Atomic Structure

Question 1

Relative mass and charge

Answer 1

Proton M= 1 C= +1
Neutron M= 1 C= 0
Electron M= 1/1835 C= -1

Question 2

atomic number definition

Answer 2

The atomic number is equal to the number of protons.

Question 3

mass number definition

Answer 3

The mass number is equal to the number of protons and neutrons combined.

Question 4

Atomic Radius size

Answer 4

determined by number of protons and electrons

Question 5

Cations & Anions

Answer 5

Positive Ions (cations) – lost electrons

Negative Ions (anions) – gained electrons

Question 6

Isotopes

Answer 6

Isotopes are atoms with the same number of protons but a different number of neutrons

Isotopes of the same element have the same chemical properties because they have the same electron configuration

Question 7

Ionisation energy definition

Answer 7

Ionisation energy is the amount of energy needed to remove a mole of electrons from a mole of atoms, in the gaseous state.
(K(g) -> K+(g) + e-)
units are kJmol-1

Question 8

Successive ionisation energies

Answer 8

1st IE Na(g) -> Na+(g) + e-
2nd IE Na+ (g) -> Na2+(g) + e-
3rd IE Na2+ (g) -> Na3+(g) + e-

Question 9

Why is the second ionisation energy of S higher than the first?

Answer 9

The second electron is removed from an ion that already has a positive charge.

Question 10

Factors influencing IE

Answer 10

-more protons, stronger attraction, more energy needed to remove outer e-
-closer e- is to nucleus, stronger attraction, more energy needed to remove outer e-
-more shells, more shielding, less energy for ionisation

Question 11

Using successive IE data to identify element

Answer 11

• The biggest jump is between the 4th and 5th ionisation energies
• the 5th electron is on the shell closer to the nucleus
• The element must have 4 electrons on its outer shell
• in period 3 this must be Silicon.

Question 12

Shells

Answer 12

Cr = 1s2 2s2 3s2 3p6 4s1 3d5
Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Ca2+ = 1s2 2s2 2p6 3s2 3p6

Question 13

Trends across period

Answer 13

• greater nuclear charge
• greater attraction between e- and nucleus
• shielding stays same

Question 14

Trends down group

Answer 14

• more shells
• more shielding
• weaker attraction between e- and nucleus
• IE decreases down group

Question 15

Explain why Aluminium has a lower 1st IE than Magnesium

Answer 15

• The first e- removed from Mg is from a 3s sub level.
• The first e- removed from Al is from a 3p sub level.
• The 3s sub level is lower in energy than 3p.
• Therefore less energy is needed to remove the electron from Al.

Applies to any element in group 3

Question 16

Explain why Sulfur has a lower 1st IE than Phosphorous

Answer 16

• The first e- removed from P is from a 3p sub level and is unpaired.
• The first e- removed from S is also from a 3p sub level, but is from a paired orbital.
• This means Sulfur has a lower ionisation energy due to electron pair repulsion.
• Therefore less energy is needed to remove the electron from S.

Applies to any element in group 6

Question 17

time of flight mass spectrometer

Answer 17

(1) Vacuum
(2) Ionisation
(3) Acceleration
(4) Ion Drift
(5) Detection
(6) Data Analysis

Question 18

Why are the sample particles ionised?

Answer 18

• So they can be accelerated towards the negatively charged plate
• So they generate a current when they hit the detector

Question 19

How is the ion accelerated?

Answer 19

• Positive ions attracted to the negatively charged plate
• All ions have the same kinetic energy

Question 20

How are ions separated in the flight tube?

Answer 20

• Ions travelling at higher speeds (small m/z) move ahead of those travelling more slowly (large m/z)

Question 21

How are the ions detected?

Answer 21

• Each ions hits the detector
• Ion gains an electron
• Generates a current
• Size of the current is proportional to the abundance of the ion

Question 22

How is a current generated?

Answer 22

e- transferred from detector to positive ion

Question 23

Electron impact

Answer 23

(also known as electron ionisation)

sample vaporised and high energy e- fired at it by an electron gun. e- knocked off to form 1+ ions

X(g) -> X+(g) + e-

Question 24

Electrospray ionisation

Answer 24

dissolve sample in volatile solvent and inject through hypodermic needle attached to positive terminal of high voltage power supply. particles gain proton as leave needle

X(g) + H+ -> XH+(g)

Question 25

Calculating relative atomic mass (Ar)

Answer 25

(Mass 1 x % 1) + (Mass 2 x % 2) + (Mass n x % n) / 100

Question 26

Mass of one atom in KG

Answer 26

mass / 6.022 x10^23 / 1000

Question 27

time of flight equations

Answer 27

KE = 0.5 x mass x velocity^2

v = d / t

t = time of flight (s)
d = length of flight tube (m)
𝑣 = velocity of the particle (m s–1)
m = mass of the particle (kg)
KE = kinetic energy of particle (J)

Question 28

Relative abundance

Answer 28