chem atomic structure

electron impact

• sample vaporised

• electron gun fires electrons at sample at high energy

• electrons knock off one electron from each particle in the sample forming positive ions

x(g) → x(g)+ + e-

electrospray ionisation

• sample is dissolved in a volatile solvent

• sample injected into a fine hypodermic needle to give a fine mist (aerosol)

• tip of the needle attached to the positive terminal of a high voltage energy supply

• particles ionised by gaining a proton (H+ ion) from the sample as they leave the needle

X(g) + H+ → XH+(g)

stage 2 acceleration

• + ions accelerated by using an electric field so that they all have the same KE

KE = ½ mv²

V = root 2KE/m

• lighter have higher vel

• heavier have lower vel

(cause they all travel at the same KE)

stage 3 flight tube

• + ions travel through a hole in the negatively charged plate into a tube

• time of flight of each particle depends on the velocity and in return it’s mass

t = d/v

t = d x root m/2KE

stage 4 detection

• + ions hit a negatively charged electric plate

• once they hit the detector plate the + ions are discharged by gaining electrons from the plate

• this generates a movement hence an electric current

• electric current is measured, size of current gives a measure of the number of ions hitting the plate

first ionisation energy definition

the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms of an element

factors effecting ionisation energy

• atomic radius

• number of protons/nuclear charge

• shielding

trend in first ionisation energies down a group

FIE decreases because

• atomic radius is increasing

• there is more shielding

• therefore weaker attraction between nucleus and outermost electron (even though nuclear charge increases)

first ionisation energies across a period

FIE increases because

• atomic radius decreases

• protons incease

• similar shielding

• therefore stronger attraction between nucleus and outermost electron

• harder to remove outermost electron

why is there a dip in FIE across a period at Al

Mg 1s² 2s² 2p^6 3s²

Al 1s² 2s² 2p^6 3s² 3p^1

• change in sub level (3s → 3p)

• when sub levels increase, energy of outermost electron increases because 3p orbitals are higher than 3s orbitals

• therefore outermost electron of aluminium is easier to remove than outermost electron of magnesium

why is there a dip in FIE at S

P 1s² 2s² 2p^6 3s² 3p³

S 1s² 2s² 2p^6 3s² 3p^4

• for sulfur the fourth outer electron has to enter an already filled orbital (3p^4)

• the result is electron pair repulsion which makes the outermost eldctron of sulfur slightly unstable, less energy required to remove the electron