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chem retake > Module 3 > Flashcards

Module 3 Flashcards

1
Q

why do elements in the same group have similar chemical properties

A
  • same outer shell electron configuration
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2
Q

how are elements classified as s,p or d blocks

A

according to which orbitals the highest energy electrons are in

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3
Q

order of blocks on the periodic table

A

s,d,p,f

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4
Q

periodicity definition

A

repeating pattern across different periods

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5
Q

trend in atomic radius as you move across a period

A
  • atomic radius decreases
  • this is because the increased number of protons creates more positive charge attraction for electrons which are in the same shell
  • similar shielding
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6
Q

first ionisation energy definition

A

the energy needed to remove an electron from each atom in one mole of gaseous atoms

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7
Q

equation example for first ionisation energy

A

H(g) = H+(g) + e-

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8
Q

factors affecting ionisation energy

A
  1. attraction of the nucleus
    - more protons = greater attraction
  2. distance of the electrons from the nucleus
    - bigger atom means outer electrons are further from the nucleus, meaning weaker attraction to the nucleus
  3. shielding of the attraction of the nucleus
    - an electron in the outer shell is repelled by electrons in complete inner shells, weakening attraction of the nucleus
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9
Q

why are successive ionisation energies larger

A

the ion formed is smaller than the atom
proton to electron ratio in the 2+ ion is greater than in the 1+ ion
attraction between electron and nucleus is therefore stronger

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10
Q

why do 1st IE decrease down a group

A
  • outer electrons are found in shells further from nucleus
  • more shielded
  • attraction to nucleus decreases
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11
Q

why does 1st IE increase across a period

A
  • electrons added to the same shell which has the same distance from the nucleus and same shielding effect
  • number of protons increases making attraction greater
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11
Q

if there is a big jump in ionisation energy between IE 2 and 3, what does this mean

A
  • the element must be in group 2
  • as the 3rd electron is being removed from an electron shell closer to the nucleus and therefore experiences more nuclear attraction
  • there must also be less shielding
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11
Q

why is there a small drop in IE from group 2 to 3

A
  • group 2 has its outer electrons in the 3s sub shell, whereas group 3 is starting to fill a 3p sub shell
  • The electrons in the 3p subshell are slightly easier to remove because
    the 3p electrons are higher in energy and are also slightly shielded by the 3s electrons
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12
Q

why is there a small drop in IE from group 5 to 6

A

in group 6 r there are 4 electrons in the 3p sub shell and the 4th is starting to doubly
fill the first 3p orbital.

When the second electron is added to a 3p orbital there is a slight repulsion between
the two negatively charged electrons which makes the second electron easier to
remove.

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13
Q

metallic bonding definition

A

the electrostatic force of attraction between
the positive metal ions and the delocalised electrons

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14
Q

what are the three main factors that affect the strength of metallic bonding

A
  1. Number of protons/ Strength of nuclear attraction.
    (The more protons the stronger the bond)
  2. Number of delocalised electrons per atom - the outer shell electrons are delocalised
    (The more delocalised electrons the stronger the bond)
  3. Size of ion.
    (The smaller the ion, the stronger the bond)
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15
Q

why has Mg got stronger metallic bonding than Na

A

Mg has stronger metallic bonding than Na and hence a higher melting point.

The Metallic bonding gets stronger because in Mg
there are more electrons in the outer shell that are released to the sea of electrons.

The Mg ion is also smaller and has one more proton. There is therefore a stronger electrostatic attraction
between the positive metal ions and the delocalised electrons and higher energy is needed to break bonds

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16
Q

explain the structure of diamond

A

MACROMOLECULAR
- Tetrahedral arrangement of
carbon atoms.
- 4 covalent bonds per atom
- high melting points because of strong covalent forces in the giant structure. It takes a lot of energy to break the many strong covalent bonds

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17
Q

explain the structure of graphite

A

MACROMOLECULAR
- Planar arrangement of carbon atoms in layers.
- 3 covalent bonds
per atom in each layer. 4th outer electron per atom is delocalised.
- Delocalised electrons between layers.
- high melting points because of strong covalent forces in the giant structure. It takes a lot of energy to break the many strong covalent bonds

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18
Q

which structures have metallic bonding

A

GIANT METALLIC LATTICES

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19
Q

properties of macromolecular substances

A

bp and mp: high due to many covalent bonds which take a lot of energy to break

solubility: insoluble

conductivity when solid: diamond and sand are poor because electrons cant move, whereas graphite is good as free electrons between layers

conductivity when molten: poor

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20
Q

properties of giant metallic substances

A

mp and bp points: high due to strong electrostatic forces between positive ions and sea of deloc electrons

solubility: insoluble

conductivity: good when solid and molten - deloc electrons can move through structure

general: malleable as the positive ions in the lattice are all identical, so the planes of ions can slide easily over one another.

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21
Q

explain the general trend of mp and bp across period one

A

or Na, Mg, Al- Metallic bonding : strong bonding – gets stronger the more electrons there are in the outer shell that
are released to the sea of electrons. A smaller positive
centre also makes the bonding stronger. High energy is needed to break bonds.

Si is Macromolecular: many strong covalent bonds
between atoms high energy needed to break covalent
bonds– very high mp +bp

Cl2 (g), S8 (s), P4 (S)- simple Molecular : weak London forces between molecules, so little energy is needed to break them – low mp+ bp

S8 has a higher mp than P4 because it has more electrons
(S8 =128)(P4=60) so has stronger London forces

Ar is monoatomic weak London forces between atoms

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22
Q

what happens to atomic radius down group 2

A
  • increases
  • atoms have more shells of electrons
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23
Q

melting point down group 2

A
  • decreases
  • metallic bonding weakens as the atomic size increases
  • distance between the + ions and deloc electrons increases
  • therefore, electrostatic
    attractive forces between the positive ions and the
    delocalized electrons weaken
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24
Q

what happens to first and second IE’s down group 2

A
  • decrease
  • The outermost electrons are held more weakly because they are successively further from the nucleus in additional
    shells
  • In addition, the outer shell electrons become more shielded from the attraction of the nucleus by the repulsive force of inner shell electron
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25
Q

second ionisation energy definition

A

the enthalpy change when one mole of gaseous ions with a single positive charge forms one mole of gaseous ions with a double positive charge

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26
Q

example equation for 2nd IE

A

Ti+ (g) = Ti2+(g) + e-

27
Q

reactivity down group 2

A
  • increases
  • As the atomic radii increase there is more shielding.
  • The nuclear attraction decreases and it is easier to remove outer electrons.
  • Cations form more
    easily
28
Q

group 2 metals reactions with oxygen

A

The group 2 metals will burn in oxygen.
Mg burns with a bright white flame
2Mg + O2 → 2MgO- Mg will also react slowly with oxygen without a flame.

  • Mg ribbon will often have a thin layer of magnesium oxide on it formed by reaction with oxygen
  • this is cleaned off by emery paper before doing reactions with Mg ribbon as Mg and MgO react at different rates
  • if testing for reaction rates with Mg and acid, an un-cleaned Mg ribbon would give a false result as both the Mg and MgO would react but at different rates
29
Q

Mg + steam

A

produces magnesium oxide and hydrogen

the Mg would burn with a bright white flame

30
Q

Mg + warm water

A

produces magnesium hydroxide and hydrogen

this is a slower reaction than with steam and produces no flame

31
Q

group 2 metals and cold water

A
  • react with increasing vigour down the group to form hydroxides

eg : Ca + 2 H2O (l) → Ca(OH)2
(aq) + H2(g)

32
Q

what are the observations for group 2 metals reacting with cold water

A
  • fizzing (more vigorous down the group)
  • the metal dissolving (faster down group)
  • the solution heating up (more down group)
  • with calcium, a white precipitate is formed (less precipitate forms down the group)
33
Q

group 2 metals reacting with acid
give an example equation

A
  • react with acids with increasing vigour down the group to form a salt and hydrogen
    for example:
    Ca + 2HCl(aq) = CaCl2(aq) + H2 (g)
34
Q

why does barium react slowly with H2SO4
give the equation

A

the insoluble barium sulfate produced will cover the surface of the metal and act as a barrier to further attack.

Ba + H2SO4 = BaSO4 + H2

the same effect will happen to a lesser extent with metals going up the group as solubility increases

35
Q

group 2 oxides + water

A
  • forms hydroxides
36
Q

why are group 2 oxides basic

A
  • the oxide ions accept H+ ions to become hydroxide ions
37
Q

what is magnesium hydroxide used for

A

used in medicine (in suspension as milk of magnesia) to neutralise excess acid in the stomach and to treat constipation.

Mg(OH)2 + 2HCl → MgCl2 + 2H2O

It is safe to use as it is weakly alkaline.

38
Q

what is calcium hydroxide used for

A
  • It is used in agriculture to neutralise acidic soils.
  • If too much calcium hydroxide is added to the soil, excess will result in soils becoming too alkaline to sustain crop growth
39
Q

appearance of the halogens

A

Fluorine (F2): very pale yellow gas. It is highly reactive

Chlorine : (Cl2) greenish, reactive gas, poisonous in high concentrations

Bromine (Br2) : red liquid, that gives off dense brown/orange poisonous fumes

Iodine (I2) : shiny grey solid sublimes to purple gas.

40
Q

mp and bp down the halogens

A

increases down the group

As the molecules become larger they have more electrons and so have larger induced dipole-dipole forces
(London forces) between the molecules. As the intermolecular forces get larger more energy has to be put into break these intermolecular forces. This increases the melting and boiling points

41
Q

reactivity of the halogens going down the group

A
  • decreases
  • as the atoms get bigger with more shielding so they less easily attract and accept electrons. They therefore form -1 ions less easily down the group
42
Q

displacement reactions of halogens

A

look at chemsheets

43
Q

disproportionation reaction definition

A

name of a reaction where an element simultaneously oxidises and reduces

44
Q

chlorine + water equation
how is it disproportionation?
what will happen is UI is added?

A

Cl2(g) + H2O(l) →HClO(aq) + HCl (aq)

Chlorine is both simultaneously reducing and oxidising. It changes from 0 in Cl2 to -1 in HCl and +1 in HClO

If some universal indicator is added to the solution it will first turn red due to the acidity of both reaction products. It will then turn
colourless as the HClO bleaches the colour.

45
Q

what are the benefits and negatives of chlorine being used in water treatment?

A

kills bacteria
toxic and irritates the respiratory system

46
Q

chlorine + cold dilute NaOH

A

Cl2(aq) + 2NaOH(aq) → NaCl (aq) + NaClO (aq) + H2O(l)

  • the colour of the halogen solution will fade to colourless
  • The mixture of NaCl and NaClO (sodium chlorate (I)) is used as Bleach and to disinfect/ kill bacteria
47
Q

chlorine + hot NaOH

A

3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O

sodium chlorate (V) is formed

48
Q

reaction of halide ions with silver nitrate
- used for?
- why is nitric acid added?
- colours?
- how can ammonia be used to help differentiate

A

used as a test to identify which halide ion is present. The test solution is made acidic with nitric acid, and then Silver nitrate solution is added drop wise

The role of nitric acid is to react with any carbonates
present to prevent formation of the precipitate
Ag2CO3 - This would mask the desired observations

Fluorides produce no precipitate
Chlorides produce a white precipitate
Ag+(aq) + Cl- (aq) →AgCl(s)
Bromides produce a cream precipitate
Ag+ (aq) + Br- (aq) → AgBr(s)
Iodides produce a pale yellow precipitate
Ag+ (aq) + I- (aq) →AgI(s)
The silver halide precipitates can be treated with ammonia
solution to help differentiate between them if the colours look similar:

Silver chloride dissolves in dilute ammonia to form a
complex ion
AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl- (aq)

Silver bromide dissolves in concentrated ammonia to form a complex ion
AgBr(s) + 2NH3(aq) →[Ag(NH3)2]+(aq) + Br - (aq)

Silver iodide does not react with ammonia – it is too insoluble

49
Q

testing for the presence of a carbonate

A
  • add any dilute acid and observe effervescence
  • fizzing due to co2 would be observed if carbonate present
  • bubble gas through limewater to test for co2- will turn limewater cloudy
50
Q

testing for presence of a sulfate
- why can sulfuric acid not be used
- equation

A
  • An acidified BaCl2 solution is used
  • white precipitate forms
  • acid is used as needed to react with carbonate impurities that are often found in salts which would form a white barium carbonate precipitate and give a false result
  • sulfuric acid cannot be used as it contains sulfate ions which would form a precipitate
  • Ba2+(aq) + SO42- (aq) = BaSO4 (s)
51
Q

how to test for an ammonium ion
- how can ammonia gas be identified

A

react with NaOH(Aq), forming NH3 gas
- ammonia gas can be identified by its pungent smell or by turning red litmus paper blue

52
Q

in what order should you test for ions

A
  1. carbonate
  2. sulfate
  3. halide
    - this prevents false tests
53
Q

is △H positive or negative in exothermic and endothermic reactions?

A

In an exothermic reaction the
∆H is negative

In an endothermic reaction
the ∆H is positive

54
Q

activation energy definition

A

the minimum energy which particles need to collide to start a reaction

55
Q

standard conditions

A
  • 100 kPa pressure
  • 298 K (room temperature or 25oC)
  • Solutions at 1mol dm-3
  • all substances should have their normal state at 298K
56
Q

enthalpy change of reaction definition

A

the enthalpy change when the number of moles of reactants as specified in the
balanced equation react together

57
Q

standard enthalpy change of formation definition

A

the enthalpy change when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all
reactants and products being in their standard states

58
Q

standard enthalpy change of combustion definition

A

the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions (298K and 100kPa), all reactants and products being in their
standard states

59
Q

enthalpy change of neutralisation definition

A

the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

60
Q

what does incomplete combustion lead to

A

soot (carbon)
carbon monoxide
water
it will be LESS exothermic than complete combustion

61
Q

errors in calorimetry experiments

A
  • energy transfer from surroundings
  • approximation in specific heat capacity of solution
  • neglecting specific heat capacity of calorimeter
  • reaction may be incomplete
  • density of solution is taken to be the same as water
62
Q

energy change equation

A

q(J) = m(g) x c x △T(K)

63
Q

errors in combustion experiments using calorimetry

A

Energy losses from calorimeter
* Incomplete combustion of fuel
* Incomplete transfer of energy
* Evaporation of fuel after weighing
* Heat capacity of calorimeter not included
* Measurements not carried out under standard conditions as H2O is gas, not liquid, in this experiment

64
Q

mean bond enthalpy

A

the enthalpy change when
one mole of bonds of (gaseous covalent) bonds is broken (averaged over different molecules)

65
Q

hess’ law definition

A

total enthalpy change for a reaction is independent of the route by which the chemical changes takes place

66
Q
A

chem retake (2 decks)

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