Atomic Structure Flashcards
describe how molecules are ionised in electrospray ionisation:
- sample dissolved in highly volatile substance
- sample injected into hypo needle at a high voltage
- sample is vaporised and molecules gain H+ ions, forming 1+ ion
chemical properties are determined by
(number of) outer electrons
ions are accelerated by
attraction to an electromagnetic field
outline how TOF mass spec is able to seperate two species m/z 104 and m/z 118 to give 2 peaks
- +ve ions accelerated by electric field
- to a constant kinetic energy
- +ve ions w/ m/z 104 have same kinetic energy as the 108 m/z and move faster
- ions with m/z of 104 arrive at detector first
state if Al or Mg have lower first ionisation energy
- Al
- outer electron in 3p subshell
- further from nucleus so easier to remove
first ionization energy
energy needed to remove one mole of electrons from one mole of atoms (in gaseous state) to form one mole of monopositive ions (also in gaseous state)
Why does FIE generally increase across a period? (4)
- nuclear charge increases
- atomic radius decreases
- attraction betw/ nucleus and outer electrons increase
- more energy required to remove outer electron
chemical properties depends on
electronic configuration
define relative atomic mass
explain how RAM of an element w/ 3 isotopes can be calculated from data obtained from mass spectrum of the element
- average mass of an atom relative to mass of one atom of 12C
- spectrum gives relative abundance
- and mass number
- multiply mass number by relative abundance
- sum these values
- divide by the sum of relative abundances
name the shape, and bond angle, and explain, for AlCl4 -
- tetrahedral
- bond angle 109.5
- equal repulsion betw/ 4 bonding pairs
Explain why perfume is volatile and cools the skin when it evaporates (4)
- solvent has weak intermolecular forces betw/ molecules
- solvent needs energy to evaporate
- takes energy from skin, cooling it
- perfume slowly spreads by diffusion
name strongest type of bonding betw/ HF and why HI doesn’t contain this bonding
- hydrogen bonding
- (diagram w/ partial charges and 3 lone pairs on F)
- dipoles result from large difference in electronegativity
- difference in electronegativity betw/ I and H is too little
- HI dipole weaker
atomic radius gets:
- smaller across period
- larger down group
describe time of flight: electron impact
1. inject
2. ionise
3. accelerate
4. drift
5. detect
- inject liquid sample into machine
-sample gets vaporised into gas - sample gets bombarded by high energy e-, e- gets knocked off from sample turning it +ve
- +ve ions attracted to -vely charged plate
- sample enters flight tube, all ions have same KE
- lower Mr ions detected first
- ion gains e-
- relative abundance is proportional to the size of the current
describe time of flight: electron spray
- sample dissolved in highly volatile solvent
- sample injected out as aerosol, at a high voltage
- sample gains a proton becoming 1+ ion
write an equation for detection of M: electron impact, electrospray ionisation
- M+(g) + e- ➜ M(g)
- MH+(g) + e- ➜MH(g)
what are the limitations to electron impact?
high energy e- can cause fragmentation
what could cause different peaks to be shown on mass spectrum:
fragments
method to ionise sample in electrospray
- sample dissolved in volatile substance
- injected through hypodermic needle coming out as aerosol
- tip of needle has high voltage
- sample gains a proton as it leaves needle
method to ionise using electron impact
- vaporise sample
- sample gets bombarded by high energy electrons
- from electron gun
- knocks off e-
Why do samples need to be positively charged in a mass spectrometer?
so they can be ACCELERATED and DETECTED.
Also so an m/z exists, because without charge, there cannot be an m/z ratio, so the different masses cannot be differentiated in the ion drift.
When is relative molecular mass used instead of relative fomula mass?
RFM is used for compounds that are ionic or giant covalent
Chromium and copper have the following electron configurations:
Cr is [Ar] 3d5 4s1 not [Ar] 3d4 4s2
Cu is [Ar] 3d10 4s1 not [Ar] 3d9 4s2
This is because the [Ar] 3d5 4s1 and [Ar] 3d10 4s1 configurations are energetically stable
The Ionisation Energy (IE) of an element
the amount of energy required to remove one mole of electrons from one mole of gaseous atoms of an element to form one mole of gaseous ions
The units of IE are
kilojoules per mole (kJ mol-1)
.
.
On an EI graph, what do the big jumps and small jumps represent?
- BIG: change in shell (eg 3s to 2p)
- SMALL: change in subshell (eg 3p to 3s)
how does Magnesiums second to third IE give evidence for sub shells in shells?
There is a huge increase from the second to the third ionisation energy, indicating that it is far easier to remove the first two electrons than the third
Therefore the valence shell must contain only two electrons indicating that magnesium belongs to group II
The large jump corresponds to moving from the 3s to the full 2p subshell
Mg: 1s2 2s2 2p6 3s2
Electronegativity only applies to
Covalently bonded molecules
why does e-tivity decrease down a group?
- no. of shells increas
- shielding increases
- atomic radius increases
- increase in nuclear charge
- decreased attraction betw/ +ve nucleus and bonding pair of e-
When working out Mr after electro spray ionisation remember to:
-1 from the number because of the added proton weight
why is the second ionisation energy for Na (1s²2s²2p⁶) greater than Ne (1s²2s²2p⁵)
- Na has a full outer orbital of electrons
- harder to remove e- than from Ne
explain How the relative atomic mass of an element can be calculated from data obtained from the mass spectrum of the element (5)
- Spectrum gives relative abundance
- and MZ
- multiply MZ by relative abundance for each isotope
- sum these values
- divide by the sum of the relative abundances
progression in bond angles from CH4 → NH3 → H2O…which is
…which is 109.5° → 107° → 104.5°.
How to work out the lone pairs in shape
group number – the number of bonds → lone pairs
All shapes with 6 areas of electron density have 90’ angle except
2 bp, 4 lp
* linear
* 180
Properties of a homologous series
- defer by a CH2
- chemically similar
- graduation in physical properties
- same functional group
Explain why the 3IE for Mg is greater than second:
(2)
- electron is removed from 2p subshell
- less shielded than 3s subshell
Name the type of bond between H-> Al and explain how it’s formed: (3)
- dative covalent
- both electrons
-
donated from H- to Al
QoL
Explain why the value of the first ionisation energy of neon is higher than that of sodium. (2)
- electron removed from a shell of lower energy (Ne) (2p rather than 3s)
- less shielding (of Ne)
Give the formula of the ion that is formed during electrospray ionisation of X:
XH+
- a small jump in FIE indicates:
- a big jump in FIE indicates:
- small -> change in subshell (eg. from 3p to 3s)
- big -> change in shell (eg 3s to 2p)
State how the relative molecular mass of a covalent compound is obtained from its mass spectrum: (1)
Mr = highest m/z value
Which molecule is not able to form a co-ordinate bond with another species?
* BH3
* CH3
* NH3
* H2O
CH4
because the carbon has 8 valence electrons
- BH3 only has 6 so can accomodate another pair
- NH3 has a lone pair so can donate
- H2O has 2 lone pairs
inorganic ionic compounds ALWAYS
form giant structures
Why does FIE increase across a period?
Across a period from left to right, the ionisation energy increases. This is due to the increase in nuclear charge having a greater pull on the electrons and therefore more energy is required to remove electrons.
The first ionisation energies of the elements in Period 2 change as the atomic number increases.
Explain the pattern in the first ionisation energies of the elements from lithium to neon. (9)
Stage 1: General Trend (Li → Ne)
1a. 1st IE increases
1b. More protons/increased nuclear charge
1c. Electrons in same energy level / shell
1d. No extra/similar shielding
1e. Stronger attraction between nucleus and outer e-
Stage 2: Deviation Be → B
2a. B lower than Be
2b. Outer electron in (2)p
2c. higher in energy than (2)s
Stage 3: Deviation N → O
3a. O lower than N
3b. 2 electrons in (2)p need to pair
3c. pairing causes repulsion
Why is it not possible to use electron repulsion theory with [CoCl4]2- (1)
Too many electrons in d-subshell
