3.1.1 Periodicity Flashcards
Classification of elements into orbital blocks
the last electron in the final orbital decides where in the periodic table the element falls, which orbital it is in decides its placement in the table
what is periodicity
repeating pattern across different periods
atomic radius across a period
decrease
increased protons means increased nuclear charge
increased nuclear attraction for electrons in the same shell
shielding stays the same
definition of first ionisation energy
energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions
equation to represent 1st ionisation energy of hydrogen
H(g) -> H+(g) + e-
factors that affect ionisation energy
- nuclear attraction
- atomic radius
- electron shielding
when does nuclear attraction increase
more protons in nucleus the greater the attraction
when does atomic radius increase
larger the atom the further away the outer electrons are from the nucleus and the weaker attraction to the nucleus
shielding of electrons
electron in outer shell is repelled by electrons in complete inner shell, weakening the attraction to the nucleus
general pattern of successive ionisation energies
increases as more electrons are removed
why does each successive ionisation energy get larger
the ion formed is smaller than the atom and the proton to electron ratio in the 2+ ion is greater than the 1+ ion
therefore the nuclear attraction is stronger
how are ionisation energies linked to electronic structure?
if there is a big jump between 2 values, can show evidence for sub shells.
big jump as the shell is then closer to the nucleus and attracted much more strongly
how can you tell which group an element is in when given successive ionisation energies
look for a big jump between two consecutive values
e.g. between 2 and 3
meaning the element must be in group 2 because the 3rd electron to be removed must be in a shell closer to the nucleus with less shielding so will have a larger ionisation energy
general trend of first ionisation energies of the elements
generally increases across the group
actual trend of first ionisation energies of the elements
increases
exception at Mg to Al and P to S
explain exceptions of first ionisation energies of the elements
Mg to Al: Al is filling 3p subshell but Mg has electrons in 3s which are more difficult to remove. Electrons in 3p are higher in energy and shielded by the 3s electrons, only one electron in the orbital so isn’t paired
P to S: sulfur has a 4th electron in the p orbital, the electron sare starting to pair with opposite spin so there is increased repulsion between the electrons and so is easier to remove
why does helium have the highest first ionisation energy
its first electron is in the first shell closest to the nucleus so has no shielding electrons, higher than hydrogen as there is one more proton
trend in ionisation energies down a group
decreases
why do first ionisation energies decrease down a group
as you go down the group the electrons are found in shells further away from the nucleus, increased shielding but the nuclear attraction decreases making it easier to remove 1 mole of electrons
why do first ionisation energies increase across a period
electrons added to the same shell shielding stays the same number of protons increases nuclear charge increases distance from nucleus stays the same nuclear attraction increases more difficult to remove electrons
why does sodium have a much lower first ionisation energy than neon
sodium has its outer shell electron in a 3s subshell further from the nucleus and is more shielded, outer electron is easier to remove
definition of metallic bonding
electrostatic attraction between positively charged ions and the sea of delocalised electrons
3 main factors that determine metallic bond strength
- Number of protons (more protons, greater charge, stronger the bond)
- Charge of metal ion/delocalised electrons (more delocalised electrons the stronger the bond)
- Size of the ion (smaller the ion the stronger the bond)
which metallic bond is stronger Mg or Na?
Mg
more electrons in outer shell that are delocalised
charge on the ion is greater
Mg ion is smaller and has one more proton
stronger force of electrostatic attraction between the positive ion and the electrons so increased energy is required to break the bond
Mg will have a higher melting point
2 examples of macromolecular compounds
diamond
graphite
structure of diamond
macromolecular
tetrahedral arrangement of carbon atoms
4 covalent bonds per carbon atom
structure of graphite
macromolecular
planar arrangement of carbon atoms
each carbon is bonded to 3 other carbons
1 electron per carbon is donated into the sea of delocalised electrons
which are located between the layers
melting and boiling points of diamond and graphite
very high
due to strong covalent bonds
require high amounts of energy to break them
boiling and melting points of macromolecular
high- because of many strong
covalent bonds
lot of energy to break the many
strong bonds
boiling and melting points of metallic
high- strong electrostatic forces between positive
ions and sea of delocalised electrons
macromolecular solubility in water
insoluble
metallic Solubility in water
insoluble
conductivity when solid macromolecular
diamond and sand: poor, because
electrons can’t move (localised)
graphite: good as free delocalised
electrons between layers
conductivity when solid metallic
good: delocalised electrons can move through
structure
conductivity when molten macromolecular
poor
conductivity when molten metallic
good
general description of giant metallic
shiny metal
Malleable as the positive ions in the lattice are all
identical. So the planes of ions can slide easily over
one another
-attractive forces in the lattice are the same
whichever ions are adjacent
general trend in melting and boiling points across a period
increase until Si then decrease
describe the trend in melting and boiling points across a period
increases Na to Al peaks at Si decreases to P increases at S decreases Cl to Ar
explain the trend in melting and boiling points across a period
Na-Al: high due to metallic bonding, high bond strength, increased energy required to break the bonds
Si: macromolecular, strong covalent bonds, more energy to break
P-Ar: simple molecular so require less energy to break the bonds
melting and boiling points simple molecular
S8
P4
Cl2
Ar
decreases due to decrease in surface area, less london forces, less energy required to break the bonds
