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Chemistry > Periodicity > Flashcards

Periodicity Flashcards

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1
Q

1st ionisation energy

A

the energy required to remove 1 electron from each atom in 1 mole of gaseous atoms

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2
Q

what are successive ionisation energies

A

the energy required to remove each electron in turn from each atom in 1 mole of gaseous atoms

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3
Q

what is the magnitude of ionisation energy influenced by

A

nuclear charge, atomic radius, shielding

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4
Q

nuclear charge

A

the greater the nuclear charge, the greater the attraction on the outer electrons so more energy is required to remove electron - larger ionisation energy

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5
Q

atomic radius

A

the further the electron from the nucleus the weaker the attraction (less energy required to remove electron - smaller ionisation energy)

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6
Q

shielding

A

full inner shells of electrons exert a repelling effect on the outer electrons as they are both negatively charged. this reduces the attraction between the nucleus and outer electron (less energy required to remove electron - smaller ionisation energy)

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7
Q

successive ionisation energies

A

steady increase in amount of energy needed to remove successive electrons

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8
Q

when an electron is removed

A

the remaining electrons are pulled closer to nucleus - the nucleus attracts remaining electrons more strongly so more energy is needed to remove them (higher i.e)

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9
Q

inner electron removed

A

there is a big jump in i.e when an electron is removed from an inner shell. the inner electron is much closer to the nucleus and is less shielded than the outer electron, so experiences a greater attraction and has a higher i.e

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10
Q

1st and 2nd i.e and 9th and 10th i.e

A

this shows clearly where an electron is being removed from an inner shell

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11
Q

1st i.e trend down a group

A
  • nuclear charge increases
  • atomic radius increases so electron is further from nucleus
  • shielding increases as number of full inner shells increases
  • increase in distance and shielding outweigh increased nuclear charge
  • nuclear attraction on outer electron decreases
    so first ionisation energy DECREASES
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12
Q

1st i.e general trend across period

A
  • nuclear charge increases
  • atomic radius decreases so outer electron is closer to the nucleus
  • shielding stays same as electrons are in same shell
  • nuclear attraction on outer electron increases
  • first ionisation energy INCREASES
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13
Q

decrease between groups 2 and 3 (Be and B)

A
  • Be: 1s2 2s2 B: 1s2 2s2 2p1
  • 2p orbitals have slightly more energy than 2s
  • s electrons provide slightly greater shielding of the p electron
  • less energy needed to remove outer electron from B than Be despite increase in nuclear charge
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14
Q

decrease between groups 5 and 6 (N and O)

A
  • N: 1s2 2s2 2p3 O: 1s2 2s2 2p4
  • spin pairing occurs in 2p subshell of oxygen
  • paired electrons in a 2p orbital of oxygen repel eachother
  • less energy needed to remove outer electron from O than N despite increased nuclear charge
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15
Q

what creates nuclear charge

A

more protons in a nucleus leading to a greater attraction between nucleus and electrons

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16
Q

what is an orbital

A

the space where an atom can hold two electrons with opposite spins

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17
Q

what did dmitri mendeleev do

A

in 1869 he noticed that if the elements are arranged in ascending order of atomic mass with new rows started at the correct points then elements with similar properties are grouped in vertical columns

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18
Q

group number

A

number of outer shell electrons

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19
Q

period number

A

number of shells of electrons

20
Q

elements in the same group…

A

have similar chemical properties because they have the same number of electrons in their outer shells

21
Q

number of period

A

gives number of highest energy electron shell

22
Q

what is periodicity

A

the repeating trends in physical and chemical properties across the periods of the periodic table

23
Q

periodicity atomic radii decrease across a period…

A
  • nuclear charge increases
  • same number of shells of electrons so no increased shielding
  • nuclear attraction increases, outer electrons attracted more strongly to nucleus, so pulled in closer and radius is made smaller
24
Q

periodicity atomic radii increases when descending a group…

A
  • more shells of electrons
  • increase in shielding outweighs increased nuclear charge
  • nuclear attraction on the outer electrons decreases so atomic radius increases
25
Q

on crossing a period what are formed

A

ions

26
Q

on crossing from na+ to si4+ and from p3- to cl-

A
  • each have same number of electrons
  • increase in nuclear charge
  • nuclear attraction on outer e- increases
  • ionic radius decreases
27
Q

anions

A

larger than atoms of the same element

- have same nuclear charge holding more electrons so nuclear attraction is weaker

28
Q

cations

A

smaller than atoms of the same element

  • same nuclear charge holding fewer electrons so nuclear attraction is greater
  • also, one less shell of electrons so shielding is reduced
29
Q

melting point of elements

A
  • depends on the strength of the bonds which have to be broken
  • the melting points of metals are relatively high because of the strong metallic bonds that have to be broken
30
Q

why do melting points increase from Na to Al

A
  • there are more outer shell electrons which can be delocalised
  • the charge density or the metal ion is increasing
31
Q

what type of structure is highest melting point

A

giant covalent lattices (eg silicon and carbon) due to many strong covalent bonds that have to be broken

32
Q

what type of structure is lowest melting point

A

simple molecular covalent

  • bonding between molecules - weak id:id attractions
  • the strength of these depends on the total number of electrons in each molecule
  • s8 > p4 > cl2 > ar
33
Q

electronegativity increases across a period

A
  • nuclear charge increasing so bonding electrons pulled closer to nucleus
  • no change in shielding
  • nuclear attraction on bonding electrons is greater
34
Q

diamond structure

A
  • carbon forms 4 covalent bonds with other carbon atoms
  • tetrahedral arrangement
  • giant covalent network
  • strong covalent bonds throughout
  • bond angle 109.5°
35
Q

diamond physical property

A
  • very high mp due to strong covalent bonds that must be broken
  • strong and hard due to covalent bonds that are strong and need to be broken (insoluble)
  • do not conduct (no free electrons or ions)
36
Q

uses of diamond

A

in tools for polishing and cutting

37
Q

graphene structure

A
  • carbon forms 3 covalent binds with other carbon atoms
  • 4th electron of each carbon is delocalised
  • single sheets of interlocked hexagons
  • one atom thick
  • strong covalent bonds throughout
38
Q

physical properties of graphene

A
  • conducts electricity as 4th electron can move

- high strength due to network of covalent bonds

39
Q

structure of buckmindsterfullerene

A

-60 carbon atoms, carbon atom at each corner of 20 hexagons and 12 pentagons

40
Q

buckminsterfullerene physical properties

A

semi conductors

41
Q

uses buckminsterfullerene

A

catalysts lubricants nanotubes

42
Q

graphite structure

A
  • carbon forms 3 covalent bonds with other C atoms
  • 4th electron delocalised
  • layers of interlinked hexagons
  • bonds 120°
  • strong covalent bonds each layer but weak london forces between these layers
43
Q

physical properties graphite

A
  • very high mp
  • strong cb to break soft solids -> layers can slide as only weak forces between them
  • insoluble strong cb need to be broken
  • conducts due to electrons able to move within layers
44
Q

uses of graphite

A 
dry lubricants (slippery)
electrodes (conducts)
45
Q

qualitative analysis of carbonate ions co32-

A
  • add dilute nitric acid
  • effervescence may mean carbonate is present
    co32- + 2h+ -> co2(g) + h2o (l)
  • bubble through limewater (saturated ca(oh)2) and will turn cloudy
    results in caco3
46
Q

qualitative analysis of sulphate ions so42-

A
  • add nitric acid and barium nitrate solution (allows halide ion test to be done)
  • white ppt of barium sulphate shows present sulphate ions
    ba2+ + so42- -> baso4
47
Q

qualitative analysis of halide ions cl- br- i-

A

add nitric acid and silver nitrate to aqueous solution

  • white for cl- (soluble in dilute nh3 aq)
  • cream for br- (soluble in conc. nh3 aq)
  • yellow for i- (insoluble in conc. nh3 aq)

Chemistry (17 decks)

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