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La Chimie H2 > Liaison chimique > Flashcards

Liaison chimique Flashcards

1
Q

Define metallic bonds

A

Strong electrostatic forces of attract n btw cations and sea of delocalised electrons in giant metallic lattice structure

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2
Q

Explain the strength of metallic bonds

A

Generally strong and large amt energy needed to break such bonds
Strength of metallic bond is directly proportional to:
- no of valence e- contributed per atom
- charge density (charge of cation divided by radius of cation) of metal cation

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3
Q

Define giant metallic lattice structures

A

Such structures consist of cations in a sea of delocalised electrons held tgt by strong electrostatic forces of attract n

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4
Q

Name and expound on properties of giant metallic substances

A
  1. High mp, bp
    bcos large amt energy required to overcome strong electrostatic forces of attract n btw cation and sea of delocalised e-
  2. Good electrical/thermal conductor
    bcos delocalised e- act as free mobile carriers to conduct electricity/heat
  3. Malleable (hammer to sheets), ductile (drawn into wires)
    Bcos when force is applied, layers of ions can easily slide over each other w/o breaking metallic bond.
    Metallic bonds easily reformed & lattice can b restored
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5
Q

Name a use of metals

A

alloy formation

  • metal tensile strength is increased due to presence of diff size of cations in metallic lattice inhibiting sliding
    eg bronze made of copper & zinc
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6
Q

Define Ionic Bonds

A

strong electrostatic forces of attract n btw oppositely charged ions in giant ionic lattice structure

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7
Q

When drawing dot-and-cross diagrams, what is and is not required?

A
  • Oni show valence e-
  • Legend not needed
  • alternate dot and cross for diff adjacent atoms/atom types
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8
Q

What is the relationship between magnitude of lattice energy and strength of ionic bond?

A

the greater magnitude lattice energy, stronger ionic bond within compound

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9
Q

magnitude of lattice energy is proportional to?

A

(q+ x q-)/(r+ + r-)

where,
+ is cation & - is anion
q is charge, r is radius

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10
Q

Strength of ionic bond is … to charge of ion and …ionic radius of ion?

A

directly proportional to charge of ion,
inversely proportional to ionic radius of ion

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11
Q

Define giant ionic lattice structures

A

such structures consist of oppositely charged ions held tgt by strong electrostatic forces of attract n

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12
Q

Explain physical properties of giant ionic lattice compounds

A
  1. High mp, bp
    bcos large amt energy needed to overcome strong electrostatic attract n btw oppositely charged ions
  2. Gd conductor of electricity in molten, aq state but non-conductor in solid state
    bcos in molten, aq state, presence of free mobile ions conduct electricity
    in solid state, ions can oni vibrate abt fixed positions, so no free mobile ions to conduct electricity
  3. Generally soluble in polar solvents eg water, insoluble in non-polar solvents
    eg
    - soluble in water means,
    energy released in forming strong ion-dipole interact n btw ions & water enough to overcome ionic bonding btw cations and anions in crystal lattice
    - insoluble in non-polar solvent means,
    due to absence of strong solute-solvent interact n, inadequate release of energy to overcome strong ionic bonding in crystal lattice
  4. Hard and brittle
    Hard:
    bcos, as ionic solid, oppositely charged ions held tgt by strong electrostatic forces of attract n
    Brittle:
    bcos stress applied on ionic lattice cause sliding of layers of ions. So, ions of similar charges come tgt and resultant repulsion shatters ionic structure
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13
Q

Name a use of ionic compound

A

used a refractories able to withstand high temp due to high mp and chemical inertness

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14
Q

Define covalent bond

A

electrostatic attract n btw shared pair of e- and +vely charged nuclei

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15
Q

What are the types of covalent bonds? Describe them

A

sigma and pi bonds

sigma bond
formed when 2 orbital overlap head-on. e- density of sigma bond is concentrated btw nuclei of bonding atoms

pi bonds
- formed when two p orbitals overlap side-on. e- density in pi bond is concentrated above & below internuclear axis
- pi bond takes place when atoms form multiple bonds (double or triple bond), where one of these bonds must be sigma, the rest are pi bonds

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16
Q

What are types of sigma bonds?

A
  1. s-s overlap
    eg H2
    head-on overlap btw 2 1s orbitals of 2 H atoms
  2. p-p overlap
    eg F2
    head-on overlap btw 2px orbitals of 2 F atoms
  3. s-p overlap
    eg HCl
    head-on overlap btw 1s orbital of H atom and 3px orbital of Cl atom
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17
Q

What are types of pi bonds?

A
  1. Double bond (1 pi, 1 sigma)
    eg O2 with p-p orbital side-on overlap between 2py orbitals
  2. Triple bond (2pi, 1 sigma)
    eg N2 with p-p orbital side-on overlap
    of 2py and 2pz orbitals of 2 N atoms respectively
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18
Q

How to count no. of sigma and pi bond?

A

All single covalent bond oni hv 1 sigma bond. Double bonds and triple bonds form pi bonds after 1 sigma bond formed

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19
Q

Strength of covalent bond directly proportional to …?
Smaller atoms have …?
Larger orbitals …?

A

extent of orbital overlap,
greater extent of overlap btw orbitals,
r more diffused -> lower accumulation of e- density

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20
Q

Stronger covalent bond, …. bond energy
so bond … during chemical rxn

A

larger,
less easily breaks

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21
Q

Bond energy depends on what factors?

A
  1. Bond order (triple bond>double bond>single bond)
  2. Bond length (shorter bond greater extent of orbital overlap)
  3. Bond polarity
22
Q

Define bond energy and bond length. Explain

A

Bond energy is energy needed to break 1mol bonds in gas phase to form gaseous atoms under standard conditions
- Multiple bonds stronger than single bonds
- more orbital overlap, more e- experience attract n of both nuclei, so stronger bond

(covalent) Bond length is distance btw nuclei of 2 bonded atoms in covalent bond
- directly proportional to size of atoms bonded
- inversely proportional to extent of overlap
- smaller atoms hv greater extent of overlap, so shorter bond length
- covalent bonds w shorter bond length are stronger

23
Q

Define dative bond and state its criteria and how to draw dot and cross diagram

A

Covalent bond in which shared pair of e- is provided by oni ONE of bonded atoms

Criteria:
- Donor atom must hv at least a lone pair of e-
- Acceptor atom must hv vacant & energetically accessible orbital to accept lone pair of e-

draw dot and cross as usual but w Arrow symbol -> used to denote dative bond, from donor to acceptor atom

24
Q

Name eg of dative bonds

A
  1. forming ammonium ion NH4 fr NH3
    N->H
  2. Dimerisation of AlCl3 to form Al2Cl6
    Cl->Al
    Al<-Cl
    *Al in AlCl3 is electron deficient as it oni hv 6 e- (2e- per Al-Cl bond)
  3. Forming ammonia-boron trifluoride compound
    H3N -> BF3
    B is e- deficient, accepts lone pair of e- from N in NH3 via dative bond to attain stable octet configuration
25
Q

Define electronegativity. What are the 3 most electronegative elements? How does electronegativity of elements change throughout the periodic table?

A

measure of tendency of atom to attract bonding pair of e-
F, O & N
electronegativity increases from left to right, bottom to top of periodic table

26
Q

How are compounds with high and low electronegativity difference (dEN) different

A

Low dEN tend hv orbital overlap form covalent bonds

High dEN tend result in transfer of e-, form ionic bonds

27
Q

What factors affect extent of covalent character in ionic bond? Explain

A
  1. Polarising power of cation
    smaller, more highly charged cation (higher charge density) -> stronger polarising power
  2. Polarisability of anion
    larger, more highly charged anion hv electron cloud that can b distorted more, so greater polarisability

If polarisation of large anion e- cloud by high charge density cation happens to enormous extent, results in distort n/ polarisat n of anion e- cloud. So, bonds hv covalent character

28
Q

Explain structure and physical properties of diamond

A

Strong, extensive covalent bonding btw atoms results in a giant 3-dimensional molecular structure

  1. Very high mp
    bcos enormous amt energy needed to overcome strong, extensive covalent bonding btw atoms in giant 3d molecular structure
  2. Non-conductor of electricity
    bcos no delocalised e- or free mobile ions to conduct electricity
  3. Hard
    thanks to strong, extensive covalent bonding btw atoms in giant 3d molecular structure
  4. Insoluble in both polar, non-polar solvents
    bcos of absence of strong solute-solvent forces to overcome strong and extensive covalent bonding btw atoms in giant 3d molecular structure
29
Q

Name a use of diamond

A

Used as abrasives due to high mp & hardness
eg diamond-tipped tools to cut, bore thru rocks
OR
jewellery

30
Q

Explain structure and physical properties of graphite

A

C atoms in hexagon flat parallel layers
C atom covalently bonded to 3 other atoms
* Adjacent layers held tgt by weak id-id attractions

  1. High mp
    bcos large amt energy needed to overcome strong, extensive covalent bonding btw atoms in giant 3d molecular structure
  2. Good electrical conductor of electricity parallel to layers but non-conductor perpendicular to layers
    bcos oni 3 of 4 valence e- of each C atom used for bonding. 4th non-bonding e- is delocalised over whole layer, thus graphite conducts electricity parallel to layers (graphite no conduct electricity perpendicular to layers as e- not mobile between adjacent layers)
  3. Soft, slippery
    as adjacent layers held tgt by weak id-id attract n, so layers can easily slide over each other
  4. Insoluble in polar and non-polar solvent
    bcos no solute-solvent forces strong enough to overcome the strong, extensive covalent bonding btw atoms in giant 3d molecular structure
31
Q

Name a use of graphite

A
  • used as lubricant eg graphite used in hot machines to reduce friction
  • used as pencil lead to write
32
Q

What are the steps to draw dot and cross diagrams?
What are guidelines to remember

A

Step 1: Determine central atom (least electronegative; written first in formula except when H is written first)

Step 2: Determine no of valence e- available for each atom present
for + charge species eg NH4+, remove e- fr least electro-ve atom
for - charge species eg NO3-, add e- to most electro-ve atom

Step 3: Link central atom to neighbour atoms w suitable no of covalent bonds via sharing of valence e- to achieve octet configurations

Step 4: Ensure uncharged molecules or polyatomic ions are correctly represented

Guidelines
- Try single bonds first, then double, triple and dative bonds to attain octet configu
* dative bond usually last resort (typically when central atom cnt hv > 8 valence e- eg NO2)
- neighbour atoms must hv 8 valence e-, but central atom can hv 8, <8 or >8 valence e- (depends on things like availability of vacant, energetically accessible d orbitals)
Use dots & crosses to represent valence e- of ALTERNATING atoms

33
Q

What are exceptions to octet rule in covalent bonding?

A
  1. Molecules w central atom of >8 valence e-
    - elements fr 3rd period onwards eg P, S, Cl can accommodate >8 e- ie can expand octet structure thanks to hving vacant, energetically accessible d orbitals to expand octet structure
    eg PBr5
  2. Molecules w central atom of <8 e-
    - Be, B and Al can b e- deficient ie can accommodate < 8 e-
    - period 2 elements (eg N and O) cnt accommodate more than 8 e-
    bcos no vacant, energetically accessible orbitals to expand octet structure
  3. Molecules w atom hving unpaired e-
    - such species named radicals, very reactive
34
Q

What are the principles of VSEPR theory?

A
  1. E- pairs (both bond pairs (bp) & lone pairs(lp) ) ard central atom arrange themselves as far as possible to minimise repulsion
  2. Strength of repulsion decrease in order:
    lp-lp repulsion> lp-bp repulsion > bp-bp repulsion
35
Q

What are steps to predict shapes of molecules?

A
  1. Draw dot and cross diagram of molecule
  2. Count total no of e- pairs ard central atom
    a. single, double, triple, dative bond counted as 1 e-pair
    b. lone pair oso counted as 1 e- pair
    *if a molecule hv >1 central atom, determine e-pairs w.r.t. each of these atoms
    eg H2O2 hv 2 O central atoms -> two bent structures
  3. Determine shape (using exact no of bp & lp ard central atom)
  4. Draw structure if needed
36
Q

List shapes of simple molecules (VSEPR theory) and bond angle

A

linear, 180

trigonal planar, 120
bent, ~119

tetrahedral, 109.5
trigonal pyramidal, 107
bent, 104.5

trigonal bipyramidal, 120 (on plane), 90 (on perpendicular axis)
distorted tetrahedral (see-saw), ~119 (on plane), <90 (on perpendicular axis)
T-shaped, <90 on perpendicular axis
linear, 180

octahedral, 90
square pyramidal, <90
square planar, 90

37
Q

What are the steps to explain shapes of molecules & bond angles?

A
  1. State no of bp & lp ard central atom
  2. To minimise repulsion & maximise stability (by VSEPR), shape is ______
  3. If bond angle required, identify repulsion
    - No lp -> equal bp-bp repulsion
    - 1 lp -> lp-bp repulsion > bp-bp repulsion
    -2 lp -> lp-lp repulsion>lp-bp repulsion>bp-bp repulsion
  4. Indicate bond angle:
    Bond angle is ______
38
Q

Define bond polarity

A

In covalent bond, bond polarity is measure of how equally e- r shared btw 2 bonded atoms. This depends on diff in e-vity btw atoms making up bond

39
Q

When is a molecule polar and non-polar?

A

Polar if hv net dipole moment (overall dipole moment ≠ 0) & non-polar if NO net dipole moment

40
Q

How does id-id compare to pd-pd imf of attract n?

A

For molecules of comparable e- cloud size (use Mr as guide) but differing polarity, pd-pd relatively stronger than id-id attract n

41
Q

What are 3 factors affecting extent of imf of attract n

A
  1. Size of e-cloud of molecule
  2. Type of imf of attract n (either id-id or pd-pd)
  3. Shape of molecule (typically applicable to organic molecules)
42
Q

How does size of e- cloud affect extent of imf of attract n?

A

greater size of e- cloud means:
- greater extent of distort n of e- cloud
- more extensive intermolecular forces of attract n (generally refer to id-id)

43
Q

How does shape of molecule affect imf of attract n?

A

Straight chain isomers of organic compounds hv higher bp than branched chain isomers as:
- straight chain hv greater surface area of interact n btw neighbouring molecule than branched chain
- greater extent of distort n of e- cloud

44
Q

Define hydrogen bond

A

Formed when hydrogen atom covalently bonded to N, O or F in one molecule is attracted to very electronegative atom (N,O or F) of another molecule w at least 1 lp of e-

45
Q

For molecules of similar size of e- cloud, how do the strengths of H bond, pd-pd and id-id attract n compare?

A

In general,
H bonding > pd-pd attract n > id-id attract n

46
Q

Why is apparent Mr of some carboxylic acids double that of actual Mr?

A

Carboxylic acids eg ethanoic acids exist as dimers due to intermolecular hydrogen bonding forming

47
Q

Describe structure of iodine

A

Within molecule, atoms bonded by strong intramolecular covalent bonds

Between molecules, separate molecules attracted by weak id-id attract n

In solid state, I2 molecules arranged in regular lattice structure held by imf of attract n

48
Q

Explain properties of iodine

A
  1. Low mp, bp
    bcos small amt energy needed to overcome weak id-id attract n
  2. Non-conductor of electricity
    Due to absence of delocalised e- or free mobile ions to conduct electricity
  3. Soluble in non-polar solvents eg benzene
    bcos energy released in forming weak id-id attract n btw iodine & benzene molecules enough to overcome weak id-id attract n btw benzene and iodine molecules respectively

Insoluble in polar solvents eg water
bcos energy released in forming weak id-id attract n btw water & iodine molecules inadequate to overcome weak id-id attract n btw iodine molecules and stronger hydrogen bonding btw water molecules

49
Q

Describe structure of ice

A
  • Intramolecular forces: strong covalent bonds hold atoms in H2O molecule
  • Intermolecular forces: strong H bonds exist btw separate H2O molecules
  • Presence of 2H atoms & 2 lp in each H2O molecule results in 3d tetrahedral structure of ice
  • Tetrahedral arrangement makes H2O not closely-packed, ensuing in open cage-like structure of ice. helps account for lower density of ice making it float on water. Prevents lakes from freezing in winter, so aquatic life can survive
50
Q

Explain properties of water and ice

A
  1. High bp
    bcos large amt of energy needed to overcome strong H bonding btw water molecules
    *otherwise water wld be gas, no oceans, lakes, rivers, rain
  2. Density of ice< density of water (at 0 C)
    When ice melts, tetrahedral arrangement is partially broken up as molecules move ard
    - molecules bcome closer
    - more water molecules per unit volume
    - density of water is higher
    *accounts for survival of marine life during winter as ice oni form on top. This ice layer insulates water below, prevent solidificat n

La Chimie H2 (25 decks)

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