3.1.1 Periodicity Flashcards

(41 cards)

1
Q

Classification of elements into orbital blocks

A

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

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2
Q

what is periodicity

A

repeating pattern across different periods

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3
Q

atomic radius across a period

A

decrease
increased protons means increased nuclear charge
increased nuclear attraction for electrons in the same shell
shielding stays the same

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4
Q

definition of first ionisation energy

A

energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions

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5
Q

equation to represent 1st ionisation energy of hydrogen

A

H(g) -> H+(g) + e-

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6
Q

factors that affect ionisation energy

A
  1. nuclear attraction
  2. atomic radius
  3. electron shielding
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7
Q

when does nuclear attraction increase

A

more protons in nucleus the greater the attraction

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8
Q

when does atomic radius increase

A

larger the atom the further away the outer electrons are from the nucleus and the weaker attraction to the nucleus

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9
Q

shielding of electrons

A

electron in outer shell is repelled by electrons in complete inner shell, weakening the attraction to the nucleus

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10
Q

general pattern of successive ionisation energies

A

increases as more electrons are removed

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11
Q

why does each successive ionisation energy get larger

A

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

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12
Q

how are ionisation energies linked to electronic structure?

A

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

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13
Q

how can you tell which group an element is in when given successive ionisation energies

A

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

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14
Q

general trend of first ionisation energies of the elements

A

generally increases across the group

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15
Q

actual trend of first ionisation energies of the elements

A

increases

exception at Mg to Al and P to S

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16
Q

explain exceptions of first ionisation energies of the elements

A

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

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17
Q

why does helium have the highest first ionisation energy

A

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

18
Q

trend in ionisation energies down a group

19
Q

why do first ionisation energies decrease down a group

A

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

20
Q

why do first ionisation energies increase across a period

A
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
21
Q

why does sodium have a much lower first ionisation energy than neon

A

sodium has its outer shell electron in a 3s subshell further from the nucleus and is more shielded, outer electron is easier to remove

22
Q

definition of metallic bonding

A

electrostatic attraction between positively charged ions and the sea of delocalised electrons

23
Q

3 main factors that determine metallic bond strength

A
  1. Number of protons (more protons, greater charge, stronger the bond)
  2. Charge of metal ion/delocalised electrons (more delocalised electrons the stronger the bond)
  3. Size of the ion (smaller the ion the stronger the bond)
24
Q

which metallic bond is stronger Mg or Na?

A

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

25
2 examples of macromolecular compounds
diamond | graphite
26
structure of diamond
macromolecular tetrahedral arrangement of carbon atoms 4 covalent bonds per carbon atom
27
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
28
melting and boiling points of diamond and graphite
very high due to strong covalent bonds require high amounts of energy to break them
29
boiling and melting points of macromolecular
high- because of many strong covalent bonds lot of energy to break the many strong bonds
30
boiling and melting points of metallic
high- strong electrostatic forces between positive | ions and sea of delocalised electrons
31
macromolecular solubility in water
insoluble
32
metallic Solubility in water
insoluble
33
conductivity when solid macromolecular
diamond and sand: poor, because electrons can’t move (localised) graphite: good as free delocalised electrons between layers
34
conductivity when solid metallic
good: delocalised electrons can move through | structure
35
conductivity when molten macromolecular
poor
36
conductivity when molten metallic
good
37
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
38
general trend in melting and boiling points across a period
increase until Si then decrease
39
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 ```
40
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
41
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