3.1 periodicity Flashcards
periodic table order
proton number
ionisation
ionisation energy is the minimum amount of energy required to removed 1 mole of electrons from 1 mole of gaseous atoms
is ionisation endo or exo?
ionisation requires energy so they are always an endothermic reaction and have a positive value
shielding
more electron shells between nucleus and outer electron means less energy is required to remove electron as there is weaker attraction
nuclear charge
more protons in nucleus means there is stronger nuclear attraction between nucleus and outer electrons meaning more energy required to remove electron
atomic radius
the larger the atom, the further away the outer electrons are from the nucleus so there is a weaker nuclear attraction and less energy is required to remove electrons
trend in ionisation energy down group
decreases
atomic radius increases down group
shielding increases down group
nuclear attraction becomes weaker therefore less energy is required to remove an electron
how do trends in ionisation energy support the Bohr model of the atom?
successive ionisation energies reflect the arrangement of electrons in shells, but doesn’t explain sub-shells
trend in ionisation energy across period
increases
proton number increases across period therefore nuclear attraction increases
shielding is similar
more energy is required to remove an outer electron
aluminium ionisation trend
outer most electron in Al sits in a higher energy sub-shell further from the nucleus than the outer electron in Mg
Mg - 1s2, 2s2, 2p6, 3s2 (3s sub-shell)
Al - 1s2, 2s2, 2p6, 3s2, 3p1 (3p sub-shell)
sulfur ionisation trend
P and S both have outer electrons in the 3p orbital so shielding is the same
Removing an electron from S involves taking it from an orbital with 2 electrons in
Electrons repel each other therefore less energy is needed to remove energy from orbital with 2 electrons than 1 electron like in P
successive ionisation
the removal of more than 1 electron from the same atom
trend in successive ionisation energy
general increase in energy as removing an electron from an increasingly more positive ion and removing electrons from shell closer to nucleus where there is stronger nuclear attraction
graphite
each C atom bonded 3 times with one delocalised electron
very high melting point due to many strong covalent bonds
layers slide as there are weak forces between layers
layers far apart in comparison to covalent bond length, low density
delocalised e- between layers allow graphite to conduct electricity as they carry charge
graphite is insoluble, covalent bonds too strong to break
diamond
each C atom bonded 4 time (tetrahedral)
tightly packed and rigid arrangement allows heat to conduct well
very high melting point due to many strong covalent bonds, also very hard
diamond doesn’t conduct electricity since it has no delocalised electrons
graphene
1 layer of graphite
1 atom thick (lightweight and transparent)
hexagonal carbon rings
delocalised electrons allows for graphene to conduct electricity as they carry charge
delocalised electrons strengthen covalent bonds, high strength property
uses of graphene
- aircraft shells
- super/highspeed computers
- smartphone screens
metallic bonding (metals)
positive metal ions are formed as metals donate electrons
electrostatic attraction between positive metal ions and negative delocalised e-
the more electrons an atom can donate, the higher melting point
good electrical conductors as delocalised electrons are mobile and can carry charge
good thermal conductors as delocalised e- can transfer kinetic energy
high melting points due to strong electrostatic attraction
solid metals are insoluble as metallic bond is too strong to break
melting point trend across period 3
general increase as metal ions have increasing positive charge
increasing number of delocalised electrons
smaller atomic radius
this means stronger metallic bond
Si melting point
highest melting point in period 3as it has a giant covalent (macromolecular) structure
many strong covalent bonds hold the silicon atoms together
large amount of energy needed to overcome these strong covalent bonds
P melting point
P4 formula
lower melting point than Si due to weaker simple molecular structure
melting point determined by weaker induced dipole-dipole forces
S melting point
S8 formula
higher melting point than P due to larger simple molecular structure
larger induced dipole-dipole forces
Cl melting point
Cl2 formula
lower melting point than P and S due to smaller simple molecular structure
smaller induced dipole-dipole forces
Ar melting point
Ar formula
lower melting point than all other elements in period 3 because it only exists as an individual atom
smaller induced dipole-dipole forces
are simple molecular/simple covalent structures soluble in water?
depends on the polarity of the molecule
polar molecules dissolve well in polar solvents like water
non-polar molecules do not
why do noble gases have low melting points?
more electrons leads to a smaller atomic radius meaning it has a higher electron charge density so the repulsion forces between the particles lower the energy needed to overcome the intermolecular forces
