a mix of AS topics Flashcards
define covalent bond
explain how they occur
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
define dative covalent bond
a covalent bond where both e- come from the same atom
Activation Energy
Minimum energy (1)
required before a reaction can occur (1)
disadvantage of dot and cross diagrams
assumes atoms within the compound are arranged along one plane only (sometimes true)
metallic structure and properties of metals
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
solubilty rules
expalin what determines the shape of a molecule
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
define electronegativity
ability of an atom to attract e- pair in a covalent bond towards itself
define enthalpy change
heat energy change measured at a constant pressure
what does Hess’s law state
enthalpy change of reaction is independent of which route it takes
define mean bond enthalpy
energy required to break a covalent bond measured over a range of molecules containing the bond
define periodicity
pattern in the change of properties of a row of elements which is repeated across the next row
standard enthalpy of combustion definition
enthalpy change when 1 mole of a substance is completely burned in excess oxygen under standard conditions
standard enthalpy of formation definition
enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions
test for ammonium, hydroxide and sulphate ions
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 CO32- ions)
- Then add BaCl2 (aq)
- A white precipitate of barium sulphate forms:
- SO42-(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
explain the trend in first IE in group 2 from Mg to Ba
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
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
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%
which element in period 3 will have the highest 2nd IE and why?
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)
explain the overall trend of 1st IE across period 3
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-
why do Al and S have a lower than expected 1st IE
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
explain the trend in electronegativity across period 3
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
explain the trend in melting points across period 3
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
explain the trend in atomic radius across period 3
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
explain trend of atomic radius down group 2
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
explain trend in 1st IE down group 2
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
explain trend in melting points down group 2
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
explain trend in reactivity with water down group 2
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.
solubility of group 2 hydroxides/sulphates in water
Uses of group 2 compounds and elements
shapes of molecules
presence of LP reduces bond angle by 2.5º
double bonds count as 1 BP
explain how you would deduce the shape of a NH3 molecule
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
ne- pairs - ncovalent bonds = nlone pairs
nlone pairs = 4-3 = 1
therefore there are 3BP and 1LP so the shape is trigonal pyramidal
what shape is a XeF2 molecule?
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
give the ideal gas equation and the units
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
electron configuration of Cr and Cu
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)
what is the electron configuration of a iron(III) ion?
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
draw the e- configuration of a vanadium atom
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
enthalpy change definition
heat energy change at a constant pressure
more exothermic
as energy released when water vapour condenses (as making bonds releases energy)
improvements to reduce heat loss in calorimetry experiments
reduce distance between flame and beaker
put sleeve around flame to reduce drafts
add lid
use copper calorimeter instead of pyrex beaker
forward reaction is exothermic
if Kc is larger this means conc of products has increased when temperature was lowered so position of eqm had shifted to the right
compare non-polar, polar, and ionic bonds
The effect of geometric isomerism on physical properties:
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
describe method to calculate the enthalpy change of neutralisation
- 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
write down the half equation for the reduction of NO3- to NO
define oxidising and reducing agent
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
write the half equation for the oxidation od SO32- to SO42-
Why is bond enthalpy measured in the gaseous state only?
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
Calculate the ΔfH 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
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
why are iodide ions better reducing agents than chloride ions
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
describe how H bonding occurs
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
explain the trend in oxidising ability of group 7
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
explain the trend in reducing ability of group 7
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
define functional group
group of atoms responsible for characteristic reactions of a compound
define homologous series
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
define molecular formula
actual number of atoms of each element in a compound
define displayed formula
shows all atoms and all covalent bonds for a compound
ionic bonds are shown using charges
define structural formula
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
define empirical formula
simplest whole number ratio of the atoms of each element in a compound
define skeletal formula
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)
order these functional groups in decreasing naming priority:
CHO, COCl, CO, COOH, NH2, COO, CN, OH, C=C
HIGHEST NAMING PRIORITY:
- COOH
- COO
- COCl
- CN
- CHO
- CO
- OH
- NH2
- C=C
LOWEST NAMING PRIORITY
explain why some ionic compounds are soluble in water whereas others are not
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
properties of molecular covalent crystals e.g ice and I2
low melting points and brittle due to weak VDW forces between molecules
do not conduct electricity as no charged particles
diamond - structure and properties
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
graphite structure and properties
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
what is the shape and bond angle of SF4 ?
is the molecule polar?
what IMF can be found between its molecules?
BF3 contains polar bonds but is a non-polar molecule
explain why
equally polar bonds are arranged symmetrically around the B atom so the polarities of the bonds cancel out
explain how VDW forces arise
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
explain trend in electronegativity down a group
decreases beause :
atomic radius/shielding increases = weaker attraction between BP of e- and nucleus
what does polarity mean?
unequal distribution of the e- density in a covalent bond
polar covalent bonds are described as having ionic character so can conduct electricity
explain how permanent dipole-dipole forces arise
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
rules for determining oxidation state
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
deduce why the bonding in nitrogen oxide is covalent rather than ionic
small electronegativity difference between N and O as both non-metals
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]
[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
define addition reaction
when a molecule is added to an unsaturated compound to form a saturated compound
state the observations you can make when:
a) Mg reacts with steam [2]
b) Na is heated with oxygen [2]
a)white solid
bright white flame
b)white solid
yellow flame
By reference to the structure of, and the bonding in, silicon dioxide, suggest why it is insoluble in water.
macromolecular with covalent bonding
water cannot supply enough energy to break strong covalent bonds
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
ionic bonding
contain O2- ions which accept H+ from water to form OH- ions
Suggest why silicon dioxide is described as an acidic oxide even though it is insoluble in water.[1]
it reacts with bases
flame tests
Na+ = yellow flame
Ca2+ = Brick red
Sr2+ = Red
Ba2+ = Pale green
why does the bpt of haloalkanes increase down group 7?
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
what has higher naming priority, alkene or haloalkane group?
alkene group
solubility of haloalkanes
insoluble in water
although CX bond is polar, large R group gets in way of H bond formation with water
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]
- 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)
what causes geometrical isomerism to occur? [2]
restricted rotation around C=C bond
different priority groups attached to the 2 C of the C=C bond
why are secondary carbocations less stable than tertiary carbocations ?
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)
in the test for halide ions,
dilute nitric acid is added before silver nitrate solution
explain why [1]
to prevent precipitation of silver compounds other than halides
which of these period 3 elements has the highest first IE?
A : Al
B: P
C: Si
D: S
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
which of these elements has the highest second IE?
A: Na
B: Mg
C: Ne
D: Ar
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
which period 3 element has the highest melting point
A: Al
B: P
C: Na
D: S
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
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
correct answer: D
B incorrect as Al has lower first IE than Mg
define elimination reaction
when a molecule lost from saturated compound to form an unsaturated compound
A → B + C
define substitution reaction
when an atom/group replaces another atom/group
AB + CD → AD + BC
a) Acid + metal→ ?
b) acid + metal oxide →
c) acid + metal/hydrogencarbonate →
d) acid + ammonia →
e) metal + water →
f) metal carbonate (when heated) →
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)
chlorine reacts with cold aqueous NaOH in the manufacture of bleach
give the reaction
Cl2 +2NaOH → NaCl + NaClO + H2O
reaction of halides with sulfuric acid e.g NaCl + H2SO4
NaCl + H2SO4 → HCl + NaHSO4
hydrogen halide + hydrogen sulfate salt
sulfuric acid acts as an oxidising agent
