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From Flashcardlet > AP Chem Ch 7-9 > Flashcards

AP Chem Ch 7-9 Flashcards

0
Q

Principle QN (n)

A

Energy level

Integer from 1-7

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1
Q

Quantum numbers

A

Four numbers that describe an electron

  1. Principle QN (n)
  2. Angular momentum QN (l)
  3. Magnetic QN (ml)
  4. Spin QN (ms)
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2
Q

Angular momentum QN (l)

A 
From 0 to n-1
When l = 0, s orgbital
1 --> p
2--> d 
3 --> f
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3
Q

Magnetic QN (ml)

A

From -l to l
So when l = 0, one possibility for S orbital
When l = 1, 3 possibilities in p orbital

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4
Q

Spin QN

A

+/- 1/2

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5
Q

Pauli exclusion principle

A

No two electrons can be in the same spot

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6
Q

Aufbau principle

A

Build up – put electrons into lowest energy orbital available

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7
Q

Hund’s Rule

A

Lowest Energy confirmation generates unpaired electron – result of energy required to pair the electron in the same orbitals.
This means first do it unpaired and then pair them up

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8
Q

Auger emission

A

When a core electron is ejected, a higher energy may relax into that electron hole, resulting in an emission of energy

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9
Q

Exceptions to normal electron configuration

A

Chromium and molybendium fill up d orbital half way and have only one S electron
Copper, silver, gold all fill up the d orbital and have one in the S
Pd has no s electrons, 10 d

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10
Q

Homogenous magnetic field

A

Orient dipoles– spins align. Observed splitting of the 5s energy level with the S going to 2 different energy states. Pauli connected this to the spin in the electrons with the +1/2 and -1/2, the 2 intrinsic states of electron.

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11
Q

Periodic trend key factors

A

Number of protons, number of electrons, effective nuclear charge, down the group, across the period, shielding, distance from the nucleus

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12
Q

Electron affinity

A

The amount of energy to add an electron
Energy associated with this rxn:
X + e- –> X-

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13
Q

Atomic radius / size periodic trend

A

Increases down and to the left. Francium the biggest, helium the smallest

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14
Q

Ionization energy

A

Amount of energy required to remove an electron
M–> M+ + e-
Increases top right
Helium has the most because very hard to remove an electron
Francium the least, easier.
Increases across the period mostly, but some exceptions as move from a metal to a non metal

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15
Q

Electronegativity

A

How much an element wants an electron
Measured in Paulings, arbitrary unit for how much an element wants the electron
Increase top right. Fluorine has the most, a 4.0

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16
Q

Effective nuclear charge

A

Number of protons - inner electrons
Aluminum is +3
Chrlorine +7

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17
Q

Why does IE increase left to right?

A

Effective nuclear charge increases because number of protons increases, so while shielding stays the same because same energy level, it is harder to remove an electron against this extra positive charge from the nucleus while the atom is smaller as well.
Decreases as go down a row because it is easier to remove an electron from a bigger atom (shielding effect in effect)

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18
Q

Why does oxygen have lower IE than nitrogen?

A

Nitrogen has the p orbital half way full, so it is more stable, so it thus harder to remove that electron than from oxygen

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19
Q

Ionic bonding

A

Involves the transfer of electrons and is the positive / attractive force that results from the charged species

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20
Q

Electrostatic interactions

A

Interactions of charged particles –> forces keeping electrons around electrons –> force that makes ionic bonds ionic.

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21
Q

Anion

A

Negative ions. Non metal

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22
Q

Cation

A

Positive ions. Metal

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23
Q

Bond energy

A

Energy required to break a bond

E= (2.3110^-19 Jnm) q1 q2 / d where d is the distance between 2 nuclei

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24
Q

Bond energy of NaCl if the bond length is 2.76 A*

A 
A* = A with the degrees symbol = 10^-10 m
E = 2.31*10^-19 (1)(-1)/.276 nm
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25
Q

Covalent bonding

A

Share electrons. More like they’re fighting over the electrons. Nonemtals and hydrogen
As you drag a hydrogen atom closer to another, lowest energy is at the bond length

26
Q

How to get delta H overall of Li (s) + 1/2 F2 (g) –> LiF(s)

A

Need to do a bunch of half reactions and add up the delta H’s
Li (s) –> Li (g) delta H sub = 161 kJ/mol
Li (g) –> Li+ (g) + e- delta H IE = 520
1/2 F2 (g) –> F (g) delta H = 154
F(g) + e- –> F- (g) delta H EA = -328
And lastly need a lattice energy–>
Li+ + F- –> LiF (s) delta H lattice = -1047
Add up all the delta H’s and done
Lattice energy dependent on size and magnitude of charge

27
Q

Polar

A

Uneven distribution of electrons

Results in a dipole with a partial positive and a partial negative end

28
Q

Non polar covalent bonds

A

When electronegativity is similar or the same

H-C or O-O or any diatomic bond is not polar

29
Q

Electron domains

A

Areas of electron density

Bonds or lone pairs around a central atom

30
Q

How many domains does a double bond count as

A

ONE!

31
Q

Formal charge

A

of valence electrons - bonding electrons / 2 - # of electrons in lone pair

32
Q

Example of formal charge of an oxygen with a double bond and 2 lone pairs vs an oxygen with a single bond and 3 lone pairs

A 
With double bond -->
6 - 4/2 -4 = 0
Single bond -->
6 -2/2 -6 = -1
0 is more favorable
33
Q

Resonance

A

Result of having more than one valid Lewis dog sfrucutre in which only difference is the placement of electrons

34
Q

Chemical bonds

A

Force that causes atoms to behave as a unit

35
Q

Sigma vs pi bond

A

Sigma bond is much stronger than pi because sigma has more orbitals overlapping.
Triple bond shortest and most strong because has sigma plus additional pi

36
Q

Another way for delta H rxn

A

Sum of the energy required to break the bonds (left side) - sum of the energy released from bonds formed (right side)

37
Q

Method for drawing Lewis dot structures

A
  1. Count valence electrons
  2. Re count valence electrons
  3. Arrange the atoms so that the central atom is the least electronegative
  4. Use the pairs of electrons to form single bonds between the atoms, the skeleton structure
  5. Spread remaining electrons on non central atoms trying to form octet
  6. Put extra electrons on central atom
  7. Check if central atom has octet
  8. Form double or triple bonds to give central atom an octet
    Note that don’t need octet. Atoms at P and larger can have 10-12 electrons and be stable
38
Q

Draw some Lewis dot structures!

A

Ok!

39
Q

Equivalent resonance structure

A

When there are the same number of single, double, and triple bonds. The bonds are meeely rearranged. Weighted equally in trying to predict actual structure

40
Q

Non equivalent resonance structures

A

Contain different numbers of single, double, and triple bonds (for example, four single bonds become one double and two single bonds). When these happen, it is possible one is more likely than the other. Determine which is more likely by looking at the formal charge. Want the one with the least variability (0,0,0 vs 2,-4,2)

41
Q

Linear

A

With 2 electron domains and 2 bonds
Like CO2
Sp

42
Q

Trigonal planar

A

3 electron domains and 3 bonds
BF3
No lone pairs
Sp2

43
Q

Bent

A

3 electron domains and 2 bonds. For example O3. Has a lone pair sp2
Also could be 4 electron domains and 2 bonds (2 lone pairs), such as H2O sp3

44
Q

Tetrahedral

A

4 electron domains and 4 bonds (no lone pairs)

CH4. 109.5 degrees sp3

45
Q

Trigonal pyramidal

A

4 electron domains. 3 bonds (1 lone pair)
NH3
Sp3

46
Q

Trigonal bipyramidal

A

5 electron domains, 5 bonds (no lone pairs)
PCl5
Sp3d

47
Q

See-saw

A

5 electron domains but 4 bonds (1 lone pair)
SF4
Sp3d

48
Q

T-shaped

A

5 electron domains and 3 bonds (2 lone pairs)
ClF3
Sp3d

49
Q

Linear with 5 electron domains

A

XeF2
2 bonds and 5 domains
Sp3d

50
Q

Octahedral

A

SF6
6 electron domains and 6 bonds
Sp3d2

51
Q

Square pyramidal

A

6 electron domains and 5 bonds

BrF5

52
Q

Square planar

A

6 electron domains and 4 bonds

XeF4

53
Q

Hybridization

A

Orbitals formed between the current orbitals to allow for equal energy bonds

54
Q

Molecular orbital theory

A

Sigma star where unpaired electrons go and sigma where bonding electrons go
If there are more or the same electrons in the sigma star than regular sigma then there isn’t a bond

55
Q

Why is the first IE of sodium smaller than magnesium but the second IE of sodium is larger than the second IE of magnesium?

A
  1. First IE of Na is smaller because sodium is bigger than magnesium and it has one less proton, so less of a positive pull from the center, so easier to remove an electron
  2. Sodiums second electron is in the next energy level, which is even closer to the nucleus as we lose an extra orbital, so it is really hard to remove this electron
56
Q

Bond order

A

(Number of bonding electrons - number of anti bonding electrons)/2

57
Q

What does bond order signify?

A

Number of bonds likely to occur. In He2, the BO=0, so He2 won’t happen.
Larger bond order means stronger interaction

58
Q

Paramagnetism

A

Attracted to an external magnetic field

UNPAIRED ELECTRON

59
Q

Diamagnetism

A

Repelled by an external magnetic field

All electrons are paired

60
Q

Relation between bond order, energy, and length

A

Increase bond order, increase bond energy, decrease bond length

61
Q

Nitrogen triple bond

A

Materials with high numbers of N atoms are explosive because their decomposition results in the formation of N2, a very stable product, thus releasing large quantities of energy

62
Q

What does HF molecular orbital diagram show us?

A

The molecular sigma orbital is closer in energy to the Florine 2p orbital, indicating that the electron will spend more time on F, supporting our theory of electronegativity and the more electronegative atom getting the electron more

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