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Chem y5 > chemical bonding 1 > Flashcards

chemical bonding 1 Flashcards

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1
Q

factors affecting strength of ionic bond and defintions

A
  1. charge of ion
  2. radius
  3. lattice energy
    - energy released when 1 mole of crystalline ionic solid is formed from its constituent gaseous ions
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2
Q

just read

A

q+ and q- are the same for both compounds. However, F- is smaller than Cl- and interionic distance for NaF is smaller than NaCl. Hence, NaF has a greater magnitude of LE and a stronger ionic bond

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3
Q

properties of ionic compounds (4)

A
  1. high bp/mp
  2. soluble in polar solvents but not in non polar solvents
  3. cannot conduct electricity in solid state as ions are held in fixed positions and cannot act as charge carriers but can in aqueous/molten state as ions are free to move to act as mobile charge carriers
  4. hard and brittle as a slight displacement along the cleavage plane can cause ions of like charges to move opposite one another and the strong repulsion causes the ionic crystal lattice to shatter
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4
Q

ionic bond

A

electrostatic attraction between cations and anions in an ionic lattice

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5
Q

giant molecular lattice

A

made up of atoms held together in an extensive network by covalent bonds

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6
Q

diamond and graphite structures

A

diamond:
- each C atom is bonded to 4 other C atoms in a 3d lattice by forming strong covalent bonds
- hence it is hard and rigid with a high melting point

graphite:
- each C atom is bonded to 3 other C atoms in a layered structure by forming covalent bonds
- there are weak intermolecular attraction between layers
- high melting point and layers can slide over each other

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7
Q

Simple molecular structure

A

made up of molecules attracted to each other by weak intermolecular forces

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8
Q

properties of simple molecular structure (3)

A
  1. low mp/bp
  2. soluble in non polar but not in polar solvents
  3. does not conduct electricity in solid/molten states but can conduct electricity in aqueous state as it can ionise in water to form ions
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9
Q

Giant metallic lattice

A

electrostatic attraction between a lattice of positive ions and delocalised electrons

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10
Q

describe metallic bond

A

metals are composed of a rigid lattice of positive ions surrounded by a sea of delocalised electrons

valence electrons are delocalised and do not belong to any particular cation but to the crystal lattice as a whole

bond is strong and non directional

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11
Q

factors affecting metallic bond

A
  1. charge of cations
  2. size of cations

these 2 affect CHARGE DENSITY, smaller cation with bigger charge will lead to greater EA

  1. number of valence electrons
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12
Q

metallic bond properties (5)

A
  1. high mp/bp
  2. high electrical conductivity even in solid state
  3. high density
    - ions are closely packed
  4. malleable and ductile
    - since metallic bond is non directional, the layers of positive ions can glide over another easily without breaking the bond
  5. good thermal conductivity
    - when heated on one end, the electrons take in thermal energy → they move faster and more randomly, colliding with other electrons and passing the energy to them
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13
Q

octet rule

A

atoms tend to lose, gain or share electrons until they have 8 electrons in their valence shell

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14
Q

covalent bonding

A

electrostatic attraction between the shared pair of electrons and 2 nuclei

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15
Q

coordinate bond

A

formed when the shared pair of electrons is provided by only one of the bonding atoms

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16
Q

VSEPR

A

check notes

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17
Q

why only period 3 can have expansion of octet

A

elements in period 3 have vacant low lying orbitals that are available fir the expansion of octet

in period 2, the 2s and 2p orbitals are occupied hence the vacant orbital with the lowest energy is 3s which is much higher in energy than 2s and 2p hence there are no vacant low lying orbitals available

in period 3, the valence 3s and 3p orbitals are occupied and the vacant 3d orbitals which are in the same electron shell are only slightly higher in energy hence there is a vacant low lying orbital and electrons in 3s or 3p can be promoted to 3d for covalent bond formation hence the expansion of octet is possible

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18
Q

why bond angle in NH3 is larger than in PH3

A

because N is more electronegative than P hence it draws the bond pair closer to itself than P draws the bond pair to itself
hence the bond pair electrons are closer to the nucleus and exert more repulsion on each other causing the bond angle to be wider

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19
Q

why non polar and polar covalent bonds are formed

A

NON POLAR
same electronegativities, atoms hence same tendency to draw the shared electron pair to itself so electron pair is equidistant between the 2 bonded atoms, electron cloud is uniformly distributed, non polar

POLAR
diff electronegativities, more electronegative atom draws shared electron pair closer to itself and electron cloud is not uniformly distributed so partial charges arise on the 2 bonded atoms and a dipole moment is formed hence it is polar

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20
Q

structure of ice

A

each oxygen atom is tetrahedrally bonded to 4 other H atoms 2 by covalent bonds and 2 by hydrogen bonds forming a rigid open 3d structure that prevents molecules from getting too close to each other

volume increases for the same mass hence ice is less dense

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21
Q

id id interactions

A

instantaneous dipole induced dipole

electron density is not symmetrical because electron cloud is not evenly distributed hence an instantaneous dipole is formed then this induces a dipole in the neighbouring electron cloud, causing an attraction between them

this id id interaction is short lived because electrons are always moving and dipoles vanish and reform

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22
Q

factors affecting id id

A
  1. size of electron cloud/no. of electrons
    - as the size of electron cloud increases, the ease of polarisability of the electron clouds increases and strength of id-id increases
  2. surface area available for intermolecular interactions
    - branching vs straight chain
23
Q

pd pd interactions

A

permanent dipole permanent dipole interactions
- electrostatic attraction between the δ+ end of one molecule and δ- end of another molecule

24
Q

factors affecting pd - pd

A
  1. how polar the molecule is
    - diff in electronegativity
25
Q

hydrogen bond explain

A

h atom covalently bonded to a small highly electronegative atom with lone electron pairs (FON)

FON cause the H atom to be highly positive and since FON is small, the positive H atom can come very close to the lone pair on FON and form strong attraction (hydrogen bond)

26
Q

why bp of H2O higher than NH3

A
  1. since O is more electronegative than N, H2O has a greater dipole moment
  2. extensiveness of hydrogen bonding between H2O molecules is greater as each water molecule forms an average of 2 hydrogen bonds while each NH3 molecule only forms 1
    - since need to break more bonds when boiling H2O, bp of H2O is higher

find no. of lone pairs + no. of hydrogen atoms and the lower one is the number of hydrogen bonds formed per molecule

27
Q

diff in bp and solubility (intramolecular hydrogen bonding)

A

due to the formation intramolecular hydrogen bonds in _______, there are less sites available for the formation of intermolecular hydrogen bonds with other _____ molecules/water molecules.
Hence, ______ is less soluble in water/ there is less extensive intermolecular hydrogen bonding in _____, leading to a lower bp

28
Q

why LE of CuO deviate more than Cu2O

A

Cu2+ has a higher charge but smaller size than Cu+ hence it has a higher charge density and has greater polarising power than Cu+ leading to greater covalent character in CuO hence will deviate from LE more

29
Q

polarising power and charge density

what is polarising power

A

greater the charge density, greater the polarising power

polarising power of cation is ability to distort the anion

30
Q

solubility of simple molecules with same type of intermolecular forces

A

solute solvent interaction is similar to solute solute and solvent solvent interaction –> favourable

solvent solvent interaction is hydrogen bonding which is stronger than solute solvent interaction
hence energy released from forming id id interaction between solute and solvent molecules is not enough to compensate for the energy needed to break the hydrogen bonds in the solvent, water

31
Q

ionic solids dissolving in water

A

ions form strong ion-dipole interactions with polar solvents and this releases enough energy to compensate for breaking the strong ionic bonds

this interaction dependent on magnitude to dipole moment, charge density of ion and size of polar molecule

32
Q

sigma and pi bonds

A

sigma bond –> head to head overlap –> electron density concentrated along nuclear axis

pi bond –> side to side overlap –> electron density concentrated above and below nuclear axis, 0 electron density along nuclear axis

sigma bond stronger due to greater degree of orbital overlap

33
Q

Factors affecting covalent bond strength (and bond energy)

A
  1. no. of shared electrons
    - triple bond stronger than double bond stronger than single bond due to increased no. of shared electrons and increased EA between bond pairs and 2 nuclei
  2. effectiveness of orbital overlap
    - greater size of atom, decreased e- density, valence orbital used in bonding is more diffuse, orbital overlap is less effective
  3. diff in electronegativity
    - greater diff, more polar the covalent bond, stronger bond
34
Q

ionic bond with covalent character

and factors that affect it

A

attraction of the cation to the valence shell electrons in the anion causes the polarisation of the electron cloud in the anion, pulling it into the region between the 2 nuclei, leading to some form of orbital overlap

  1. polarising power of cation (greater charge density, power increases)
  2. polarisability of anion (greater e- cloud size, power increases)
35
Q

AlF3 and AlCl3

A

AlF3 is ionic and AlCl3 is covalent (simple molecular)

AlF3 –> electron cloud size of F- is so small it is hard to be distorted

AlCl3 –> electron cloud size is large so e- cloud can be distorted by Al3+ with high charge density such that electron sharing becomes predominant
–> difference in electronegativity between Al and Cl is small enough for it to exist as a covalent compound

36
Q

explain hybridisation + definition

A

on hybridisation, a set of atomic orbitals are mixed to generate a set of equivalent hybrid orbitals with the same shape and energy

  • to form a normal covalent bond, a orbital with a single electron is required (orbitals overlap to form the covalent bond)
  • Carbon in its ground state only has 2 singly occupied orbitals (can only form 2 covalent bonds)
  • Hence a 2s electron needs to be promoted to the empty 2p orbital in order to form 4 covalent bonds
37
Q

hybridisation in CH4

A
  • tetrahedral in shape, all H-C-H bonds have a bond angle of 109.5˚
  • carbon must provide 4 equivalent hybrid orbitals for head on overlap with the 1s orbitals of the 4 H atoms to form 4 identical C-H bonds
  • 4 equivalent hybrid orbitals for carbon are generated by mixing one 2s orbital with three 2p orbitals, resulting in the formation of four sp3 hybrid orbitals
38
Q

draw sp3 sp2 and sp hybridisation

A

check notes

39
Q

hybridisation in C2H4

A
  • trigonal planar in shape with all H-C-H bonds having a bond angle of 120˚
  • carbon needs to provide 3 equivalent hybrid orbitals for the head on overlap with the 1s orbitals of the 2 H atoms and another hybrid orbital of the other C atom to form 3 sigma bonds
  • the unhybridised 2pz orbital remains for the formation of the pi bond (side on overlap)
  • 3 hybrid sp2 orbitals are formed from the mixing of 1 2s orbital and 2 2p orbitals
40
Q

how is the unhybridised p orbital located

A

the unhybridised p orbital is perpendicular to the plane containing the sp2 hybridised orbitals/to the line containing the sp hybridised orbitals

41
Q

draw structure bonding thing of C2H4 (overlap of orbitals)

A

check notes

42
Q

hybridisation in C2H2

A
  • linear shape with the H-C-H bonds having a bond angle of 180˚
  • carbon needs to provide 2 equivalent hybrid orbitals for the head on overlap with the 1s orbital of the H atom and with another hybrid orbital of the other C atom to form 2 sigma bonds
  • the side on overlap of the 2 unhybridised 2p orbitals result in the formation of the 2 pi bonds (C- C triple bond total)
43
Q

draw structure bonding thing of C2H2 (overlap of orbitals)

A

check notes

44
Q

just read

A

for ethyne (C2H2)
1 C-C sigma bond from the sp-sp head on overlap
2 C-H sigma bonds from the sp-s head on overlap
2 C-C pi bonds from the 2p-2p side on overlap

45
Q

number of regions of electron density for sp3 sp2 and sp hybridisation + shape

A

4, 3, 2
tetrahedral, trigonal planar, linear

46
Q

effect of s character of the hybrid orbital on bond

A
  1. hybrid orbital is less diffuse
  2. orbital overlap is more effective
  3. bond strength increases, bond length decreases
  4. the more tightly the shared electrons are held by the nuclei

sp is more spherical and less diffuse compared to sp3 which is more elongated and more diffuse

also applicable for overlap between different kinds of orbitals (eg. sp-sp3 orbital overlap more effective than sp3-sp3 orbital overlap due to higher s character)

47
Q

what are resonance structures

A

structures that have the same placement of atoms but different arrangement of electrons

differ in placement of pi bonds and non bonding electrons while the placement of atoms and sigma bonds stay the same

48
Q

when does a molecule exhibit

A

when there is continuous side on overlap of p orbitals over at least 3 adjacent atoms, allowing for delocalisation of the pi electrons

49
Q

why is resonance hybrid more stable

A

because it delocalises electron density over a larger volume

50
Q

explain resonance in O3 + draw

A

check notes for drawing

  • resonance structures suggest that ozone is unsymmetrical with the O=O bond being shorter than the O–> O bond
  • actual structure is a hybrid of the resonance structures (resonance hybrid)
  • delocalisation of pi electrons over the 3 adjacent oxygen atoms via the continuous side on overlap of the p orbitals that are perpendicular to the plane of the molecule
51
Q

explain resonance in NO3- + draw

A

check notes for drawing

  • delocalisation of the pi electrons over the adjacent N and O atoms via the continuous side on overlap of the p orbitals that are perpendicular to the plane of the molecule
  • negative charge is delocalised over the 3 O atoms
52
Q

explain resonance in benzene + draw

A

check notes for drawing

  • delocalisation of the pi electrons above and below the plane of the benzene ring via the continuous side on overlap of the 6 unhybridised p orbitals that are perpendicular to the plane of the molecule
  • the delocalisation of the pi electrons results in stabilisation
53
Q

why silicon lower mp than diamond

A

valence orbital of silicon is larger so the Si-Si orbital overlap is less effective than the C-C orbital overlap hence Si-Si bond is weaker than C-C bond

54
Q

why graphite higher mp than diamond

A

in diamond, carbon atoms are sp3 hybridised. in graphite, carbon atoms are sp2 hybridised.

Due to higher s character in the hybrid orbitals of carbon in graphite, they are less diffuse and sp2-sp2 orbital overlap is more effective than sp3-sp3 orbital overlap hence the C-C bond in graphite is stronger

also, each carbon atom in graphite has an unhybridised p orbital containing one electron each. the p orbital of adjacent C atoms overlap side on, allowing for the delocalisation of p electrons above and below the plane of the carbon atoms in graphite. thus, there is additional electron density between the carbon atoms, holding the nuclei more closely, hence strengthening the C-C bond

Chem y5 (8 decks)

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