Atomic structure electron configuration + ionisation energy Flashcards
which electrons are involved in chemical reactions
Only electrons in the outer shell are involved in chemical reactions
How to write shorot hand electron configuration of sodium
Look at position of sodium in periodic table - group 1 period 3
Look at the noble gas in the period before sodium - neon
Compare how electron configuration between the two elements
Inner shells of sodium have the same electron configuration as neon
[Ne] 3s1
Explain why for d block elements, we show the d subshell in the shorthand electron configuration
We show the d subshell in the shorthand electron configuration even though these electrons are not in the outer shell
This is because electrons in the d subshell can be involved in chemical reactions
Electron configuration of ions
Once the 4s subshell contains electrons, it now has a higher energy than the 3d subshell
Means that when d block elements lose electrons, the electrons are always lost from the 4s subshell before the 3s subshell
write the equation for when an electron is removed from a magnesium atom
Mg —> Mg+ + e -
To remove an electron from an atom requires energy
This is called the first ionisation energy
What is meant by first ionisation energy
The first ionisation energy is the energy needed to remove one mole of electrons from one mole of atoms in their gaseous state to form one mole of 1+ ions (also in their gaseous state)
write out the equation for the first ionisation energy of magnesium
Mg (g) —-> Mg(g) + + e-
M (g) ——> M+ (g) + e-
Once we’ve removed one electron we can continue to remove electrons and measure the ionisation energy each time
e.g. removing one electron from a 1+ ion - the energy required to do this is called the second ionisation energy
State the definition of the second ionisation energy
The second ionisation energy is the energy required to remove one mole of electrons from one mole of 1+ ions in their gaseous state to form one mole of 2+ Ions ( also in their gaseous state)
What is meant by successive ionisation energy
Once we’ve removed one electron we can continue to remove electrons and measure the ionisation energy each time
this is called successive ionisation energy
To remove an electron - to form an ion - what must occur
To remove an electron to form an ion, the attraction between it and the nucleus must be overcome
Energy must be put in
Therefore this is an endothermic process
electrons in an atom are attracted to the positive protons in the nucleus
The greater the attraction between the outer electrons and the nucleus, the greater the ionisation energy
IE depends on attraction
state the fifth ionisation energy of magnesium
Mg (g) 4+ —-> Mg 5+ (g) + e-
describe the factors that affect ionisation energy
The first factor is the distance between the nucleus and the outermost electrons
This is called the atomic radius
As the atomic radius increases, the force of attraction between the positive nucleus (contains positively charged protons) and the outer electrons decreases
So IE. Is decreased
The second factor is the charge on the nucleus
Remember that the electron are attracted to the positively charged protons in the nucleus
The greater the number of protons, the greater the force of attraction between the outer electrons and the nucleus
So I.E increases
The final factor is shielding
Electrons in the outer shell are repelled by electrons in inner shells
This shielding effect reduces the attractions between the outer electrons and the nucleus
So I.E. decreases
explain what successive ionisation energy tell us about how electrons are arranged in atoms
Successive ionisation energies of oxygen
There is a gradual increase in I.E. as we remove the first six electrons
This is because each time we remove an electron, the remaining electrons in the outer shell are pulled slightly closer to the nucleus
This means that there is a greater attraction between the outer electron and the nucleus
This causes the ionisation energy to gradually increase
There is a massive increase in ionisation energy when we remove the seventh electron
This is because the first six electrons are found in the second electron shell
But once we’ve removed those electrons, the seventh electron is removed from the first electron shell
Compared to the second electron shell, the first electron shell is closer to the nucleus and electrons in the first shell experience much less shielding
This means that the electrons in the first shell have a greater attraction to the nucleus compared to the electrons in the second shell
This explains why the ionisation energy is much greater for the 7th and 8th electron compared to the first six electrons
Describe and explain how IE varies down a group e.g. group 1
The first ionisation energy decreases as you go down the group
This is due to two factors
As you move down the group, the atomic radius increases
This means that the outer electron shell is further away from the nucleus
Secondly, going down the group the number of internal energy levels also increases
This means that there is more shielding between the nucleus and the outer electrons
Both of these factors mean that going down a group, the attraction between the nucleus and the outer electrons decreases
This causes the first ionisation energy to fall
The nuclear charge also increases as we go down the group
However the effect of this is offset by the two factors - atomic radius and shielding
Describe and explain how the first ionisation energy varies across a period
+ draw the graph
As we move across the period, first ionisation energy tends to increase
However there are two cases where first ionisation energy decreases Be to B and N to O (group 2 to 3 and 5 to 6)
Why it generally increases across a period
As we move across the period, the nuclear charge increases, as the number of protons increases
This increases the attraction between the nucleus and the electrons
Because of this, the atomic radius decreases across a period
Both the increased nuclear charge and the decreases atomic radius mean that the outer electrons are more attracted to the nucleus
And this causes the first ionisation energy to increase across the period
In all of these elements we are removing an electron from the same electron shell which in this case is the second electron shell
This means that the shielding effect due to the inner electron shell is similar for each element
Exceptions
For beryllium and lithium an electron is being removed from 2s subshell
However in boron, the outer electron is now in the 2p subshell
The 2p sub-shell has a higher energy than the 2s sub shell
This means that it takes less energy to remove the outer electron of boron compared to the outer electron of beryllium
Therefore has a lower first ionisation energy than beryllium
IE falls again at oxygen
In nitrogen, each electron is in a separate 2p orbital
However with oxygen, one of the orbitals contains a pair of electrons
These electrons repel each other
This means that it takes less energy to remove one of these electrons than if the electrons were in separate orbitals
So because of this the first ionisation energy of oxygen is less than that of nitrogen
Same with period 3 but referencing shell 3
Mass spectrometry used for
Used to find the abundance and mass of each isotope of an element and hence the relative atomic mass
type of mass spectrometry
Time of flight mass spectrometry
Stages of Time of Flight mass spectrometry
1)Ionisation
2) Acceleration
3) Flight tube
4) Ion drift
4) Detection
State the two methods of ionisation
Electron impact
Electrospray ionisation
State how electron impact works
An electron is knocked off each particle by the high energy electrons to form 1+ ions
This is done using an electron gun (hot wire filament) - which is a coil of metal which emits electrons at high energy
Harsh technique which uses high energy electrons
Fine for element but not for moleucles
Would cause molecules to split apart
General equation for electron impact
M(g) + e- –> M+ (g) + 2e-
an electron is used to hit an electron that is knocked off
so you end up with 2 electrons
Describe electrospray ionisation
Sample in a volatile solvent is inserted into a hypodermic needle attached to the positive terminal of high voltage power supply
Particles gain a proton (H+ ion from the acid in the solvent) as they leave the needle
A fine mist is formed as the MH+ ions repel one another
This technique is less harsh than electron impact
State the ionisation equation for electrospray ionisation
M + H+ –> MH+
Describe what happens in the acceleration stage
Only ionisaed molecules can interact with and be accelerated by the electric field
Positive ions are accelerated by a negative electric plate
K.E. = q x V (charge on ion and applied voltage of the electric field)
q and V are the same for all ions regardless of their mass
So all ions have the SAME K.E.
Describe what happens in the flight tube stage
All of the different mass ions set off along the flight tube at the same time
Lighter ions travel fastter and start to separate out
The fastest ions reach the detector first
in ToF spectrometer - this is also called the drift region
Drift region is under vacuum and is free of any electric field
Ions separate according to how long it takes them to travel across the drift region, that is their time of flight
length of flight tube = distance travelled by ALL the ions
v = d/t
E.K. = 1/2 m v^2
Describe what happens in the detection stage
When the ions reach the ion detector (a negatively charged electric plate) they accept an electron and generate an electric current (non-ionised molecules will not do this)
The size of the current is proportional to the number (abundance) of ions hitting the detector plate
A computer interprets this electric signal to produce a mass spectrum ( a plot of m/z - mass/charge against relative abundance)
The time taken to move down the drift chamber is used by the machine to determine the mass of the isotope
Size of the current produced when each isotope hits the detector is used to determine the abundance of each isotope
