C1: periodic table 1 Flashcards
down the same group (vertical), elements have?
- same no. of valence electrons
- increasing no. of electron shells
across the same period (horizontal), elements have?
- same no. of occupied electron shells
- increasing proton (atomic) no.
- increasing no. of valence electrons
atomic radius increases down the group
- although no. of protons in nucleus increases, e are further & shielded by more e shells
- valence e are attracted less strongly to nucleus = not held so tightly
atomic radius decreases across the period
- nuclear charge becomes more +ve as no. of protons inc
- although no. of e also inc, outermost e are in same e shell, hence experience relatively similar shielding effect
- outermost e are attracted more strongly to nucleus
ionic radius __ down the group, __ across the period
ions with same no. of e called __ ie. Li+, Be2+, B3+
inc, dec
isoelectronic
cation is __ than their neutral atom, while anion is __
smaller, bigger
why atomic radius of S < Al?
(S and Al are found in the same period)
S has more protons and electrons than Al
-> outermost electrons are in the same shell with relative similar shielding effect
-> higher number of protons attracts the outermost electrons more strongly to the nucleus, reducing the atomic radius.
why ionic radius of S2- > Al3+?
(S and Al are found in the same period) S2- has 2 electrons added in the outermost shell to form anion -> it has 3 electron shells -> Al3+ has 3 electrons removed from the outermost shell to form cation, leaving it with 2 electron shells -> S2- has one more electron shell than Al3+
across the period, 1st ionization energy __
inc
harder to remove e
…
atomic radius smaller -> outermost e attracted more strongly -> higher energy required
down the group, 1st ionization energy __
dec
easier to remove e
…
outermost e further from nucleus -> inc in shielding effect -> weaker attraction to nucleus -> less energy required to remove
electronegativity __ down the group, __ across the period
dec, inc
S or Cl has a larger 1st ionisation energy
(S and Cl are found in the same period)
Cl
higher nuclear charge than S which attracts the outermost electrons more strongly to the nucleus = harder to remove
Mg or Ca has a lower electronegativity value
(Mg and Ca are found in the same group) Ca
bonding electrons further away from the nucleus = dec attraction for bonding e
grp2 metals = reducing agents (form cations)
Ba highest tendency to be oxidised = strongest reducing agent
Be = weakest reducing agent
grp2 elements:
react with O2 to form __
react with halogens to form __
metal oxides
2Mg(s) + O2 (g) -> 2MgO (s)
metal halides
Mg (s) + Cl2 (g) -> MgCl2 (s)
grp 2 elements:
burns in steam to produce __
react with warm water to produce __
Mg (s) + H2O (g) -> MgO (s) + H2 (g)
Mg (s) + 2H2O (l) -> Mg(OH)2 (s) + H2 (g)
grp2 rxn with cold water to form __
Other Group 2 elements (except Be and Mg) react with cold water to form metal hydroxides and hydrogen gas
Ca (s) + 2H2O (l) -> Ca(OH)2 (aq) + H2 (g)
grp2 hydroxides become more __ down the grp
soluble
Mg(OH)2 is classified as insoluble in water and Ca(OH)2 is considered reasonably soluble in water
grp2 decomposition:
carbonates are __ thermally stable down the grp
more
MgCO3 (s) -> MgO (s) + CO2 (g)
grp2 decomposition:
metal ions are __ down the grp but has the same charge
bigger
charge density is reduced down the grp
- metal cation is able to polarize carbonate anion
identify grp2 elements rxn with H2O
X: colourless solution formed
Y: no rxn
Z: solution with white precipitate
Ca = reacts with water to form Ca(OH)2 which dissolves in water
Be = does not react with water
Mg = reacts with warm water to form Mg(OH)2 which is insoluble in water and appears as a white precipitate
Ca(NO3)2 has a higher or lower decomposition temperature than Mg(NO3)2
higher
Ca has a lower charge density and weaker polarising power = nitrate ion is less polarised = more thermally stable
down the grp:
grp2 elements __ reactive
grp17 elements __ reactive
more, less
down grp17, MPBP __
inc
(non-polar molecules)
molecules become larger = more e move around to set up temporary dipoles = stronger van der Waals forces = intermolecular forces get stronger = more energy required to break forces
