chapter 4: bonding Flashcards

1
Q

how does the strength of a coordinate bond differ from a covalent bond?

A

the strength of a coordinate bond is the same as other bonds as once the coordinate bond is formed, it is the same as any other covalent bond

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2
Q

why is ionic bonding referred to as ‘non-directional’?

A

the electrostatic forces of attraction between the oppositely charged ions in the giant lattic structure are in all directions and not just between 2 atoms like in a covalent bond.

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3
Q

bonding of: propane, propanol, and propanoic acid

A

propane: alkane, is symmetrical, has LDF
propanol: can form H-H bonds
propanoic acid: carboxylic acids, can form H-H bods stronger than propanol

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4
Q

bond angle of O-Si-O in SiO2?

A

SiO2 has a tetrahedral giant structure, cannot be linear, bond angle is 109 degrees

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5
Q

benzene structures

A
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6
Q

how does a compound contain both ionic and covalent bonds?

A

non-metal bonded to a polyatomic ion

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7
Q

compounds that break the octet rule

A

SCl4, BCl3

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8
Q

characteristics of square planar

A

6ED, 2LP

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9
Q

a. explain why you would expect a BeF2 molecule to be linear but an SF2 molecule to be bent

b. why is the H-N-H angle in ammonia smaller than the H-N-H angle in the ammonium ion?

c. what is the bond angle and hybridisation in the C-C-C bond in benzene, C6H6?

d. what is the bond angle and hybridisation in the C-C-C bond in cyclohexane, C6H12?

A

a. BeF2 has 2 electron domains with no lone pairs, so it has a linear shape w bond angles of 180 degrees

SF2 has 4 electron domains w 2 lone pairs, so it has a bent shape w bond angles of about 105 degrees

b. ammonia molecule has a trigonal pyramidal shape due to its 4 electron domains and lone pair. the lone pair-bonding pair electron repulsion is stronger than the bonding pair-bonding pair electron repulsion and therefore pushes the bonding pair closer to each other. hence the H-N-H angle in ammonia is 107 degrees

ammonium ion has 4 electron domains, so the ion has a tetrahedral shape. the angle is 109.5 degrees

c. the carbon atoms have 3 electron domains and are sp2 hybridised, so the C-C-C bond angle in benzene is 120 degrees

d. the carbon atoms are sp3 hybridised and have 4 electron domains, so the C-C-C bond angle in cyclohexane is 109.5 degrees

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10
Q

explain why the 2 C-O bond lengths in propanoic acid, C2H5COOH, are different and yet the 2 C-O bond lengths in the propanoate anion, C2H5COO-, are the same length

A

in propanoic acid, the C-O bond lengths are different as one is a single covalent bond while the other is a double covalent bond

in propanoate ion, the C-O bond lengths are the same due to resonance resulting in equivalent C-O bonds where the negative charge is spread over the O-C-O bonds

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11
Q

why is the light required to break the O-O bond in oxygen of a higher frequency than the light required to break the O-O bond in ozone?

A

the O-O bond in oxygen is a double covalent bond whereas the O-O bond in ozone is between a double covalent bond and a single covalent bond due to its resonance. hence, O-O bond is stronger in oxygen than in ozone and requires more energy to break. higher frequency = higher energy, so higher frequency light required

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12
Q

why does penta 1,4-diene H2C=CH-CH2-CH=CH2 not contain any delocalised electrons whereas penta-1,3-diene, H2C=CH-CH=CH-CH3 does contain delocalised electrons?

A

the pi bonds in penta-1,4-diene are not next to a carbon atom with no free p orbitals and cannot be delocalised

the pi bonds in penta-1,3-diene are separated by a single covalent bond and can be delocalised

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13
Q

explain why at room temperature, F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid

A

halogen molecules are non-polar, so halogens have LDF
size of halogens increases down the group, so the number of electrons increases. as the number of electrons increases, the LDF increases, hence the melting and boiling points increases

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14
Q

butane (-1 degrees C), propanone (56 degrees C), and ethanol (78 degrees C) have similar molar masses but different boiling points, explain the difference in boiling points between the three molecules

A

they have similar molar masses, so they have similar LDF strength.

butane is non-polar, only has LDF, low esfoa which require a low amount of energy to overcome, so lowest boiling point

propanone is polar, has dipole-dipole forces, higher boiling point

ethanol is polar and has hydrogen bonding, high esfoa which require a large amount of energy, highest bp

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15
Q

why does boiling point change down the group?

A

down the group, the number of valence electron shells and protons increases, so the esfoa due to the van der waals’ forces increases. hence, a larger amount of energy is required to overcome the bonds held together by these forces. so the boiling point increases.

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16
Q

why is the boiling point of water higher than expected when H2S has a boiling point of -60 deg. C?

A

water is a polar bond with hydrogen bonding which is stronger than va der waals’ forces, so a large amount of energy is required to overcome the bonds.

17
Q

shape and electron distribution of water molecule
state and explain the value of the HOH bond angle
state the shape and explain using VSEPR theory of CO2 and H2S
state the H-N-H bond angle in an ammonia molecule and an ammonium ion
predict and explain the shape and bond angle of SCl2 and C2Cl2
bond angle in NCl3
state the shape and bond angle of H3O+ and C2H4
in HCOOH (methanoic acid), the bond angle around the carbon atom and the oxygen atom bonded to the hydrogen atom

A

shape of molecule: tetrahedral
shape of e distribution: bent

105 degrees, lone pairs repel each other more than bonding pairs

CO2 is linear, 2 bonds w no lone pairs, lone pairs repel each other more than bonding pairs
H2S is bent, 2 bonds and 2 lone pairs

ammonia molecule = 107 degrees
ammonium ion = 109.5 degrees

SCl2 has 2 bonding pairs and 2 lone pairs, bent shape, 105 degrees
C2Cl2 has 2 bonds around C, is linear, 180 degrees

NCl3 has a bond angle of 107.5

H3O+: trigonal pyramidal, 106 to 109.5 degrees
C2H4: trigonal pyramidal, 120 degrees

HCOOH:
carbon atom: 120 degrees
oxygen bonded to the hydrogen atom: 109.5 degrees

18
Q

explain why the bonds in silicon tetrachloride (SiCl4) are polar but the molecule is not

A

Cl has an electronegativity of 3.2, Si has an electronegativity of 1.9, the electronegativity difference between both elements is more than 0.5, so they have polar bonds.

However, SiCl4 is symmetrical, so there is no net dipole movement, so the molecules are not polar.

19
Q

explain the high volatility of iodine compared to sodium and sodium iodide

A

iodine has a simple covalent structure, so iodine has weak bonds, a low amount of energy is required to overcome the weak intermolecular foa btwn the I2 molecules (LDF) as compared to Na and NaI which has strong bonds so a large amount of energy is needed to overcome the bonds in Na and NaI, hence I2 is more volatile

20
Q

outline the principles of VSEPR theory

A
  • find number of e pairs in valence shell of central atom, electrons repel one another to positions of minimum repulsion for maximum stability
  • pairs forming a double or triple bond act as a single bond
  • non-bonding pairs repel more than bonding pairs
21
Q

explain why aluminium has a higher melting point than sodium

A
22
Q

what are molecules consisted of

A

simple covalent bonding and small difference in electronegativity calues

23
Q

describe the structure and bonding of diamond, graphite and fullerene

A

diamond:
- giant covalent structure
- carbon atoms arranged in a tetrahedral structure
- 1 C atom covalently bonded to 4 other C atoms

graphite:
- giant covalent structure
- hexagonal structure arranged in layers in which the separated layers are held together by LDF
- 1 carbon atom covalently bonded to 3 C atoms

fullerene:
- giant covalent structure
- 12 pentagons and 20 hexagons

24
Q

c

A

higher no. of bonds = increase in bond strength = decrease in bond length

25
Q

properties of metallic bonding

A
26
Q

number of bonds and the type of bonding in them

A
  • single bond = 1 sigma bond
  • double bond = 1 sigma 1 pi bond
  • triple bond = 2 sigma 1 pi bond