T1: Atomic Structure Flashcards
The Bohr Model
- p and n found in the centre of the atom -> nucleus.
- p and n also called nucleons.
- e- are held in shells / energy levels.
Subatomic Particles Relative Charge
p = +1, n = 0, e- = -1
Subatomic Particles Relative Mass
p = 1, n = 1, e- = 1/1840
Atomic number
Z, equal to the number of protons.
Mass number
A, equal to the number of protons and neutrons combined.
X
the symbol of the element
Cations
+ Atom lost electrons so there are more protons than electrons p > e-
Anions
- Atoms gained electrons so there are more electrons than protons p < e-
Ions definition
An atom equal of neutral charge (no charge) where the number of protons (Z) is equal to the number of electrons
Isotopes definition
Atoms with the same number of protons but a different number of neutrons.
Chemical properties of Isotopes
isotopes of the same element have the same chemical properties because they have the same electron configuration.
ER: He 2+ ions went through foil and arrived at point P
He concluded that most of the atom has empty space
ER: very small number of He 2+ ions detected at point Q
He concluded that the atom must have a small positive nucleus.
ionisation energy meaning
the amount of energy needed to remove a mole of electrons from a mole of atoms, in the gaseous state. units = kJmol-1
EXAMPLE 1st IE of K
K(g) -> K+(g) + e-
3 things that will influence IE
Nuclear Charge (No of protons in the nucleus), Distance from the nucleus, Shielding
Nuclear Charge (No of protons from the nucleus)
More protons = stronger attraction
Distance from the nucleus
Closer = stronger attraction
Shielding
e- in inner shell give outer e- slight repulsion
IE Boron graph outer shell
IE 1-3 increase -> each electron removed from a more positive ion each time
IE boron graph p and e- attraction
Attraction of p and e- increases -> less electrons being attracted by same number of protons
IE boron graph inner shell
4th e- removed from shell closer to the nucleus = more stronger attraction to nucleus
why 2nd IE is higher than the 1st
The 2nd e- is removed from an ion that already has a positive charge.
MA: successive IE for element identification
- The largest increase is between 4th and 5th IE
- the 5th e- is on the shell closer to the nucleus
- element must have 4 electrons on its outer shell
- in period 3 this must be silicon
Bigger radius
Lower ie
why Li is a bigger atom than Be
- Both atoms have the same number of shells
- they have the same shielding
- but Be has more protons
- so it attracts the outer most electrons more strongly so its smaller
why He has the highest 1st IE
it has more p than H and only 1 shell so the same shielding as H
why Li is a bigger atom than He
- Li has an extra e- shell and is further away from the nucleus
- the outer e- is more shielded
- the outer e- is less strongly attracted to the nucleus
Why is Li a bigger atom than F
Both atoms have the same number of shells
They have the same shielding
But F has more protons
So it attracts the outer most electrons more strongly
Why is Li+ a smaller ion than F-
Li+ ion has only 1 shell
The electrons are closer to the nucleus so there is less shielding
So outer electrons are more strongly attracted
Predicting trends in 1st IE across a period
The 1st IE will increase
There are more protons in the nucleus
The shielding remains the same
So the attraction between the nucleus and the outer electrons increases
No of e- in shell 1
2
No of e- in shell 2
8
No of e- in shell 3
2, 10, 8 = 18
No of e- in shell 4
2, 10, 8, 14 = 32
S orbital shape
Spherical shape
P orbital shape
Dumbbell shape
S subshell
1 s orbital
P subshell
3 p orbitals
D subshell
5 d orbitals
F subshell
7 f orbitals
Chromium electron configuration
1s2 2s2 2p6 3s2 3p6 4s1 3d5
Copper electron configuration
1s2 2s2 2p6 3s2 3p6 4s1 3d10
1st IE trend across a period
general trend increases
no. of protons increase (nuclear charge)
same amount of shielding
greater attraction between the nucleus and outer electron
1st dip in 1st IE (s2 and p1)
- 1st e- removed from Be is from 2s sb lv
- 1st e- removed from B is from 2p sb lv
- 2s sb lv is lower in energy than 2p
- therefore less energy is needed to remove the electron from B
2nd dip in 1st IE (p3 and p4)
- 1st e- removed from N is from 2p sb lv + unpaired
- 1st e- removed from O is from 2p sb lv + a paired orbital
- O has a lower IE due to electron pair repulsion
- therefore less energy is needed to remove the electron from O
how does IE change down the group
- atoms get bigger
- more shielding
- weaker attraction from nucleus to e- in the outer shell
- IE decreases down the group
trend in atomic radius down the group
atomic radius increases
no. of shells and amount of shielding increases
weaker attraction between nucleus and outer e-
trend in atomic radius across the period
atomic radius decreases
no. of protons increases but amount of shielding stays the same
greater attraction between e- and nucleus
what 2 factors does the mass spectrometer measure
relative abundance
mass/charge ratio (m/z)
mass spectrometer vacuum
entire machine in vacuum to prevent any particles being tested colliding with molecules from the air
mass spectrometer ionisation
2 methods
electrospray ionisation
electron impact ionisation
sample particles gain a positive charge in both
mass spectrometer acceleration
the positive ions are attracted to the negatively charged plate and accelerate towards it. amount of acceleration depends on m/z ration of the ion. high m/z ratio ions accelerate to lower speeds than low m/z ration ions once accelerated all ions will have the same kinetic energy
mass spectrometer ion drift
some ions will pass through a hole in the negatively charged plate. they form a beam of particles to travel along tube towards detector. particles travel at different speeds so drift apart as slow particles cant keep up with faster ones
mass spectrometer detection
different m/z ratio ions arrive at the detector at different times due to different velocities. TOF recorded as each ion hits detector it gains an electron. generates current the size of which is proportional to the number of each type of atom (abundance)
mass spectrometer data analysis
the signal from the detector is passed to a computer which generates a mass spectrum.
electron impact ionisation
- sample is vaporised
- high energy electrons are fired at it from an electron gun
- knocks off 1 electron from each particle forming 1+ ion
electron impact ionisation equation
X(g) -> X+(g) + e-
for electron impact ionisation the molecular ion…
breaks down into smaller fragments
electrospray ionisation
- sample dissolved in a volatile solvent
- injected through a fine hypodermic needle
- the needle is attached to the positive terminal of high voltage
- particles are ionised by gaining a proton H+ ion
electrospray ionisation equation
X + H+ -> XH+
for electrospray ionisation fragmentation…
rarely takes place
Ar =
(Mass1 x Abundance1) + (Mass2 x Abundance2) + (Mass n x Abundance n) / ∑ Abundances
KE =
(m x v^2) / 2
m =
2KE / v^2
v^2 =
2KE / m
v =
√v^2
t =
d / v
t is
time of flight (s)
d is
length of flight tube (m)
v is
velocity of the particle (m s^-1)
m is
mass of the particle (kg)
KE is
kinetic energy o particle (J)
mass of an atom
(mass number of element / 6.022 x 10^23) / 1000
