2.., 12.. Flashcards

1
Q

what is the nucleus composed of

A

protons and neutrons

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2
Q

what occupies the space outside the nucleus

A

electrons

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3
Q

atomic number

A

number of protons
the bottom number on the element (Z)

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4
Q

mass number

A

number of protons and number of neutrons (A)
top number on element

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5
Q

what do atoms of the same element always have

A

same number of protons

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6
Q

ions

A

charged particles
same number of protons but different number of electrons

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7
Q

Isotopes

A

atoms of the same element with the same atomic number (protons) but different mass number (protons+neutrons)

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8
Q

Isotopes chemical and physical properties

A

same chemical properties - same number of electrons
different physical properties - bc mass number is different

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9
Q

radio isotopes

A

used in
- Nuclear medicine
o Diagnostics
o Treatment
o Research
* Tracers in biomedical and pharmaceutical research
* Geological and archaeological dating

e.g. C14

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10
Q

mass spectrometer

A

instrument used to determine the relative atomic mass of an element and it’s isotopic composition
mass spectrum is a plot of the relative abundance of each isotope versus the mass number
Ar=%abundance x Ar + %abundance x Ar

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11
Q

the further the energy level from the nucleus…

A

the higher its integer number (n) and the higher it

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12
Q

sub energy levels…

A

contain a fixed number of atomic orbitals, or regions of space where there is a high probability of finding an electron

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13
Q

how many electrons can each sub energy level hold

A

s - 2
p - 6
d - 10
f - 14

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14
Q

how many sub levels and names of sub levels do each principle energy level

A

n=1 - 1 - s
n=2 - 2 - s,p
n=3 - 3 - s,p,d
n=4 - 4 - s,p,d,f

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15
Q

shape of s atomic orbital

A
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16
Q

shape of p orbitals

A
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17
Q

principles of electronic configuration

A

Aufbau principle – electrons must always occupy the lowest energy levels first.
Pauli exclusion principle – two electrons cannot occupy the same orbital unless they have opposite spins, denoted by upwards and downwards arrows.
Hundu’s maximum multiplicity principle – when filling degenerate orbitals (orbitals of equal energy / orbitals of the same sub-level), the electrons fill the orbital singly first to avoid repulsion.

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18
Q

Electron configuration

A

show how the electrons are arranged in an atom

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19
Q

order of filling sub-levels

A

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p10 7s2 5f14 6d10 7p6

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20
Q

orbital diagram

A
21
Q

electron configuration of Copper

A

1s2 2s2 2p6 3s2 3p6 4s1 3d10-> half filled 4s1 is relatively stable; more stable than 3d9 and 4s2
.

22
Q

chromium electron configuration

A

1s2 2s2 2p6 3s2 3p6 4s1 3d5 -> half filled 3d5 and 4s1 are relatively more stable than 3d6

23
Q

electron configuration of ions

A

Ions are formed when atoms gain or loose electrons from the valence orbitals (outer energy level in the electronic configuration – energy level with highest n value).

24
Q

electromagnetic spectrum

A
25
Q

Wavelength (lambda, λ)

A

the distance between two crests in a wave (in metres, m).

26
Q

Frequency (f)

A

the number of waves that pass a point in one second (in hertz, Hz, or s-1).

27
Q

how to calculate frequency or wavelength

A

c = f x λ (DB.1)
Where c = speed of light (DB.2)

28
Q

shorter wavelenght

A

higher energy, higher frequency

29
Q

lower wavelength

A

lower energy, lower frequency

30
Q

Emission spectra

A

produced when photons are emitted from atoms as excited electrons return to a lower energy level
- when an electron is excites, it moves to a higher energy level for a short time
- when the electron falls back down to a lower energy level, it emits a photon (discrete amount of energy)
- this photon correspons to a particular wavelength depending on the energy difference between the two energy levels

31
Q

continuous spectrum

A

shows all wavelengths of visible light
VBGYOR

32
Q

line spectrum

A

shows discrete wavelenghts of visible light

33
Q

what electron ‘drop’ causes uv radiation

A

anything dropping to n=1

34
Q

what electron ‘drop’ causes visible light

A

anything dropping to n=2

35
Q

what electron ‘drop’ causes IR radiation

A

anything dropping to n=3

36
Q

energy levels diagram

A

energy levels converge at higher energies

37
Q

what is The Limits of Convergence

A
  • Where the lines appear to meet is called the limit of convergence
  • The convergence limit is the frequency at which the spectral lines converge
38
Q

ionisation energy and the emission spectrum

A
  • the energy require for an electron to escape the atom, or reach the upper limit of convergence, is the ionisation energy
  • the frequency of the radiation in the emission spectrum at the limit of convergence can be used to determine the first ionisation energy
39
Q

how to calculate first ionisation energy

A
  • When dealing with the Lyman series the largest transitions represent the fall from the infinite level to n=1
    In reverse, it can be considered to be equal to the ionisation energy (note that ionisation energy is given per mole of atoms)
  • Therefore, the first ionisation energy (IE1) of an atom can be calculated using the frequency (or wavelength) of the convergence limit
  • do this by using the following equations
    ΔE = h ν
    c = ν λ
  • first calculate frequency using given data
    c = ν λ
    as
    ν = c ÷ λ
  • Once we know the frequency, we can use this to calculate the ionisation energy
40
Q

The convergence limit for the hydrogen atom has a wavelength of 91.16 nm. Calculate the ionisation energy for hydrogen in kJ mol−1.

A

Step 1: Calculate the frequency of the convergence limit, converting wavelength into m (nm to m = × 10−9)

      c = ν λ

      ν = c ÷ λ

      ν = 3.00 × 108 ÷ 91.16 × 10−9

      ν = 3.29 × 1015 s−1

Step 2: Substitute into the equation to calculate IE1 for one atom of hydrogen in J mol−1

ΔE = h ν

     IE1 = 6.63 × 10−34 × 3.29 × 1015

     IE1 = 2.18 × 10-18 J atom−1

Step 3: Calculate IE1 for 1 mole of hydrogen atoms

      IE1 = 2.18 × 10−18  × 6.02 × 1023

      IE1 = 1 313 491 J mol−1

Step 4: Convert J mol−1 to kJ mol−1

      IE1 = 1313 kJ mol−1

So the first ionisation energy (IE1) of hydrogen has been calculated as 1313 kJ mol−1

41
Q

Ionisation energy trends across a period

A

grp 1 metals have low ionisation energy
noble gases have high ionisation energy
- nuclear charge increases
- atomic radius decreases, outer shell pulled closer to nucleus
- harder to remove electron as you move across a period

42
Q

what factors affect first ionisation energy

A

The size of the first ionisation energy is affected by four factors:
- Size of the nuclear charge
- Distance of outer electrons from the nucleus
- Shielding effect of inner electrons
- Spin-pair repulsion

43
Q

ionisation energy trends

A

First ionisation energy increases across a period and decreases down a group

44
Q

what are dips in the trend of first ionisation energy

A
  • slight decrease in IE between Beryllium and Boron as the fifth electron in boron is in the 2p subshell, which is further away from the nucleus than the 2s subshell of beryllium
  • slight decrease in IE between nitrogen and oxygen due to spin - pair repulsion in the 2px orbital of oxygen
45
Q

IE from one period to the next

A
  • large decrease in ionisation energy between the last element in one period and the first element in the next period
  • There is increased distance between the nucleus and the outer electrons as you have added a new shell
  • There is increased shielding by inner electrons because of the added shell
  • These two factors outweigh the increased nuclear charge
46
Q

what can successive ionisation data be used for

A
  • Predict or confirm the simple electronic configuration of elements
  • Confirm the number of electrons in the outer shell of an element
  • Deduce the group an element belongs to in the Periodic Table
47
Q

How to deduce group # of an atom from its ionisation energy

A
  • look for big jumps
  • big jumps = jump up in period - from grp 1 to grp 8
48
Q

calculating relative atomic mass using percentage abundance

A

first find the mass of 100 atoms by multiplying the percentage abundance by the mass of each isotopethen, divide by 100 to find the average atomic mass

49
Q

draw a mass spectrometer

A