3.1.1 - ATOMIC STRUCTURE Flashcards
state the definition of an orbital
- a region of space that an electron is likely to be found
how many orbitals (therefore how many electrons can fit) in each sub level?
- s has 1 orbital and can fit 2 electrons
- p has 3 orbitals and can fit 6 electrons
- d has 5 orbitals and can fit 10 electrons
- f has 7 orbitals and can fit 14 electrons
state electronic configuration for Copper, Cu
[Ar] 4s1 3d10
what is time of flight mass spectrometry used for?
- abundance and mass of each isotope in an element sample
- allows for its relative atomic mass to be found (r.a.m or Ar)
state the stages of TOF mass spectrometry
1 - ionization
2 - acceleration
3 - flight tube
4 - detection
state the 2 forms of ionization and what type of substances they are used for in TOF mass spec
- electron impact ionistion - substances with a low formula mass, e.g. small molecules and individual atoms
- electrospray ionisation- substances with a higher molecular mass, e.g. proteins
describe electron impact ionisation
- sample is vapourised
- high energy electrons fired at it (from electron gun)
- these knock of an electron from each particle, forming a +1 ion.
Ex) X(g) + 2e —> X+(g) + 2e
describe electrospray ionistion
- sample dissolved in volatile solvent
- injected through a fine hypodermic needle to give a fine mist/aerosol
- needle attached to positive terminal of power supply
- particles gain a proton (H+ ion) from the solvent as they leave the needle
Ex) X(g) + H+ —> XH+(g)
describe the acceleration stage of TOF mass spec
- what happens with kinetic energy
- what determines velocity of a particle
- positive ions are accelerated using an electric field
- ensures all particles have the same kinetic energy
- lighter particles have faster velocities
- heavier particles have slower velocities
- describe the flight tube stage in TOF mass spec
- what type of particles reach detector first
- positive ions travel through hole in the negatively charged plate into the flight tube
- the lightest ions reach the detector first, as they have the fastest velocities
describe the ionisation stage in TOF mass spec
- particles either lose an electron or gain a proton to form a +1 ion
- positive ions are then attracted to a negatively charged electron plate, where they are accelerated
describe the detection stage in TOF mass spec
- positive ions hit the negatively charged electron plate
- ions are discharged by gaining electrons from the plate
- generates a movement of electrons, therefore a current can be measured
- size of current is proportional to abundance
what does the size of current tell us in TOF mass spec?
- measures the number of ions hitting the plate
- is proportional to the abundance of that isotope
- what does a mass spectrum show
- a computer shows the mass to charge ratio (m/z) and abundance of each isotope
what does the peak on the greatest mass to charge (m/z) value tell us?
- the Mr, molecular mass
- e.g. 84 for Kr84
what are peaks bigger than the Mr caused by?
isotopes of atoms in molecules
what are peaks smaller than the Mr caused by?
- fragmentation
- occurs during electron impact ionisation only
what can be calculated from mass spectrum graphs?
- the relative atomic mass (R.A.M or Ar)
- due to the relative abundance and mass being provided
what are the isotopes of chlorine and their abundances?
- Cl 35 = 75% abundance
- Cl 37 = 35% abundance
why are there peaks on the lower m/z end of the chlorine mass spectra?
- due to fragmentation of Cl2 molecules during impact ionisation
state the electron configuration of Chromium, Cr
[Ar] 4s1 3d5
explain why the electron configuration for Copper is NOT [Ar] 4s2 3d9, but instead [Ar] 4s1 3d10
- it reduces repulsion
- as pairs of electrons in the 4s orbit cause more repulsion than pairs of electrons in 3d orbits
explain why the electron configuration for Chromium is NOT [Ar] 4s2 3d4, but instead [Ar] 4s1 3d5
- reduces repulsion
- as pairs of electrons in the 4s orbit cause a high level of repulsion, so it is easier to move an electron from the 4s orbit to an empty 3d orbit
Sc’s electronic configuration is [Ar] 4s2 3d1.
state the electronic configuration for its ion, Sc 2+
[Ar] 4s0 3d1
when writing electronic configurations of ions, where are electrons removed from first and why?
- the 4s orbit
- as once filled with electrons, the 4s orbit becomes higher in energy than the 3d orbit
state the 2 elements that have anomalous electronic configurations
- Chromium, Cr
- Copper, Cu
define ionisation energy
the minimum energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous +1 ions
state 3 factors that determine ionisation energy
- atomic radius
- nuclear charge
- shielding
state the effect of atomic radius on ionisation energy
- distance between outermost e- and the nucleus
- larger atomic radius, lower force of attraction
- e- is easier to remove
- LOWERS IE
state the effect of nuclear charge on ionisation energy
- no. of protons in the nucleus
- larger nuclear charge, greater attraction between e- and + nucleus
- INCREASES IE
state the effect that shielding has on ionisation energy
- when outershell e-‘s are repelled by inner shell e-‘s
- reduces attraction between outer e- and +nucleus
- LOWERS IE
why do successive IE’s increase?
- due to overall decrease of repulsion
- e-‘s are more attracted to + nucleus
how do successive IE graphs provide evidence of electron shells?
- sudden large increases indicate a change in energy level
- e-‘s closer to nucleus require more energy to remove, due to greater attraction
describe the trend in IE across period 3
- generally increases
- dips from Mg to Al
- dips from P to S
why does IE generally increase across a period?
- increase in atomic number
- therefore INCREASES NUCLEAR CHARGE
- no. of electrons shells stay the same, so no effect on shielding
why does IE decrease from Mg to Al
- Mg has outermost e- in 3s sub shell
- Al has outermost e- in 3p sub shell
- electrons in 3p sub shell have more energy, so easier to remove
- LOWER IE FOR AL
why does IE decrease from P to S?
- P has no pairs of electrons in an orbit
- S has 1 pair of electrons in its 3p orbit
- e-‘s in same orbit repel each other, so require less energy to remove
- S HAS LOWER IE THAN P
explain the trend in IE going down group 2
- IE DECREASES
- shielding increases, due to more electron shells
- atomic radius increases, greater distance between + nucleus and e-
