Chemical Bonding and Structure Flashcards
1
Q
structure of ionic compound
A
+ve and -ve ions attracted to each other by strong electrostatic forces, building up into strong crystal lattice
2
Q
polyatomic ions
A
ions formed form >1 element where charge = often delocalised over whole ion. -ve ions = known as acid radicals bc formed when acid loses 1/+ H+ ions
3
Q
hydroxide
A
(OH)-
4
Q
nitrate
A
(NO3)-
5
Q
hydrogencarbonate
A
(HCO3)-
6
Q
phosphate
A
(PO4)3-
7
Q
sulfate
A
(SO4)2-
8
Q
hydrogensulfate
A
(HSO4)-
9
Q
carbonate
A
(CO3)2-
10
Q
ethanoate
A
(CH3COO)-
11
Q
what EN diff. needed for ionic compounds to form?
A
> 1.8
12
Q
what is ionic bond?
A
sum of all electrostatic attractions and repulsions w/in the lattice
13
Q
describe the properties of ionic compounds:
A
- high MP bc large amount of E to break lattice
- soluble in H2O bc hydration E of ion provides E to overcome lattice enthalpy
- solid ionic compounds don’t conduct bc ions = fixed positions
- molten ionic compounds conduct, bc ions = free to move and carry charge
14
Q
covalent bond
A
electrostatic attraction between shared pair of electrons and nuclei of atoms making bond
15
Q
Lewis structure
A
shows all valence electrons of molecule
16
Q
coordinate/ dative bond
A
when electrons in shared pair originate from same atom
e.g. sulfur dioxide, sulfur trioxide
17
Q
how does bond length and strength vary?
A
single bonds = longest and weakest
triple bonds = shortest and strongest
18
Q
bond polarity
A
when one end of molecule = more electron rich bc atom w/ > EN exerts > attraction for electron pair. small diff. in charge. > EN diff. = > polarity
19
Q
describe bond polarity in diatomic molecules:
A
no polarity bc electron pair shared equally bc both atoms exert identical attraction
20
Q
VSEPR theory
A
Valence Shell Electron Pair Repulsion theory states that pairs of electrons arrange themselves around central atom to be as far away as possible. > repulsion between non-bonded pairs than bonded electron pairs
mostly refers to domains bc a triple bond counts as one domain/ pair
21
Q
octet rule
A
tendency of an atom (except H) in molecule/ ion to have 8 valence electrons
exceptions inc. phosphorus, sulfur + boron
22
Q
shape + bond angle for 2 domains:
A
linear, 180
23
Q
shapes + bond angle for 3 domains:
A
- trigonal planar (3 bonding pairs), 120
- bent/ v-shaped (2 bonding pairs, 1 lone pair), <120
24
Q
shapes + bond angles for 4 domains:
A
- tetrahedral (4 bonding pairs), 109.5
- trigonal pyramidal (3 bonding pairs, 1 lone), <109.5 e.g. 107
- bent/ v-shaped (2 bonding, 2 lone), «109.5 e.g. 105
25
shapes + bond angles for 5 domains:
- trigonal bipyramidal (5 bonding pairs), 120
- seesaw/ distorted tetrahedral (4 bonding pairs), <180
- T-shaped (3 bonding pairs), <<180
- linear (2 bonding pairs), 180
26
shapes + bond angles for 6 domains:
- octahedral (6 bonding), 90, 180
- square pyramidal (5 bonding), <90, 180
- square planar (4 bonding), 90, 180
27
resonance hybrids
- extreme forms of the true structure, which lies somewhere in between i.e. when >1 Lewis structure can be drawn for a molecule
- e.g. ozone, bc bond length between O atoms = between double bond + single bond
28
allotropes
occur when an element can exist in diff. crystalline forms
29
allotropes of carbon
diamond, graphite (graphene), buckminsterfullerene
30
diamond
- each C atom covalently bonded to 4 other C atoms = giant covalent structure
- bonds = equally strong w/ no plane of weakness so v. hard
- electrons = localised, can't conduct electricity
31
graphite
- each C atom strong bonds to 3 others = layers of hexagonal rings w/ weak bonds between layers
- excellent lubricant bc layers slide over each other
- delocalized electrons = good conductor
32
graphene
- single layer of hexagonally arranged C i.e. 1 atom thick of graphite
- v. light
- semiconductor
- 200x stronger than steel
- magnetic form = graphone
33
buckminsterfullerene
- 60 C atoms in hexagons/ pentagons = geodesic spherical structure
- led to nanotechnology
34
what does molecular polarity depend on?
- EN of atoms
- shape of molecule
- if individual bonds = polar doesn't always mean molecule = polar bc resultant dipole may cancel out individual dipoles
35
van der Waals' forces
-general term for intermolecular forces inc. dipole-dipole, dipole-induced dipole, LDF
36
London Dispersion Forces
- weakest intermolecular forces
- instantaneous dipole-induced dipole forces existing between any atom
- this is bc electrons can be unevenly spread, producing temporary dipoles, leading to weak attraction between particles
- increases w/ increasing mass
37
Dipole-dipole forces
-when polar molecules = attracted to each other by electrostatic forces
38
H bonding
- strongest intermolecular force
- occurs when H bonded to small v. EN element e.g. O, N, F
- electron pair drawn away from H by EN element, proton remains
- proton attracts lone pair electrons from F, N, O = v. strong dipole-dipole attraction
39
how does solubility depend on bonding type?
- polar solvents dissolve polar substances
- organic molecules have polar head, non polar C tail
- as length of chain increases, solubility in water decreases
- ethanol = good solvent for many substances bc contains both polar and non-polar ends
40
how does conductivity depend on bonding type?
- needs electrons/ ions that are free to move
- metals and graphite hav delocalised electrons = v. good conductors
- molten ionic salts also conduct electricity but chemically decomposed in the process
41
why is BP a good indicator of strength of intermolecular force but not MP?
- when liquid -> gas, attractive forces between particles broken
- when solid -> liquid, crystal structure broken down, but still attractive forces between particles - they show strength but also determined by way particles pack in crystal state and presence of impurities (impurities weaken structure = lower MP)
42
describe the MP and BP for covalent structures:
-cov. bonds = so macromolecular covalent structures = v. high MP + BP
-also depends on type of attraction forces between molecules:
H bonding > dipole-dipole > LDF
43
MP and BP of ionic compounds
- relatively high MP and BP bc ionic attractions
| - doesn't necessarily mean that compounds w/ smaller highly charged ions = higher MP and BP
44
describe metallic bonds:
- close packed lattice of +ve ions in sea of delocalised electrons
- metallic bond = attraction that 2 neighbouring particles have for the electrons between them
- malleable + ductile bc close packed layers of +ve ions can slide over each other without breaking bonds
45
alloy
--metallic solid solution- made of >1 metal but some have carbon e.g. steel = Fe + C
46
properties of alloys in comparison to pure metals:
- -lower MP
- -added metals hav diff. radius/ even charge, distorting structure of original metal so bonding = < directional
- - < ductile, malleable bc impurities disturb lattice
- -harder as well
47
examples of alloys:
- small amounts of C added to Fe = steel w/ high tensile strength
- if Cr is added, produces stainless steel, w/ increased resistance to corrosion
48
what does MP of metals depend on?
- size of ion
- charge of ion
- way in which atoms = arranged in solid metal
49
describe MPs of metals:
- trend in grp 1: smaller metal ion = stronger bond, higher MP, also they melt below 181
- most metals have high MPs but mercury = liquid
50
sigma bond
- 1st bond between 2 atoms
| - when 2 atomic orbitals on diff atoms overlap head on bet. s, p and both
51
pi bond
- 2nd, 3rd bond
| - 2 p orbitals overlap sideways on, above and below plane of nuclei
52
what happens instead of resonance?
- instead of forming double bond, electrons can delocalise over all atoms = energetically more favourable
- delocalisation can occur when alternate and single bods occur bet. C atoms
53
Formal Charge
FC = (no. valence electrons) - (no. lone electrons) - 1/2(no. bonding electrons)
54
what assumption is formal charge based on?
assumes that all atoms in a molecule have the same EN
55
what is formal charge used for?
- to determine which out of several potential Lewis structures = preferred when several = possible
- even if total formal charge is equal, preferred structure is when individual atoms have lowest formal charges
56
describe bonds in ozone:
-'one and a half' bond between oxygen atoms, weaker than double bond in O2
57
why is the ozone layer important?
diff. in bond enthalpies helps protect us from sun's harmful UV radiation. rate of ozone production = equal to rate of ozone destruction- this process absorbs a lot of UV light- known as steady state
58
describe how ozone protects us:
- stratospheric ozone in equilibrium to O2, continually formed and decomposed.
- strong double bond in O2 = broken by high E UV to form O atoms called radicals- possess unpaired electron and v. reactive so radicals can react w/ O2 to form ozone
- weaker O3 bonds need lower E UV to break them
- broken into O2, and O radical, which can then react w/ O3 to form 2 O2 molecules
59
how to find wavelength of UV light necessary for O2 and O3 dissociation:
ʎ < 242 nm (highest energy)
UV radiation of the right frequency can break bonds
use E = hv and c = ʎv to find wavelength
60
what causes destruction of ozone:
-CFCs and nitrogen oxides create holes in ozone
61
how do CFCs catalyse ozone destruction?
- high E UV in stratosphere -> homolytic fission of C-Cl to produce chlorine radicals
- radicals break ozone -> more radicals so process continues
- estimated that 1 CFC can catalyse breakdown of up to 100K molecules of ozone
62
how do nitrogen oxides catalyse ozone destruction?
- catalytically decompose ozone by radical mechanism
| - oxygen radicals generated by breakdown of NO2 in UV
63
hybridisation
-mixing of atomic orbitals to form new orbitals for bonding
64
describe hybridisation in methane (CH4)
- bond angles = 109.5
- orbitals = far apart = same E (degenerate)
- however, on C, 2s = lower E than 2p
- hybridisation combines orbitals to produce 4 degenerate ones
- therefore 1 electron promoted from 2s orbital to 2p orbital to form 4 sp3 orbitals
65
describe hybridisation in ethene (C2H4)
- C bonds to 3 atoms, so only 2s and 2 of 2p orbitals used
- leaves unaffected p orbital
- formation of double bond
- 3 sp2 orbitals = far apart so 120, trigonal planar
- unaffected p orbital then overlaps to form pi bond
66
describe sp hybridisation and give an example:
- 2s orbital hybridises w/ 1 2p orbital to form linear sp hybrid w/ 180
- remaining p orbitals overlap to form 2 pi bonds
- e.g. ethyne, N2
67
how to predict type of bonding and hybridisation in carbon compounds:
- each single bond = sigma
- each double bond = 1 sigma and 1 pi
- each C either side of a double bond = sp2 hybridised
68
what can hybridisation show?
shape and bond angles of molecules
