3.1.1.Atomic structure

Question 1

John Dalton

Answer 1

-Indivisible solid spheres
-Diffiferent elements made up of different spheres

Question 2

Thompsons plum pudding model

Answer 2

-Discovery of electrons
-Proved atoms are divisible
-Positively charged pudding with negative electrons

Question 3

Rutherford gold foil experiment

Answer 3

-Firing +ve alpha particles at thin gold sheet of foil in a vacuum
-Thought it would deflect most particles but most passed straight through with a few deflecting at large angles
-Indicates atoms are mostly empty space
-Those that deflected had presence of tiny dense, +ve charged nucleus. Electrostatic repulsion.
-Proposed -ve electrons must orbit nucleus in cloud to maintain neutrality.

Question 4

Modification of Rutherfords nuclear model

Answer 4

-Discover of neutron and protons

Question 5

Bohr

Answer 5

-Rutherford unable to explain why electrons wouldn’t simply collapse
-Discovered electrons orbit nucleus in specific energy levels and can jump between shells by absorbing or emitting light of particular wavelengths

Question 6

Mass no. (Top no.)

Answer 6

-Tot. no. of protons and neutrons

Question 7

Atomic no. (bottom no.)

Answer 7

-No.of protons=no.of electrons

Question 8

Isotope

Answer 8

-Same no. of protons but different no. of neutrons.
-Same chemical properties (electron config) but different physical properties as it depends on atomic mass

Question 9

Relative atomic/Isotopic/molecular mass

Answer 9

-The average mass of an atom/Isotope/ Molecule of an element compared to 1/12 the mass of a carbon-12 atom
-Ar=sum of (isotopic mass x abundance) /Sum of all abundance

Question 10

Mass spectrometer

Answer 10

-used to measure mass to charge ratio (m/z) of ions
1.Ionisation: unknown sample mixed with polar solvent, inserted at high pressure into +ve gaseous ion
-Electron impact-firing the vaporised sample with high energy from hot cathode/electron gun which knocks off an electrons forming +ve ion
-Electrospray ionisation- dissolving sample in volatile solvent and spraying through charged fine needle, adding a proton(H+ ion) from solvent to each molecule

2.Acceleration: +vely charged ions accelerate towards -ve plate by an electric field giving them same KE. Lighter ones move faster
KE=1/2MV*2

3.Ion drift: Ions enter flight tube where they drift at a steady speed related to their mass. Lighter ions(lower m/z ratio) move through tube quicker
t=d X ^m/2KE
d=m
t=s
m=Kg
KE=Kg

4.Detection: As ions hit detector, gain electron which creates an electric current. Size of current is proportional to abundance of isotopes

Question 11

Electron configuration

Answer 11

-Cr: 1s2, 2s2, 2p6, 3s2, 3p6, 3d5,4s1
-Cu: 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

Question 12

1st ionisation

Answer 12

-The energy required to remove 1mole of electrons from each atom in 1mole of gaseous atoms to form 1mole of gaseous 1+ ions under standard conditions
-IE is always +ve value

Question 13

Factors affecting IE

Answer 13

1.Nuclear charge- Atoms with more protons have stronger +ve charge which creates stronger electrostatic attraction between nucleus and outer electrons
2.Atomic radius- Electrostatic attraction reduces with increasing distance. Larger distance= weaker attraction
3.Electron shielding- Inner electron shells shield outermost electrons from full attractive force of nucleus, reducung the nuc. charge experienced. electrons are -vely charged, inner ones repel outer ones reducing attraction
-low shielding and small atomic size=high IE.

Question 14

Trends in 1st IE

Answer 14

1.Down group:
IE decreases as
-Nuclear charge increases=more protons +increasing attraction for electrons
-Atomic radius increases=more shells
-Electron shielding increases= less attraction
-Nuclear attraction decreases

2.Across periods :
IE increases as
-Nuclear charge increases
-Atomic radius decreases as extra electrons added to same shell
-Electron shielding Decreases/ stays similar due to no extra inner shells

Question 15

Drops between groups

Answer 15

1. Drop between groups 2+3:
   -in group 3 electron is removed from p orbital rather the s orbital like in group 2
   -P orbital has additional shielding therefore less energy required to remover outermost p electron from group 3 compared to removing s from group 2.
   Ie. Al has lower 1st IE than Mg

2.Drops between groups 5+6:
-in group 5 electron is removed from singly occupied orbital but group 6 is removed from orbital with 2electrons.
-Pairs electron is group 6 experience greater electron electron repulsion. Therefore less energy needed to remove one of these paired electrons in group 6
Ie. Sulfur has lower 1st IE than phosphorus

Question 16

Successive ionisation energies

Answer 16

-2nd IE is energy needed to remove 1electron from each ion in 1mole of gaseous 1+ ions to form 1mole of gaseous 2+ions
-The remaining electrons experience greater electrostatic attraction to the increasingly +ve nucleus. Increased nuc. attraction requires more energy to remove next electron from the shell
-Large jumps on successive IE between shells as attractions to nucleus is much greater for inner shell electrons closer to the nucleus