- Created by: sydjow17
- Created on: 14-12-19 14:11
- Sample dissolved in a volatile solution (e.g. methanol, water) and injected through a fine hypodermic needle as a fine spray into a vacuum in the ionisation chamber.
- High voltage applied to the end of the needle where the spray emerges (the needle is positively charged).
- The particles gain a proton and become ions as a fine mist.
- The solvent evaporates leaving 1+ ions.
- The detector is a negatively charged plate - a current is produced when the ions hit the plate; the more ions that hit the detector the bigger the current.
- The mass of the ion hitting the detector can be calculated from the time of flight.
- The mass spectrum shows the number of particles (abundance) of each mass that hit the detector.
- The horizontal axis is actually the mass to charge ratio (m/z) of the particles that hit the detector, but as the charge is usually +1, the m/z ratio is effectively the mass.
- In an electrospray ionisation mass spectrum, the main peak is usually at Mr+1 and so the Mr is 1 less than the molecular ion peak (e.g. if peak is at 505, the Mr is 504).
Rules for allocating electrons to atomic orbitals.
1. Atomic orbitals of lower energy are filled first (Aufbau principle).
2. Atomic orbitals of the same energy are filled singly before pairing starts (Hund's rule). This is because electrons repel each other.
3. No atomic orbital can hold more than 2 electrons.
Chromium and copper.
Chromium, Cr (24 electrons):
Copper, Cu (29 electrons):
This is a slightly lower energy arrangement as the reduced electron repulsion makes up for the fact that one electron is in a slightly higher energy level.
Removing electrons one by one.
- The first electron needs the least energy to remove it, because it is being removed from a neutral atom. This is the first IE.
- The second electron needs more energy to remove than the first, because it is being removed from a +1 ion. This is the second IE.
- The third electron needs even more energy to removem because it being removed from a +2 ion. This is the third IE.
And so on. These are called successive ionisation energies.
Trends in ionisation energies across a period.
Ionisation energies generally increase across a period, because the nuclear charge is increasing (due to more protons), making it more difficult to remove an electron.
- From magnesium to aluminium the ionisation energy decreases, because the outer electron in aluminium is in a 3p orbital, which is of a slightly higher energy than the 3s orbital, it therefore needs less energy to remove it.
- From phosphorus to sulfur the ionisation energy decreases. In phosphorus, each of the three 3p orbitals contains just one electron, while in sulfur, one of the 3p orbitals must contain two electrons. The repulsion between these paired electrons makes it easier to remove one of them.
Both these cases, which go against the expected trend, are evidence that confirms the existence of s- and p- sublevels.
Trends in ionisation energies down a group.
There is a general decrease in ionisation energy down a group, because the atomic radius increases so the outer electron gets further from the nucleus in each case. Also, there is increasing shielding, which is repulsion by electrons in the inner shells between the nucleus and outer electron. Both of these reduce the positive charge 'felt' by an electron in the outer shell from the nuclear charge.