Topic 3: Periodicity Flashcards
(25 cards)
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
Which group is the exception?
Decrease due to increasing quantum number n
Group 13,
Al to Ga: expected decrease not observed (same ionisation energy)
as increased Z eff less effectively shielded by 3d orbitals
In to Tl: expected decrease not observed (increase) as increased Z eff less effectively shielded by 4f orbitals
Electronegativity
he power [ability] of an
atom in a molecule to attract
electrons to itself.
Pauling’s values range from 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
the energy released when 1 mole of electrons is added to 1 mole of gaseous atoms in their ground (or standard) state
– Ea is +ve for exothermic
– Ea is –ve for endothermic
(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
Increases across a period due to increasing Z eff and decreasing atomic radii
First row generally less favourable
compared with the second row due to the small size of the first row elements so introduction of large
inter-electron repulsions
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
