Atomic Structure Flashcards
4 stages of mass spectrometry
Electrospray ionisation
Acceleration
Ion drift
Detection
Describe the electrospray ionisation process in mass spectrometry
The sample is dissolved in a polar solvent and pushed through a small nozzle at a high pressure. A high voltages is applied to the sample, causing the particles to lose an electron and become positively charged. The ionised particles are separated from the solvent, leaving a gas made up of positive ions
Describe the acceleration process that occurs in mass spectrometry
The positive ions are accelerated by an electric field. The electrics field gives the same kinetic energy to all the ions
Why must the particles in the acceleration stage of mass spectrometry be positive?
They need to be positively charged in order to be accelerated by the electric field
What happens to ions with a lower mass/charge ratio in the acceleration stage of mass spectrometry?
They experience a greater acceleration because of the fact that they receive as much energy as heavier ions
Describe the ion drift process of mass spectrometry
The ions leave the electric field with a constant speed and constant kinetic energy. They enter a region with no electric field and drift through it at the same speed as they left the electric field
Describe the detection process of mass spectrometry
When the ions hit the detector, a current is created and the detector records how long they took to pass through the spectrometer. This data is then used to calculate the mass/charge values needed to produce a mass spectrum
Calculating relative atomic mass from a mass spectrum
Multiply the relative isotopic abundance by the relative atom mass
Add the totals
Divide answer by the total relative abundance (100 if % was used)
electron configuration order
1s (x2) 2s (x2) 2p (x6) 3s (x2) 3p (x6) 4s (x2) 3d (x10) 4p (x6) 5s (x2) 4f (x14)
Why does the 4s sub shell fill up before the 3d sub shell?
It has a lower energy level, despite the fact that its principle quantum number is bigger
Explain noble gas shorthand notation
Put the noble gas before the element in square brackets then write out the remaining electrons using electron configuration
Give the electron configuration of silicon
atomic number - 14
1s2, 2s2, 2p6, 3s2, 3s2
Give the electron configuration of potassium
atomic number - 19
1s2, 2s2, 2p6, 3s2, 3p6, 4s1
Give the noble gas shorthand notation of calcium
atomic number - 20
[Ar] 4s2
Give the noble gas shorthand notation of iodine
atomic number - 53
[Kr] 5s2, 4d10, 5p5
Give the electron configuration of chromium
atomic number - 24
1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5
Give the electron configuration of copper
atomic number - 29
1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10
Why are the electron configurations of copper and chromium different from what is expected?
They each donate 1 electron from the 4s sub shell to the 3d sub shell because it makes them more stable
Which groups are in the s block of the periodic table?
Groups 1 and 2
What bonds are formed by the s block elements and why?
Ionic bonds because the atoms are in groups 1 and 2 so easily lose their outer electrons (which are found in the s sub shell)
What 3 things affect ionisation energy?
Nuclear charge
Distance from nucleus
Shielding
How does nuclear charge affect ionisation energy?
The more protons there are in the nucleus, the more positively charged the nucleus is and the stronger the attraction for the electrons
How does nuclear distance affect ionisation energy?
Attractions falls off very rapidly with distance. An electron close to the nucleus will be much more strongly attracted than one further away
How does shielding affect ionisation energy?
As the number of electrons between the outer electrons and the nucleus increases, the outer electrons feel less attraction to the nucleus. This lessening of the pull of the nucleus thanks to the inner electron shells is called shielding
General equation for ionisation energies
X(n-1)+ —> Xn+ + e-
Describe the ionisation energy trend down group 2
The first ionisation energy down group 2 decreases
Explain the ionisation energy trend down group 2
As you go down group 2, the ionisation energy decreases. This is because shielding increases so the attraction between the positive nucleus and outer negative electron decreases; so the outer electron is lost easier. Due the increased shielding, the distance between the outer electron and the nucleus also increases. This also decreases the attraction forces between the nucleus and outer electron so it is lost easier
Describe the general trend of ionisation energies
Across a period, the general trend of ionisation energies increases
Explain the general trend of ionisation energies across a period
Across the period, the number of protons increase. This increases the nuclear charge which increases the attraction forces between the positive nucleus and negative outer electron; so it is harder to lose the outer electron
Explain the drop in ionisation energies between magnesium and aluminium, across period 3
Aluminium has a lower ionisation energy that magnesium. This is because the outer electron of aluminium is in the 3p orbital whereas the outer electron of magnesium is in the 3s orbital. The 3p orbital is further away from the nucleus than the 3s orbital. This means the attraction forces between aluminium’s outermost electron is weaker than that of magnesium’s so it is lost more easily. Shielding is also increased by the fact that aluminium’s outer electron is in the 3p orbital. This too lowers the attraction forces between the nucleus and the outer electron, making it easier to lose.
How does repulsion affect the first ionisation energy?
Two electrons in the same orbital repel one another. This offsets the attraction of the nucleus, causing the electrons to be lost more easily
Explain the drop in ionisation energies between phosphorus and sulfur, across period 3
Sulfur has a lower first ionisation energy than phosphorus. This is because the electron being removed from sulfur occupies an orbital containing two electrons. These electrons repel one another which offsets the attraction of the nucleus and makes it easier to lose the electrons. The electron being removed from phosphorus is in a single occupied orbital so doesn’t experience this repulsion
What is Avogadros constant?
6.02x10^23
Number of particles
Number of moles x avogadros constant
Ideal gas equation
pV=nRT
Units of all parts of ideal has equation
P = Pa V = m3 R = JK-1mol-1 T = K n = miles
The value of gas constant
8.31 JK-1mol-1
1 dm3 to m3
1x10-3
1 cm3 to m3
1x10-6
