Atomic structure

Question 1

There is a general trend for an increase in ionisation energy across Period 3.
Give one example of an element that deviates from this trend.
(3 marks)

Answer 1

M1 Aluminium / Al
M2 Outer electron in 3p orbital / sub-shell
M3 3p higher in energy / slightly more shielded than 3s
or
M1 Sulfur / S
M2 Outer electrons in 3p orbital begin to pair
M3 Repel

Question 2

Give an equation, including state symbols, to represent the process that occurs when the third ionisation energy of sodium is measured.
(1 mark)

Answer 2

M1 Na2+(g) → Na3+(g) + e−

Question 3

Define the term relative atomic mass.
[2 marks]

Answer 3

M1 The average mass of an atom of an element
M2 Compared to 1/12th the mass of an atom of carbon-12

Question 4

Define the term 1st Ionisation energy
[2 marks]

Answer 4

M1 The amount of energy required to remove 1 mole of electrons from one mole of gaseous atoms

Question 5

Explain why there is a general increase across a period.
[2 marks]

Answer 5

M1 Proton number/nuclear charge increases across a period
M2 But the amount of shielding stays the same

Question 6

Explain why Aluminium has a lower 1st IE than Magnesium
[3 marks]

Answer 6

M1 An electron is removed from the 3p subshell of Al
M2 An electron is removed from the 3s subshell of Mg
M3 The 3p subshell is higher in energy than the 3s

Question 7

Explain why Oxygen has a lower 1st IE than Nitrogen
[3 marks]

Answer 7

M1 The first electron removed from Nitrogen is in a 2p orbital and is unpaired
M2 The first electron removed from Oxygen is in a 2p orbital and is paired
M3 Paired electrons repel, so less energy is required to remove the electron from Oxygen

Question 8

The trend in 2nd ionisation energy across period 3 follows the same pattern. Suggest which elements deviate from the general increase in ionisation energy when the 2nd ionisation energy is measured.
[2 marks]

Answer 8

M1 Si
M2 Cl

Question 9

Sodium has a lower 1st ionisation than both Ne and Mg, explain for each element why this is the case. [4 marks]

Answer 9

M1 Mg has more protons AND the same shielding as Na.
M2 Therefore the outer electron in Mg is more strongly attracted to the nucleus and requires more energy to remove it
M3 Ne has its outer electron on a shell closer to the nucleus, so there is less shielding
M4 Therefore, the outer electron in Ne is more strongly attracted to the nucleus and requires more energy to remove it

Question 10

Explain how the ions are detected
[1 mark]

Answer 10

M1 The positive ions hit the detector and gain an electron.

Question 11

Explain how the ion abundance is measured
[1 mark]

Answer 11

M1 This induces a current proportional to the abundance.

Question 12

Explain which two variables are measured by a mass spectrometer?
[1 mark]

Answer 12

M1 Mass to charge ratio (m/z) and Abundance

Question 13

Explain why does the flight tube have to be in a vacuum
[1 mark]

Answer 13

M1 So ions don’t hit and deflect off molecules in air. This could affect their time of flight.

Question 14

Write an equation to represent what occurs when:
Br is ionised
[1 mark]

Answer 14

M1 Br(g) -> Br+(g) + e−

Question 15

Write an equation to represent what occurs when: K+ is detected
[1 mark]

Answer 15

M1 K+(g) + e− -> K(g)

Question 16

Suggest why there is no significant difference between the first ionisation energies of 16O, 17O, 18O.
[2 marks]

Answer 16

M1 Each isotope has the same number of protons and the same amount of shielding.

Question 17

How do you calculate relative atomic mass

Answer 17

RAM =
(isotopic mass x % abundance) / 100

Question 18

Calculate the relative atomic mass of tellurium from the following abundance data:
124-Te relative abundance 2;
126-Te relative abundance 4;
128-Te relative abundance 7;
130-Te relative abundance 6

Answer 18

R.A.M = [(124x2) + (126x4) + (128x7) + (130x6)] / 19 = 127.8

Question 19

Describe electron impact

Answer 19

• The sample being analysed is vaporised and then high energy electrons are fired at it.
• The high energy electrons come from an ‘electron gun’ which is a hot wire filament with a current running through it that emits electrons.
• This usually knocks off one electron from each particle forming a 1+ ion.

X(g) -> X + (g) + e–

• The 1+ ions are then attracted towards a negative electric plate where they are accelerated

Question 20

Describe electrospray ionisation

Answer 20

• The sample X is dissolved in a volatile solvent (eg water or methanol) and injected through a fine hypodermic needle to give a fine mist (aerosol).
• The tip of the needle is attached to the positive terminal of a high-voltage power supply.
• The particles are ionised by gaining a proton (ie an H + ion which is simply one proton) from the solvent as they leave the needle producing XH+ ions (ions with a single positive charge and a mass of Mr + 1).

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

• The solvent evaporates away while the XH+ ions are attracted towards a negative plate where they are accelerated.

Question 21

Calculate the mass, in kg, of one atom of 46Ti
The Avogadro constant L = 6.022 × 1023 mol−1

Answer 21

46 g mol−1 = 0.046 kg mol−1
0.046 / (6.022 × 1023) = 7.639 × 10−26 kg