a mix of AS topics

Question 1

define covalent bond

explain how they occur

Answer 1

strong electrostatic attraction between a shared pair of e- and the nuclei of the bonded atoms

unpaired e- in an orbital of an atoms shared with an unpaired e- in another atom’s orbital

atoms may promote e- into previously unoccupied orbitals in the same energy level to form more covalent bond

Question 2

define dative covalent bond

Answer 2

a covalent bond where both e- come from the same atom

Question 3

Activation Energy

Answer 3

Minimum energy (1)
required before a reaction can occur (1)

Question 4

disadvantage of dot and cross diagrams

Answer 4

assumes atoms within the compound are arranged along one plane only (sometimes true)

Question 5

metallic structure and properties of metals

Answer 5

lattice of postive metal ions surrounded b sea of delocalised e-

heat conductivity due to delocalised e- being able to carry heat energy and passing it through metal cations

high density due to cations packing closely together

Question 6

solubilty rules

Answer 6

Question 7

expalin what determines the shape of a molecule

Answer 7

determined by arrangement of e- pairs around the central atoms which is influenced by the repuslion between the e- pairs (depends if e- pairs are BP or LP)

e- pairs repels each other to be as far apart as possible

all BP repel each other equally

LP repel more than BP because they are held closer ti the central atom

LP repel LP more than they repel BP

molecules take up the shape that minimises repulsion

Question 8

define electronegativity

Answer 8

ability of an atom to attract e- pair in a covalent bond towards itself

Question 9

define enthalpy change

Answer 9

heat energy change measured at a constant pressure

Question 10

what does Hess’s law state

Answer 10

enthalpy change of reaction is independent of which route it takes

Question 11

define mean bond enthalpy

Answer 11

energy required to break a covalent bond measured over a range of molecules containing the bond

Question 12

define periodicity

Answer 12

pattern in the change of properties of a row of elements which is repeated across the next row

Question 13

standard enthalpy of combustion definition

Answer 13

enthalpy change when 1 mole of a substance is completely burned in excess oxygen under standard conditions

Question 14

standard enthalpy of formation definition

Answer 14

enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions

Question 15

test for ammonium, hydroxide and sulphate ions

Answer 15

positive ion tests:

• test for ammonium ions
  
  • Add dilute NaOH (aq) to sample in a test tube and gently heat
  • Dip damp litmus paper (so ammonia gas can dissolve) into test tube
  • NH₃ gas produced in positive test which turns litmus paper blue as it is alkaline
  • NH₄⁺ (aq)+ OH^- (aq)⟶ NH₃ (g)+ H₂O (l)

test for negative ions

• Hydroxide ions
  
  • as they are alkaline, either a pH meter or litmus paper can be used
  • red litmus paper turns blue in presence of OH^-
• sulphate ions
  
  • Add dilute HCl (to eliminate presence of CO₃^2- ions)
  • Then add BaCl₂ (aq)
  • A white precipitate of barium sulphate forms:
  • SO₄^2-(aq)+Ba²⁺(aq)⟶ BaSO₄(s)
• carbonate ions
  
  • add dilute HCl
  • Bubble gas formed through test tube of limewater
  • effervescence and limewater turns cloudy in positive result

Question 16

explain the trend in first IE in group 2 from Mg to Ba

Answer 16

decreases because:

• atomic radius increases - greater distance between outermost e- and nucleus
• shielding increases
• both these things decrease the strength of the electrostatic atraction between the -vely charged outermost e- and the +vely charged nucleus = less energy required to overcome the attraction

Question 17

A sample of Boron has a relative atomic mass of 10.8. the sample contains the isotopes ¹⁰B and ¹¹B. calculate the percentage abundance of ¹⁰B in this sample

Answer 17

let the abundance of 10B be x. this means that the abundance of 11B is (100-x)
set up an equation to find x: (10x+11(100-x)) / 100 = 10.8
so 10x+1100-11x=1080
so x=20
Therefore the abundance of ¹⁰B is 20%

Question 18

which element in period 3 will have the highest 2nd IE and why?

Answer 18

Na because the e- being removed from the Na⁺ is in a 2p orbital (all other elements have their outer e- in a higher energy level from 3s and above)

Question 19

explain the overall trend of 1st IE across period 3

Answer 19

increases because:

• no. of protons increases
• same shielding
• decreased atomic radius
• results in stronger electrostatic attraction between outer e- and nucleus
• more energy required to remove e-

Question 20

why do Al and S have a lower than expected 1st IE

Answer 20

the outer electron of Al is in a 3p orbital rather than 3s (like Na and Mg )which is further from nucleus = more shielding from inner electrons = less energy needed to remove this electron

the 3p⁴ electron configuration in S means that electron lost from S is from an orbital containing a pair of e- = more repulsion between electrons in S= less energy required to remove electrons

Question 21

explain the trend in electronegativity across period 3

Answer 21

electronegativity increases

Nuclear charge increases= more protons and same amount of shielding= decreasing atomic radius= increased electrostatic attraction between nucleus and the electron pair in the covalent bond

Question 22

explain the trend in melting points across period 3

Answer 22

from Na to Al, ionic radius decreases and more delocalised electrons= stronger electrostatic attraction between delocalised electrons and positive metal ions = more energy required to overcome

Si has a very high melting point due to macromolecular structure - many strong covalent bonds - lots of energy needed to overcome

S₈ has a higher melting point than P₄ because it is a bigger molecule= more electrons= stronger Van der Waals’ forces between S₈ molecules= more energy needed to separate the molecules

Ar has a very low melting point as it is monoatomic = very weak Van der Waals’ forces between the atoms= little energy needed to separate Ar atoms

Question 23

explain the trend in atomic radius across period 3

Answer 23

decreases because:

• bigger nuclear charge and shielding stays the same= stronger electrostatic attraction between outer electrons and nucleus= outer electrons drawn in more closely to nucleus

Question 24

explain trend of atomic radius down group 2

Answer 24

Atomic radius increases down the group because:

there are more filled shells between the nucleus and the outer electrons

the outer electrons are more shielded from the attraction of the nucleus = weaker electrostatic attraction between -ve e- and +ve nucleus

so outer electrons are further from the nucleus

Question 25

explain trend in 1st IE down group 2

Answer 25

First ionisation energy decreases down the group because:

Atomic radius increases= each element has one more electron shell than the previous one=more shielding = weaker electrostatic attraction between nucleus and outer electrons = less energy required to remove outer electron

Question 26

explain trend in melting points down group 2

Answer 26

Melting point decrease down the group because:

Ionic radius increases=delocalised electrons are further away from the positive nucleus= weaker electrostatic attraction between delocalised electrons and positive 2+ ions= weaker metallic bond= less energy required to break metallic bonds

Mg is an exception to this trend because of its metallic crystalline structure - higher MP than expected

Question 27

explain trend in reactivity with water down group 2

Answer 27

Be- no reaction, Mg reacts very slowly to produce Mg(OH)₂ + H₂, Ca, Sr and Ba react vigorously to form metal hydroxides + H₂

With steam, Mg reacts to form MgO + water (because Mg(OH)₂ is less stable at high temps)

In the reactions of Ca, Sr and Ba with water the bubbles of H₂ can be seen as effervescence and a white precipitate will form. The reaction is exothermic so the solution would feel hot.

Question 28

solubility of group 2 hydroxides/sulphates in water

Answer 28

Question 29

Uses of group 2 compounds and elements

Answer 29

Question 30

shapes of molecules

Answer 30

presence of LP reduces bond angle by 2.5º

double bonds count as 1 BP

Question 31

explain how you would deduce the shape of a NH₃ molecule

Answer 31

central atom is N

N is in group 5 so has 5 outer e-

add an e- for each covalent bond - 5+3 = 8 e-

add e- if you have negative ion, subtract e- if you have positive ion - NH₃ is not an ion so ignore this step

divide by two to find number of e- pairs - 8/2 = 4

n_{e- pairs} - n_{covalent bonds} = n_{lone pairs}

n_{lone pairs} = 4-3 = 1

therefore there are 3BP and 1LP so the shape is trigonal pyramidal

Question 32

what shape is a XeF₂ molecule?

Answer 32

linear 180º

2BP and 3LP

basic shape is trigonal bipyramidal but LP repel more than BP so shape takes up this shape to minimise repulsion

Question 33

give the ideal gas equation and the units

Answer 33

pV = nRT

p = pressure in Pa

V = volume in m³ (to convert cm³ to m³ x 10^-6, dm³ to m³ x10^-3)

n = moles

R = gas constant - 8.31 JK^-1mol^-1

T = temp in K

Question 34

electron configuration of Cr and Cu

Answer 34

Cr - 1s²2s²2p⁶3s²3p⁶3d⁵4s¹

Cu - 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

explanation for Cr= one of the 4s² e- moves to a 3d sub-orbital, this means there is one e- in each orbital of the 3d sub-shell. atom is more stable in this form due to symmetry around nucleus (half-full and full shells more stable)

explanation for Cu - one of the 4s² e- moves to a 3d sub-orbital. having 10 e- in the 3d sub-shell makes the atom more stable due to the symmetry around the nucleus (full shells are more stable)

Question 35

what is the electron configuration of a iron(III) ion?

Answer 35

iron atoms have 26 e-

4s e- lost first

iron(III) ions have 23 e- , 2e- lost from 4s and 1 e- from the 3d

this gives:

1s²2s²3s²3p⁶3d⁵

Question 36

draw the e- configuration of a vanadium atom

Answer 36

4s occupied before 3d - because 3d is at a higher energy level than 4s (electrons fill lowest energy sub-shells first)

however still write 3d before 4s

overall e- configruation = 1s²2s²2p⁶3s²3p⁶3d³4s²

Question 37

enthalpy change definition

Answer 37

heat energy change at a constant pressure

Question 38

Answer 38

more exothermic

as energy released when water vapour condenses (as making bonds releases energy)

Question 39

improvements to reduce heat loss in calorimetry experiments

Answer 39

reduce distance between flame and beaker

put sleeve around flame to reduce drafts

add lid

use copper calorimeter instead of pyrex beaker

Question 40

Answer 40

forward reaction is exothermic

if K_c is larger this means conc of products has increased when temperature was lowered so position of eqm had shifted to the right

Question 41

compare non-polar, polar, and ionic bonds

Answer 41

Question 42

The effect of geometric isomerism on physical properties:

Answer 42

E isomers have higher melting points because:

E isomers pack together more closely than Z isomers so have stronger intermolecular forces=more energy needed to melt them

Z isomers have higher boiling points because:

• Z isomers are polar; E isomers are non-polar

When alkyl groups are attached to the C=C bond, they release e- which makes the C atoms in the C=C bond δ+

E isomers still contain polar bonds but overall they are non-polar as the position of the atoms cancel out the δ charges

Z isomers will form permanent dipole-dipole forces as well as VDW forces between molecules which are stronger=more energy required to vaporise

Question 43

describe method to calculate the enthalpy change of neutralisation

Answer 43

• place equal volumes and concs of acid and alkali (at least 1 mol dm^-3) in separate beakers
• allow each to reach equal temps
• add both acid and alkali to polystyrene cup with lid (to minimise heat loss)
• record temp at regular intervals until max temp reached

Question 44

write down the half equation for the reduction of NO₃^- to NO

Answer 44

Question 45

define oxidising and reducing agent

Answer 45

oxidising agent - accepts e- (so oxidation state decreases) so causes another substance to lose e- thus oxidising it

reducing agent = loses e- (so oxidation state increases) so causes another substance to gain e- thus reducing it

Question 46

write the half equation for the oxidation od SO₃^2- to SO₄^2-

Answer 46

Question 47

Why is bond enthalpy measured in the gaseous state only?

Answer 47

The gaseous state is the only state in which the intermolecular forces are negligible. We only want to measure the strength of the covalent bonds, so we don’t want any interference of the value from intermolecular forces, which are stronger in solid and liquid state

Question 48

Calculate the Δ_fH of CCl₄ (l) given the following data:

CCl₄ (l) ⟶ CCl₄ (g) = +31 kJmol-1

C (s) ⟶ C (g) = +715 kJmol-1

Bond enthalpy (Cl-Cl) = +242 kJmol-1

Bond enthalpy (C-Cl)= +336 kJmol-1

Answer 48

First write down the reaction for standard enthalpy of formation of CCl4:

C (s) + 2Cl₂ (g) ⟶ CCl₄ (l)

Work out the enthalpy change of the reaction using the bond enthalpies. Add on the extra energy needed to convert the C and CCl4 into gaseous state

ΔH= [2(242) – 4(336)] + 31 + 715= -114 kJ mol-1

Question 49

why are iodide ions better reducing agents than chloride ions

Answer 49

iodide ions have more energy levels than chloride ions so more shielding

so weaker electrostatic attraction between outermost e- in iodide ions and positive nucleus than in chloride ions

outermost e- more easily removed from iodide ions

Question 50

describe how H bonding occurs

Answer 50

occurs when one of the most electronegative atoms (O,F or N) are bonded to Hydrogen

H bonds are the electrostatic attractions between a LP on the electronegative atom (in a molecule) and the δ+ H in another molecule

Question 51

explain the trend in oxidising ability of group 7

Answer 51

decreases down group because:

atomic radius and shielding increases = outer e- further from nucleus = weaker electrostatic attraction between outer e- and nucleus

more energy required to gain an e- so atom less likely to be reduced

Question 52

explain the trend in reducing ability of group 7

Answer 52

increases down group

ionic radius increases and increased shielding = weaker electrostatic attraction betwen nucleus and outer e- being lost

halide ion more likely to lose an e- because less energy is required to remove it

Question 53

define functional group

Answer 53

group of atoms responsible for characteristic reactions of a compound

Question 54

define homologous series

Answer 54

group of compounds which :

• have same general formula
• differ from successive members by a -CH₂ group
• similar chemical properties
• show a gradation in physical properties
• have same functional group

Question 55

define molecular formula

Answer 55

actual number of atoms of each element in a compound

Question 56

define displayed formula

Answer 56

shows all atoms and all covalent bonds for a compound

ionic bonds are shown using charges

Question 57

define structural formula

Answer 57

shows arrangement of atoms in a compound without showing bonds

each carbon is written separately followed by the atoms/groups attached

when groups are attached, a bracket is use to show the group is not part of the main carbon chain

Question 58

define empirical formula

Answer 58

simplest whole number ratio of the atoms of each element in a compound

Question 59

define skeletal formula

Answer 59

diagrammatic representations of compounds which do not show C or H atoms attached to the carbon chain but do show other atoms (e.g. N,O, and H attached to something other than a C)

Question 60

order these functional groups in decreasing naming priority:

CHO, COCl, CO, COOH, NH₂, COO, CN, OH, C=C

Answer 60

HIGHEST NAMING PRIORITY:

• COOH
• COO
• COCl
• CN
• CHO
• CO
• OH
• NH₂
• C=C

LOWEST NAMING PRIORITY

Question 61

explain why some ionic compounds are soluble in water whereas others are not

Answer 61

in most ionic comound, water molecules can knock off ions from the lattice and surround them

in some ionic compounds, the electrostatic attraction between opp charged ions are so high water molecules cannot break up the lattice

ionic substances do not dissolve in non-polar solvents

Question 62

properties of molecular covalent crystals e.g ice and I₂

Answer 62

low melting points and brittle due to weak VDW forces between molecules

do not conduct electricity as no charged particles

Question 63

diamond - structure and properties

Answer 63

macromolecular substance

each C is bonded to 4 others in a tetrahedral arrangement

very high melting pt due to many strong covalent bonds

does not conduct heat/electricity as there are no charged particles that can move

Question 64

graphite structure and properties

Answer 64

macromolecular structure

each C atom bonded to 3 others in a hexagonal arrangement with a bond angle of 120º,

layered structure with weak VDW forces between so layers can slide over each other = soft and lubricating

4th e- of each C atom becomes delocalised between layers

high melting point due to many strong covalent bonds between C atoms which require lots of energy to overcome

Question 65

what is the shape and bond angle of SF₄ ?

is the molecule polar?

what IMF can be found between its molecules?

Question 66

BF₃ contains polar bonds but is a non-polar molecule

explain why

Answer 66

equally polar bonds are arranged symmetrically around the B atom so the polarities of the bonds cancel out

Question 67

explain how VDW forces arise

Answer 67

e- orbiting the nuclei of atoms are in constant rapid motion and at any time they may be distributed more on one side of the molecule than another

this creates a temporary induced dipole

adjacent molecule attracted to the first molecule, inducing a temporary dipole in the next molecule and so on

induced dipoles are continuously forming and disappearing

Question 68

explain trend in electronegativity down a group

Answer 68

decreases beause :

atomic radius/shielding increases = weaker attraction between BP of e- and nucleus

Question 69

what does polarity mean?

Answer 69

unequal distribution of the e- density in a covalent bond

polar covalent bonds are described as having ionic character so can conduct electricity

Question 70

explain how permanent dipole-dipole forces arise

Answer 70

occur betwen polar molecules with permanent dipoles

e- density in covalent bond pulled towards more electronegative atom, 1 atom is delta - the other deta + (known as permanent dipole)

the delta positive end of one molecule attracts the delta negative end of the dipole of another molecule

Question 71

rules for determining oxidation state

Answer 71

O is -2 except in peroxides it is -1 (e.g H2O2, K2O2, Na2O2) and if OF2 it is +2

H is +1 except for hydrides it is -1

in simple compounds containing 2 elements, the more electronegative element has the negative oxidation state

Question 72

deduce why the bonding in nitrogen oxide is covalent rather than ionic

Answer 72

small electronegativity difference between N and O as both non-metals

Question 73

2A + B ⇌ 3C + D

justify the statement that adding more water to the eqm mixture will lower the amount of A in the mixture [3]

Answer 73

[A], [B], [C] and [D] all decrease

pos of eqm shifts to right as more moles on right

so [C] & [D] increase and [A] decreases

Question 74

define addition reaction

Answer 74

when a molecule is added to an unsaturated compound to form a saturated compound

Question 75

state the observations you can make when:

a) Mg reacts with steam [2]
b) Na is heated with oxygen [2]

Answer 75

a)white solid

bright white flame

b)white solid

yellow flame

Question 76

By reference to the structure of, and the bonding in, silicon dioxide, suggest why it is insoluble in water.

Answer 76

macromolecular with covalent bonding

water cannot supply enough energy to break strong covalent bonds

Question 77

Some Period 3 oxides have basic properties. State the type of bonding in these basic oxides. Explain why this type of bonding causes these oxides to have basic properties

Answer 77

ionic bonding

contain O2- ions which accept H+ from water to form OH- ions

Question 78

Suggest why silicon dioxide is described as an acidic oxide even though it is insoluble in water.[1]

Answer 78

it reacts with bases

Question 79

flame tests

Answer 79

Na⁺ = yellow flame

Ca²⁺ = Brick red

Sr²⁺ = Red

Ba²⁺ = Pale green

Question 80

why does the bpt of haloalkanes increase down group 7?

Answer 80

if number of carbons increases and/or X atom gets bigger :

bigger Mr so more e- = stronger VDW forces

if same number of carbons look at electronegativity of X atom

higher electronegativity = higher polarity of C-X bond = stronger VDW forces

Question 81

what has higher naming priority, alkene or haloalkane group?

Answer 81

alkene group

Question 82

solubility of haloalkanes

Answer 82

insoluble in water

although CX bond is polar, large R group gets in way of H bond formation with water

Question 83

there are 3 separate solid samples. 1 is sodium carbonate, another is sodium fluoride and the last is sodium chloride

outline a logical sequence of test-tube reactions to identify each compound

include observations and equations with state symbols for any reactions

[6]

Answer 83

• dissolve each solid in same volume of water in separate test tubes
• add HNO₃ to each solution (used as nitrate ions never form ppts)
• test tube with sodium carbonate would show effervescence
  
  • Na₂CO₃(s) + 2HNO₃ → 2NaNO₃ (aq) + CO₂(g) +H₂O(l)
• add AgNO₃ (aq) to the other two test tubes
• no visible change for test tube with sodium fluoride
• test tube with sodium chloride forms white ppt
  
  • Cl^-(aq) + Ag⁺ (aq) → AgCl(s)

Question 84

what causes geometrical isomerism to occur? [2]

Answer 84

restricted rotation around C=C bond

different priority groups attached to the 2 C of the C=C bond

Question 85

why are secondary carbocations less stable than tertiary carbocations ?

Answer 85

3º carbocations have 3 R groups attached directly to the positive C atoms

2º carbocations have 2 R groups directly attached to the positive C atom

therefore 2º carbocations have a weaker positive inductive effect ( the e- density is less repelled toward the positive C atoms so the C⁺ is less stabilised)

Question 86

in the test for halide ions,

dilute nitric acid is added before silver nitrate solution

explain why [1]

Answer 86

to prevent precipitation of silver compounds other than halides

Question 87

which of these period 3 elements has the highest first IE?

A : Al

B: P

C: Si

D: S

Answer 87

correct answer : B (Phosphorus)

increasing first IE across period but S is an exception

first IE of S lower than first IE of P due to 3p⁴ configuration

Question 88

which of these elements has the highest second IE?

A: Na

B: Mg

C: Ne

D: Ar

Answer 88

correct answer: A

Na outer electron is in 3s orbital

Na+ outer electron removed from full 2 shell

as changing from 3rd energy level to 2nd energy level (all other e- have their second e- removed from the shame shell as first e-) large amount of energy required

Question 89

which period 3 element has the highest melting point

A: Al

B: P

C: Na

D: S

Answer 89

answer: A

as Al has metallic bonding

metallic bonding in Al stronger than in Na

P₄ and S₈ has simple molecular structure with weak VDW forces so lower melting points

Question 90

which elements are shown in increasing order of the stated property:

A: atomic radius : Phosphorus, Sulfur, Sodium

B: first IE : Sodium, Magnesium, Aluminium

C: electronegativity : Sulfur, Phosphorus, Silicon

D: melting point : Argon , chlorine, sulfur

Answer 90

correct answer: D

B incorrect as Al has lower first IE than Mg

Question 91

define elimination reaction

Answer 91

when a molecule lost from saturated compound to form an unsaturated compound

A → B + C

Question 92

define substitution reaction

Answer 92

when an atom/group replaces another atom/group

AB + CD → AD + BC

Question 93

a) Acid + metal→ ?
b) acid + metal oxide →
c) acid + metal/hydrogencarbonate →
d) acid + ammonia →
e) metal + water →
f) metal carbonate (when heated) →

Answer 93

a) salt + hydrogen
b) salt + water
c) salt + water + carbon dioxide
d) ammonium salt
e) metal hydroxide + hydrogen gas
f) metal + carbon dioxide (thermal decomposition)

Question 94

chlorine reacts with cold aqueous NaOH in the manufacture of bleach

give the reaction

Answer 94

Cl₂ +2NaOH → NaCl + NaClO + H₂O

Question 95

reaction of halides with sulfuric acid e.g NaCl + H₂SO₄

Answer 95

NaCl + H₂SO₄ → HCl + NaHSO₄

hydrogen halide + hydrogen sulfate salt

sulfuric acid acts as an oxidising agent