * C→ Si follows trend in increasing due the higher n. * Si→ Ge shows only a very slight increase as the Ge valence containing poorly shielding d-orbital. This increases Zeff to almost overcome the increasing n. * Ge→ Sn again shows return to trend with increasing n . * Sn → Pb shows strong reversal by increasing radii. This is due to Pb valence now containing poorly shielding f-orbitals and this is enough to allow Zeff to overcome increasing n