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AP Chemistry > Unit 2- Molecular & Ionic Compounds Structures & Properties > Flashcards

Unit 2- Molecular & Ionic Compounds Structures & Properties Flashcards

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1
Q

Ionic bond

A

Metal transfer electrons to non-metals

Electrostatic bonds result due to attraction of opposite charges

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2
Q

Covalent bond

A

Non-metal atom shares one more more pair of electrons with another non-metal
Results in a full octet of electrons

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3
Q

Molecules

A

2 or more non-metals bonded together to form compound

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4
Q

Bonds in molecules are usually

A

nonpolar or polar covalent

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5
Q

Electronegativity

A

Element’s ability to attract bonding electrons in a bond

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6
Q

Electronegativity increases as atomic radius

A

Decreases

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7
Q

For two atoms in a chemical bond, the electrons are more attracted to the

A

positive nucleus closest to them

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8
Q

Coulomb’s Law

A 
F = k * (q1q2)/d^2
q1 = charge of ion
q2 = charge of ion
d = distance
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9
Q

Bonds can be classified according to differences in

A

electronegativity

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10
Q

Non-polar covalent bonds have EN differences between

A

0 and 0.5

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11
Q

Polar covalent bonds have EN differences between

A

0.5 and 1.9

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12
Q

Ionic bonds have EN differences between

A

1.9 and 3.5

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13
Q

All bonds have characteristics of

A

Other bond types

Classification is a continuum

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14
Q

Non-polar covalent bond example

A

F - F

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15
Q

Polar covalent bond example

A

H - F

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16
Q

Ionic bond example

A

Na+ - F-

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17
Q

In polar covalent bonds, the more electronegative atom has …

A

A partial negative charge

Higer probability of electrons

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18
Q

In polar covalent bonds, the less electronegative atom has …

A

A partial positive charge

Lower probability of electrons

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19
Q

Partial charges will increase as EN differences

A

increases

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20
Q

Sum of partial charge is equal to

A

overall charge

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21
Q

H - H
EN of H=2.1
What type of bond exist?

A

Non-polar covalent

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22
Q

H - C
EN of H=2.1
EN of C=2.5
What type of bond exist?

A

Non-polar covalent

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23
Q

H - H
EN of H=2.1
EN of Br=2.8
What type of bond exist?

A

Polar covalent

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24
Q

O - C - C
EN of O=3.5
EN of C=2.5
What type of bond exist?

A

Polar covalent b/t O and C

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25
Q

Li - F
EN of H=1.0
EN of C=4.0
What type of bond exist?

A

Ionic

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26
Q

Does type of bonds solely depend on EN?

A

NO - need to examine other properties of substance to determine whether the bond between atoms are ionic or covalent

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27
Q

Rule of thumb for ionic vs covalent

A

Ionic - between a metal & non-metal

Covalent - between two non-metals

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28
Q

Strength of ionic bonds

A

Very endothermic - HIGH levels of energy required to break bonds

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29
Q

Write the chemical equation of NaCl(s) breaking apart.

A

NaCl(s) -> Na+(g) + Cl-(g) Delta H-Lattice=+788 kJ/mol

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30
Q

Melting point of ionic substances are impacted by

A

Charge of ions & distance

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31
Q

Higher melting points are found in substances that have

A

Greater charges

Smaller distances

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32
Q

Determine which substance has the higher melting point

Li - F vs Li - I

A

Li - F Li - I
+1 -1 +1 -1 same charges
Li - F -> smaller distance
Therefore, Li - F has the higher melting point.

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33
Q

Determine which substance has the higher melting point

Mg - Cl2 vs Mg - O

A

Mg - Cl2 Mg - O
+2 -2 +2 -2 same charges
Mg - O -> smaller distance
Therefore, Mg - O has the higher melting point.

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34
Q

Determine which substance has the higher melting point

Na - F vs Mg - I2

A

Na - F Mg - I2
+1 -1 +2 -2 different charges
Therefore, Mg - I2 has the higher melting point.

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35
Q

Potential energy of valence electrons decreases as nucleus begin to

A

Approach each other

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36
Q

Bond energy

A

Energy released during formation of a bond

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37
Q

Same amount of energy must be added to

A

Break bond

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38
Q

Energy required to break bond is equal to

A

Energy required to form bond

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39
Q

If it takes 432kJ to break HCl, how much energy is needed to form HCl?

A

432kJ

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40
Q

As atomic radii increase of bonding atoms, bond energy

A

Decreases

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41
Q

Cations in an ionic bond is

A

metal positive ion

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42
Q

Anions in an ionic bond is

A

non-metal negative ion

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43
Q

What happens to EN as we move down a group?

A

EN decreases because successive element has one more shell - increased distance

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44
Q

What happens to EN as we move across a period?

A

EN increases as more protons are added to nucleus & valence electrons are in the same shell - greater force of attraction exerted by nucleus on electrons

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45
Q

As atomic radii of bonding atoms increase, bond length

A

Increases

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46
Q

Bonds can be though of as

A

Springs

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47
Q

At 0.074 nm, the bond length of H2, the energy is at its

A

lowest point

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48
Q

Explains what happens to these properties as atomic radii increase:
Bond length
PE
Bond energy

A

Bond length increases
PE increases
Bond energy decreases

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49
Q

Potential energy decreases as attractions between nuclei & valence elctrons

A

pull the atoms closer together

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50
Q

Which molecule has the highest BE? Lowest BE?
C - Cl
C - Br
C - I

A

Highest BE -> C-Cl

Lowest BE -> C-I

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51
Q

Which molecule has the longest bond length? Shortest bond length?
C - Cl
C - Br
C - I

A

Longest bond length -> C-Cl

Shortest bond length -> C-I

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52
Q

As the number of bonds between two atoms increases, what happens to the bond length, bond energy, and PE?

A

Bond length decreases
Bond energy increases
PE decreases

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53
Q

Why does bond length decreases and bond energy increases as number of bonds increases?

A

As electron density b/t positive nuclei increases, attractive forces b/t protons and bonding electrons increase

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54
Q

Bond order

A

number of bonds b/t two atoms

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55
Q

When bond order increases

A

PE decreases
Bond energy increases
Bond length decreases

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56
Q

Which bond type corresponds with which bond order?

A

Single -> 1
Double -> 2
Triple -> 3

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57
Q

Compare BE & PE in following molecules
C2H2
C2H4

A

C2H2
Lower PE, Higher BE
C2H4
Higher PE, Lower BE

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58
Q

Space-Filling Model

A

Shows differences in atomic radii of bonded atoms & relative bond length

Does not show 3D positions very well & number of bonds b/t atoms

59
Q

Ball-and-Stick Model

A

Shows 3D positions well & single, double, and triple bonds

Balls are not proportional to size of atoms
Sticks are nor proportional to bond length

60
Q

Properties of Ionic Solids

A

Strong bonds
Cleave along places
Soluble in polar solvents
Conduct electricity when molten/dissolves

61
Q

Why do ionic solids have strong bonds

A

Very strong Coulombic forces of attraction between cations & anions

62
Q

Strong bonds in ionic solids contributes to

A

High melting points
Low volatility
High hardness

63
Q

Why does ionic solids cleave along planes

A

Ions line up in repetitive pattern that maximizes attractive forces & minimizes repulsive forces. Once struck, it causes like charges to line up which then, are repulsed

64
Q

Cleaving along planes causes ionic solids to be

A

Brittle

Not malleable or ductile

65
Q

Why do ionic solids conduct electricity once molten or dissolved?

A

Charge particles of the solids are free to move

Higher concentration of ions in a solution -> Higher electrical conductivity

66
Q

Most ionic solids are soluble in

A

polar solvents

67
Q

Properties of covalent compounds

A

Lower melting & boiling points
Covalent solids are usually soft & flexible
Do not conduct electricity when dissolved in water

68
Q

Polyatomic ions

A

Combination of non-metals or metals and non-metals bonded together

69
Q

Bonds in polyatomic ions are

A

Either non-polar covalent or polar covalent

70
Q

Crystalline solids

A

Order that ionic substances take

Ions are arranged in an orderly fashion that follows a pattern of repetition in three dimensions

71
Q

Unit cells

A

Segments that repeat in 3D

72
Q

Crystalline solids usually have

A

Flat surfaces that make definite angles to one another

73
Q

Ions in an ionic solid are arranged in order to

A

Maximize Coulombic forces of attraction between cations and anions

Minimize repulsive forces between ions with like charges

74
Q

The way in which ions are arranges depends on:

A

– the relative size of the cations and anions, and

– the ratio of cations to anions.

75
Q

Possible arrangements of ions in an unit cell

A

1 ion/unit cell
2 ions/unit cell
4 ions/unit cell

76
Q

Metallic bonding

A

Attractions between nuclei and declocalized valence electrons moving throughout structure

77
Q

Bond strength in metallic solids increases as

A

number of bonding electrons increases

78
Q

Electron Sea Model

A

Nuclei and inner core electrons are localized while valence electrons are free to move throughout solid

79
Q

Characteristics of Metallic solids

A

Conduct electricity
Conduct heat
Malleable and Ductile
Lacking directional bonds

80
Q

Solution

A

Homogeneous mixture of two or more substances

81
Q

Solvent

A

Substance that is more plentiful in a solution

82
Q

Solute

A

Substance that is less plentiful in a solution

83
Q

Alloy

A

solid solution composed of two or more metals,

or one or more metals and one or more non-metals

84
Q

Interstitial Alloys

A

Atoms with a small radius occupies the spaces between atoms with a larger radius.

85
Q

Example of interstitial alloy

A

Steel -> carbon fills some spaces between iron atoms

86
Q

Substitutional Alloys

A

Radii of solute and solvent atoms are similar

Alloys remain malleable and ductile

87
Q

Example of substitutional alloy

A

Brass -> zinc atoms substituted for some copper atoms

88
Q

Properties of steel

A

Pure iron lacks directional bonds

Steel is MORE rigid, less malleable, and less
ductile than pure iron, as a result of the STRONG
directional bonds that form between carbon and
iron atoms.

The DENSITY of steel is GREATER than that of pure
iron, as interstitial atoms do not expand the lattice
by much

89
Q

Lewis Diagrams

A

Provide visual representation of location of atoms and relative distribution of electrons

90
Q

Steps to draw lewis structures

A

1 - Count total number of valence electrons in molecule
2 - Put least electronegative atom in center & connect terminal atoms to it with single bonds
3 - Complete octets for all terminal atoms except H
4 - Add up electrons used and subtract them from total number of valance electrons
Attach leftover as lone pairs

91
Q

If we complete the steps above and there is an atom without full atom, what do we do

A

5 - Make multiple bonds to complete the octet of central atom
Double or triple bonds may be needed

92
Q

Lewis Structure for polyatomic ion

A

need to take overall charge of ion into consideration

Parentheses around ion with charge written outside

93
Q

What are the exceptions to octet rule?

A

Incomplete octets

Expanded octets

94
Q

Incomplete octets

A

Occurs when an element has less ability to attract electrons to fill in octet

Exp -> Boron has only 5 protons and can’t attract electrons away from fluorine

95
Q

Expanded octets

A

Atoms in periods 3 through 7 can bond with
other atoms in such a way that they end up with
more than eight electrons in their octets.

96
Q

Formal charges

A

Calculated to identify most stable or likely structure to form

Neutral molecules -> sum of formal charges = 0

Polyatomic ion -> sum of formal charges = overall charge of structure

97
Q

Formal charges rules

A

1) The more likely Lewis Structure will have formal
charges that are closer to or equal to zero.

2) If there are negative formal charges, they should
reside on the more electronegative elements in
the structure.

98
Q

Calculating formal charge

A

Number of valence electrons assigned to neutral atom - number of electrons assigned to atom in structure

99
Q

Number of electrons assigned to atom in structure is equal to

A

of lone electrons around atom + # of bonding electrons / 2

100
Q

Resonance Structures

A

For many molecules, double or triple bonds are
located between different atoms.

This can results in two or more possible Lewis
structures that are equally valid.

101
Q

In resonance structures, the bonds being averaged are

A

the same length

102
Q

To calculate the effective number of bonds,

A

Divide number of bonds but the number of resonance structures

103
Q

Resonance structures with higher bond orders has

A

shorter bonds -> greater attraction

more bond energy

104
Q

Limitations of Lewis Structure Model

A

1 - Many bonds are actually partial (1.5 or 1.33)
2 - Octet rule fails when there are odd numbers of valence electrons
3 - Accepted Lewis structures for compounds of boron are not accurate
4 - Expanded octets also fail octet rule

105
Q

VSEPR Theory

A

• Charge clouds repel each other due to Charge clouds repel each other due to
Coulombic forces.
• Terminal atoms move as far away from one
another as possible.
• Results in distinctive geometric shapes

106
Q

To predict the geometric shapes of a molecule, you have to

A

Count the number of charge clouds, bonds, and lone pairs around the central atom

107
Q

What is considered a single charge cloud?

A

• One single bond (consisting of 2 electrons)
• One double bond (consisting of 4 electrons)
• One triple bond (consisting of 6 electrons)
• One lone pair (consisting of 2 electrons)
• One single unpaired electron (consisting of 1 lone
electron)

108
Q

What is considered a bond?

A
  • A single bond (consisting of 2 electrons)
  • A double bond (consisting of 4 electrons)
  • A triple bond (consisting of 6 electrons)
109
Q

What is considered a lone pair?

A

• One lone pair (consisting of 2 electrons)
• One single unpaired electron (consisting of 1
electron)

110
Q

Linear

A

Charge clouds - 2
Bonds - 2
Lone pair - 0
Bond angle - 180

111
Q

Trigonal Planar

A

Charge clouds - 3
Bonds - 3
Lone pair - 0
Bond angle - 120

112
Q

Bent

A

Charge clouds - 3
Bonds - 2
Lone pair - 1
Ideal Bond angle - 120 across from lone pair

113
Q

Tetrahedral

A

Charge clouds - 4
Bonds - 4
Lone pair - 0
Bond angle - 109.5

114
Q

Trigonal Pyramidal

A

Charge clouds - 4
Bonds - 3
Lone pair - 1
Bond angle - 109.5

115
Q

Bent

A 
Charge clouds - 4
Bonds - 2
Lone pair - 2
Ideal bond angle - 109.5
Actual bond angle - 104.5
116
Q

Trigonal Bypyrmidal

A

Charge clouds - 5
Bonds - 5
Lone pair - 0
Bond angle - 90 from side and 120 from top

117
Q

Seesaw

A

Charge clouds - 5
Bonds - 4
Lone pair - 1
Bond angle - 90 from side and 120 from top

118
Q

T-Shaped

A

Charge clouds - 5
Bonds - 3
Lone pair - 2
Bond angle - 90

119
Q

Linear

A

Charge clouds - 5
Bonds - 2
Lone pair - 3
Bond angle - 180

120
Q

Octahedral

A

Charge clouds - 6
Bonds - 6
Lone pair - 0
Bond angle - 90

121
Q

Square Pyramidal

A

Charge clouds - 6
Bonds - 5
Lone pair - 1
Bond angle - 90

122
Q

Square Planar

A

Charge clouds - 6
Bonds - 4
Lone pair - 2
Bond angle - 90

123
Q

How to predict shape in compounds with multiple central atoms

A

1 - look at each central atom on its own and everything it is bonded to is considered to be a terminal atom
2 - Count the charge clouds and bonds around it
3 - Predict the shape around it
4 - Isolate the next central atom and repeat steps 1 to 3

124
Q

Hybrid Orbitals

A

atomic orbitals around the central atom in a molecule must hybridize in order for bonding to occur.

125
Q

If there are 2 charge clouds around central atom, then the hybridization and ideal bond angle are

A

Hybridization - sp

Ideal bond angles - 180

126
Q

If there are 3 charge clouds around central atom, then the hybridization and ideal bond angle are

A

Hybridization - sp^2

Ideal bond angles - 120

127
Q

If there are 4 charge clouds around central atom, then the hybridization and ideal bond angle are

A

Hybridization - sp^3

Ideal bond angles - 109.5

128
Q

Double bonds consist of

A

one sigma bond

one pi bond

129
Q

Triple bonds consist of

A

one sigma bond

two pi bonds

130
Q

Single bond

A

sigma bonds are able to spin on an axis

131
Q

Double bond

A

pi bonds prevent sigma bonds from spinning on this axis

132
Q

Cis isomers

A

Carbon chain follows the same side

133
Q

Trans isomers

A

Carbon chain follows opposite sides

134
Q

Sigma and pi bonds involve

A

overlapping of atomic oribitals

135
Q

Sigma bonds contain

A

more bond energy so overlap between orbitals is stronger

136
Q

Extended pi bonding

A

as found in benzene or C6H6
• CH in every corner
• Alternating single and double bonds that can flip-flop
• Each p-orbital can overlap with two different p-orbitals
• Leads to delocalization of electrons
• Can be used to explain resonance in Lewis structures

137
Q

Shared electrons spend more time around

the

A

most electronegative element in the chemical bond.

138
Q

More electronegative element have a

A

slightly negative charge

139
Q

Less electronegative element have a

A

slightly positive charge

140
Q

Greater electronegativity differences lead to

A

greater partial charges and greater bond dipoles.

141
Q

If a molecule is polar, it must have a

A

dipole moment

142
Q

To know if a molecule is polar, you must know if

A

the bonds are polar and the overall shape

143
Q

Two diatomic molecules that contain atoms from

the same groups in the same proportions will have the

A

same shape and both be either polar or non-polar

144
Q

Replacing an element from one group with another

from the same group could lead to a

A

new substance

with similar properties.

AP Chemistry (9 decks)

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