Atomic Structure Flashcards
Describe the Bohr model:
- Protons and neutrons are found in the centre of the atom called the nucleus.
- Electrons orbit around the nucleus in shells or energy levels.
Mass of subatomic particles:
Protons: 1
Neutrons: 1
Electrons: 1/1840
Charge of subatomic particles:
Protons: 1+
Neutrons: 0
Electrons: 1-
Definition of atomic number:
Number of protons
Definition of mass number:
Total number of protons and neutrons
Definition of an Ion:
An atom that has lost or gained an electron so has a +/- charge
Cation: + ion
An atom has lost electrons so that there are more protons than electrons
Anion: - ion
An atom has gained electrons so there are more electrons than protons
Definition of an Isotope:
Isotopes are atoms of the same element with the same number of protons but a different number of neutrons.
Why do isotopes of the same element have similar chemical properties?
Isotopes of the same element have the same chemical properties because they have the same electron configuration
Evidence to support Bohr’s model:
Rutherford’s experiment
-Rutherford fired He2+ ions at a sheet of gold foil
-When the He2+ ions arrived at the back of the atom he concluded that most of the atom was empty space.
-A very small number of the He2+ ions where detected at the side of the atom
-He concluded that the atom must have a small positive nucleus.
Definition of Ionisation Energy:
Amount of energy needed to remove a mole of electrons from a mole of atoms, in the gaseous state.
Units: kJ mol-1
Eg: Write down the equation for the 3rd ionisation energy of Potassium
(2m)
K 2+ (g) → K 3+ (g) + e-
INCLUDE STATE SYMBOLS FOR 2ND MARK
Factors that will influence IE:
1) Nuclear Charge
-More protons
-Stronger attraction to nucleus
-More energy to remove outermost electron
Factors that will influence IE:
2) Distance from the nucleus
-Electron closer to nucleus
-Stronger attraction to nucleus
-More energy to remove outermost electron
Factors that will influence IE:
3) Shielding
-Electron on shell further from the nucleus
-More shielding
-Weaker attraction
-Less energy to remove outermost electron
Successive ionisation energies:
A single atom can be ionised to form a 1+ ion. This ion can then be ionised again to form a 2+ ion. This process can be repeated until all the electrons have been removed from the atom.
Describing IE levels:
Eg: Boron - Group 4
-IEs 1-3 increase because each electron is being removed from a more positive ion each time. (The attraction between the electrons and the protons becomes stronger as there are less electrons being attracted by the same number of protons)
-The 4th electron is removed from a shell closer to the nucleus and so is much more strongly attracted to the nucleus therefore a significantly higher amount of energy is required to remove the 4th electron
Why is the second ionisation energy of Boron higher than the first?
The second electron is removed from an ion that already has a positive charge
Eg: The following data shows the first seven successive ionisation energies of a period 3 element. State which element it is and explain your reasoning.
Data: IE
1st - 2947
2nd - 3683
3rd - 4837
4th - 5929
5th - 14839
6th - 15538
-the largest increase is between the 4th and 5th ionisation energies
-the 5th electron is on the shell closer to the nucleus
-The element must have 4 electrons on its outer shell
·-in period 3 this must be Silicon
Why is Li 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
Why is Li a bigger atom than He?
Li has an extra electron shell and is further away from the nucleus
The outer electron is more shielded
The outer electron 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-?
A Li+ ion only has one shell
Its electron are closer to the nucleus and there is less shielding
So the outer electrons are more strongly attracted
Which element has the highest IE?
Helium has the highest 1st IE of all elements because:
-It has more protons than Hydrogen and only one shell so has the same shielding as Hydrogen
General trend in 1st IE across a Period:
The 1st IE will increase
There are more protons in the nucleus (nuclear charge increases)
The shielding remains the same
Electron shells:
The 1st shell is closest to the nucleus, and each shell further is numbered 2nd, 3rd, 4th etc in succession
Within each shell there are sub-shells (s, p, d), and within each sub-shell we find orbitals.
Maximum number of electrons in each shell:
1st - 2
2nd - 8
3rd - 18
4th - 32
Definition of an orbital:
An orbital is a region within an atom that can hold up to two electrons with opposite spins.
S and P orbital shape
S - Spherical shape
P- Dumbbell shape
Highest energy electron
Outer most electron on an atom or ion is the highest energy electron because it is the furthest from the nucleus
Electron Configuration:
Eg - Xenon
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6
Electron Configuration meaning
First number represents shell/energy level of electron
Letter represents sub shell of electron
Last number refers to number of electrons in that sub shell
When electrons surround a nucleus, there are 3 rules for how to allocate electrons to orbitals:
1) Orbitals of lower energy are always filled first
2) Atomic orbitals of same energy fill single first before electrons pair up
3) No orbital can have more than 2 electrons
Rule for exception for electron configuration
d block elements are more stable when they have a full or exactly half full sub shell
Exception 1 for electron configuration: Chromium
Expected: 1s2 2s2 2p6 3s2 3p6 4s2 3d4
Correct: 1s2 2s2 2p6 3s2 3p6 4s1 3d5
Exception 2 for electron configuration: Copper
Expected: 1s2 2s2 2p6 3s2 3p6 4s2 3d9
Correct: 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Exception 3 for electron configuration: d block ions
When d block elements form positive ions, electrons are removable from the 4s sub shell first and not the 3d sub shell
Example: Fe(2+) ion
Iron: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Iron(2+): 1s2 2s2 2p6 3s2 3p6 3d6
(2 e- REMOVED FROM 4S SUB SHELL FIRST)
Transition metal ions: Electron configuration
4S
First in
First out
Trend in ionisation energy across a period
Eg: Period 2
1) General increase in IE across Period 2
2) this is because across Period 2 there is a greater nuclear charge (more protons)
3) and same amount of shielding
4) so a greater attraction between nucleus and outer electron
Period 2
Be + B configuration
1) first electron removed from Be is from a 2s sub level
2) first electron removed from B is from a 2p sub level
3) 2s sub level is lower in energy than 2p
4) therefore less energy is needed to remove the electron from B
Period 2
N + O configuration
1) first electron removed from N is from a 2p sub level and is unpaired
2) first electron removed from O is also from a 2p sub level but is from a paired orbital
3) this means O has a lower IE due to electron pair repulsion
4) therefore less energy is needed to remove electron from O
Trends in IE down a group
IE decreases down the group
Atomic radius increases as number of shells increases
More shielding
Weaker attraction between nucleus and outer electron
Why does atomic radius decrease from left to right across a period
Nuclear charge (proton number) increases
Greater attraction between electrons and nucleus
Amount of shielding stays the same
What is a mass spectrometer
A machine that can be used to analyse elements or compounds to accurately determine Ar of atoms or Mr of molecules
2 factors that mass spectrometer measure:
1) Relative abundance
2) Mass/charge ratio (m/z)
Time of flight mass spectrometer: Steps
Step 1: Ionisation
Step 2: Acceleration
Step 3: Ion Drift
Step 4: Detection
TOF mass spectrometer:
MP1) Vacuum
Entire machine is a vacuum inside to prevent any of the particles being tested colliding with molecules from the air
TOF mass spectrometer:
MP2) Ionisation
Two Methods:
a) Electron impact
b) electrospray ionisation
in both methods, sample particles gain a positive charge
TOF mass spectrometer:
MP3) Acceleration
Positive ions are attracted to negatively charged plate and accelerate towards it.
High m/z ratio ions will accelerate to lower speeds. Low m/z ratio ions will accelerate to higher speeds
Once accelerated, all ions will have same kinetic energy
TOF mass spectrometer:
MP4) Ion drift
Some ions will pass through a hole in negatively charged plate. They form a beam of particles and travel along flight tube towards detector.
Due to particles travelling at different speeds, ions drift apart as slower particles can’t keep up with faster ones
TOF mass spectrometer:
MP5) Detection
Different m/z ratio ions arrive at detector at different times due to different velocities
As each ion hits the detector, it gains an electron
This generates a current
Size of current is proportional to abundance of ion
TOF mass spectrometer:
MP6) Data analysis
Signal from detector is passed to a computer which generates a mass spectrum
Why are sample particles ionised
1) so they can be accelerated towards negatively charged plate
2) so they generate a current when they hit the detector
How is the ion accelerated
1) positive ions attracted to negatively charged plate
2) all ions have same kinetic energy
How are ions separated in flight tube
Ions travelling at higher speeds (small m/z ratio) move ahead of those traveling more slowly (large m/z ratio)
How are ions detected
1) each ion hits the detector
2) ion gains an electron
3) generates a current
4) size of current is proportional to abundance of ion
Ionisation method 1:
Electron impact
Sample is vaporised
High energy electrons are fired from electron gun at sample
An electron is knocked off forming 1+ ion
Ion is attracted towards negatively charged plate and accelerated
Equation:
X (g) → X+ (g) + e-
Ionisation method 2:
Electrospray ionisation
Sample is dissolved in a volatile solvent and injected through fine hypodermic needle to give a fine mist
Tip of needle is attached to positive terminal of a high voltage power supply
Particles gain a proton
Ions are attracted to negatively charged plate and are accelerated
Equation:
X (g) + H+→XH+ (g)
Equation for Ar:
Ar = (Mass 1 × Abundance 1) + (Mass 2 × Abundance 2) ÷ Sum of Abundances
Equation for Kinetic energy:
Kinetic Energy (J) = 1/2 × Mass (kg) × Velocity^2 (m/s2)
Rearranging KE equation to find:
a) mass
b) velocity^2
m = KE ÷ (1/2 × v^2)
v^2 = KE ÷ (1/2 × m)
Equation for time of flight
t = d ÷ v
t = d√(m ÷ 2KE)
t - time of flight (s)
d - length of flight tube (m)
v - velocity of particle (m/s-1)
m- mass of particle (kg)
KE - kinetic energy of particle (J)
Working out mass of an ion
mass ÷ (6.022 × 10^23)
mass units: gmol-1 / kgmol-1
1 gmol-1 = 0.001 kgmol-1
