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chemCHEMMMMM Flashcards

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AP® Chemistry flashcards Periodic Table flashcards
1
Q

how to calc number of molecules

A

x moles by avogadros constant

  • x 6.02 x10^23
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2
Q

state what is meant by term ionic bond

A
  • electrostatic attraction between positive and negative ions
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3
Q

how to calc number of ions

A

x moles by avogadros constant

  • x 6.02 x10^23
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4
Q

whats def electronegativity

A
  • ability of an atom to attract electrons in covalent bond
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5
Q

ionic lattice def

A
  • repeating pattern of oppositely charged ions
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6
Q

first ie def

A

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

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7
Q

equation 1st ie

A

H(g)  H+(g) + e-

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8
Q

y r successive ie always larger

A

The second ionisation energy of an element is always bigger than the first ionisation energy.
This is because the ion formed, is smaller than the atom and the proton to electron ratio in the 2+ ion is greater than in the 1+ ion.
The attraction between nucleus and electron is therefore stronger

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9
Q

Why do first ionisation energies decrease down a group?

A

As one goes down a group, the outer electrons are found in shells further from the nucleus and are more shielded so the attraction of the nucleus becomes smaller

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10
Q

Why is there a general increase in first ionisation energy across a period?

A

As one goes across a period the electrons are being added to the same shell which has the same distance from the nucleus and same shielding effect. The number of protons increases, however, making the effective attraction of the nucleus greater.

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11
Q

Why is there a small drop from Mg to Al?

A

Al is starting to fill a 3p sub shell, whereas Mg has its outer electrons in the 3s 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 from P to S?

A

With sulphur 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

wt r 3 main factors affecting ionisation energies

A

1.The attraction of the nucleus (The more protons in the nucleus the greater the attraction)
2. The distance of the electrons from the nucleus (The bigger the atom the further the outer electrons are from the nucleus and the
weaker the attraction to the nucleus)
3. Shielding of the attraction of the nucleus
(An electron in an outer shell is repelled by electrons in complete inner shells, weakening the attraction of the nucleus)

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14
Q

wt happens to atomic radius moving down group 2

A

Atomic radius increases down the Group.
As one goes down the group the atoms have more shells of electrons making the atom bigger

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15
Q

wt happens to melting point as move down group 2

A

Melting points decrease down the group.
The metallic bonding weakens as the atomic size increases.
The distance between the positive ions and delocalized electrons increases.
Therefore the electrostatic attractive forces between the positive ions and the delocalized electrons weaken.

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16
Q

2nd ie

A

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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17
Q

2nd ie equation

A

Ti+ (g)  Ti2+(g) + e-

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18
Q

what happens to reactivity of group 2 metals as down group

A

reactivity increases down the group.
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.

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19
Q

reaction group 2 + o2

A

The group 2 metals will burn in oxygen.
Mg burns with a bright white flame
2Mg + O2  2MgO

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20
Q

reaction group 2 w water

A

Magnesium reacts in steam to produce magnesium oxide and hydrogen.
The Mg would burn with a bright white flame
Mg (s) + H2O (g) MgO (s) + H2 (g)

other ones made hydroxides & h2

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21
Q

group 2 + acid

A

The group 2 metals will react with acids with increasing vigour down the group to form a salt and hydrogen
Ca + 2HCl (aq) CaCl2 (aq) + H2 (g)
Sr + 2 HNO3 (aq) Sr(NO3)2 (aq) + H2 (g) Mg + H2SO4 (aq)MgSO4 (aq) + H2 (g)

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22
Q

ph of diff group 2 and water

A

ca(oh)2 = 12 this used agriculture to neutralise acid soils

Mg(OH)2 = 9 this used for heartburn/neutralise stomach acid

23
Q

halogen colours F,CL,BR,I

A

f2= pale yellow high reactive
cl2= greenish reactive
br2= red liquid gives off orange poisonous fumes
i2= shiny grey solid sublimes to purple gas

24
Q

wt happens to melting point as move down group7

A

increase coz electrons increase so stronger dipole dipole London forces
as intermol forces increase more energy required to overcome

25
Q

what happens to reactivity as move down group 7

A

decreases down the group 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

26
Q

how do group 7 react

A

displacement so higher one in group 7 displaces lower one

27
Q

cl + h2o disproportionation

A

hclo and icl

If universal indicator is added> will first turn RED due to the acidity of both reaction products.
then turn COLOURLESS as the HClO bleaches the colour.

28
Q

cl/br/i2 + COLDDDD NaOH

A

Cl2(aq) + 2NaOH(aq)–» NaCl(aq) + NaClO(aq) + H2O(l)

makes NaCl and NaClO bleach
colour of halogen fades to colourless

29
Q

cl/br/i2 + WARM NaOH

A

diff disportionation

forming sodium chlorate(V)
3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O

30
Q

how to identify halide ions

A

nitric acid, and then Silver nitrate solution is added drop wise.

Cl2 produce white precipitate
Ag+(aq) + Cl- (aq) –> AgCl(s)

Br2 cream precipitate
Ag+(aq) + Br- (aq) –> AgBr(s)

I2 produce a pale yellow precipitate
Ag+(aq) + I- (aq) –> AgI(s)

31
Q

why nitric acid used for halide ion test

A

to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observation

32
Q

wt test if halide test look similar results

A

using ammonia

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

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

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

33
Q

test 4 anions carbonate and result

A
  • add dilute acid
  • see effervesence
  • bubble w limewater- turns cloudy if co2 present for carbonate
    -fizzing coz of co2
34
Q

test 4 anion sulfate

A
  • acidified bacl2
  • white precipitate baso4
35
Q

test for cation
NH4+

A
  • react w warm NaOH
  • NH3 gas fomerd
  • turns red litmus paper blue
36
Q

wt do enthalpy diagram of exo look like

A
  • energy exiting from system to surroundings
  • products lower than reactant
37
Q

wt do enthalpy diagrams of endo look like

A
  • energy is transferred from the surroundings to the system (chemicals)
  • so products higher than reactants
38
Q

wt r standard conditions of enthalpy changes

A
  • 100 kPa pressure
  • 298 K (room temperature or 25oC)
  • Solutions at 1mol dm-3
  • all substances should have normal state at 298K
39
Q

enthalpy change reaction def don’t need to men

A

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

40
Q

Enthalpy change formation need 2 know

A

standard enthalpy change of formation of a compound is 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
Symbol fH

41
Q

wt to know ab enthalpy formation

A

enthalpy form for elements 0

42
Q

Enthalpy change combustion def

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

43
Q

will incomplete combustion be more or less exo than compl combustion

A

less

44
Q

enthalpy change neutralisation

A

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

45
Q

bond breaking…

A

absorbs energy
becomes +

46
Q

bond making

A

releases energy
becomes negative

47
Q

when does mean bond energies def come into use

A

when the
substances start and end in the gaseous state.

48
Q

equation for mean bond enthalpies
SBSM

A

sum bond energies BROKEN - sum bond energies MADE
SBSM
- if all substances r GASES

49
Q

issues w mean bond energies

A

mean bond energies r not exact so becomes problem

50
Q

average bond enthalpy def

A

avg enthalp change when 1 mole of covalent bond broken in gaseous conditions

51
Q

RULES OXIDATION NUMBERS

A
  1. Group 1 metals = +1
  2. Group 2 metals = +2
  3. Al = +3
  4. H = +1 (except in metal hydrides where it is –1 eg NaH) F = -1
  5. Cl, Br, I = –1 except in compounds with oxygen and fluorine
  6. O = -2 except in peroxides (H2O2 ) where it is –1 and in compounds with fluorine.
52
Q

How many electrons do following hold

S
P
D
F

A

S=2
P=6
D=10
D=14

53
Q

relative atomic mass of element def

A

Weighted mean mass of an atom compared with 1/12th mass of carbon-12

54
Q
A

wrong chem q (2 decks)

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