S1.3 Electron configurations

Question 1

how are emission spectra produced

Answer 1

by atoms emitting photons when electrons in excited states return to lower energy levels

Question 2

continuous spectrum

Answer 2

• in the visible region
• all colours
• all wavelengths
• all frequencies

Question 3

line spectrum

Answer 3

• Only has specific wavelengths
• Only has specific frequencies
• Only has specific energies of light
• Tells us that emitted light from atoms can only be at fixed frequencies → quantised
• Electrons can only have certain energy amounts

Question 4

radio waves

Answer 4

low energy
long wavelength
low frequency

Question 5

gamma rays

Answer 5

high energy
short wavelength
high frequency

Question 6

relationship between frequency and wavelength + the supporting equation

Answer 6

Speed of light (c) = vλ → frequency and wavelength are inversely proportional

Question 7

hydrogen line emission spectrum

Answer 7

• Provides evidence for the existence of electrons in discrete energy levels, which converge at higher energies
• As energy increases (frequency increases, wavelength decreases), lines converge so the electron is reaching a maximum amount of energy → ionisation energy
• The visible lines correspond to the electron jumping from higher levels to n=2

Question 8

n∞ → n=3

Answer 8

• paschen
• infrared region
• low energy

Question 9

n∞ → n=2

Answer 9

• balmer
• visible region
• medium energy

Question 10

n∞ → n=1

Answer 10

• lyman
• UV region
• high energy

Question 11

how is energy related to frequency and wavelength

Answer 11

• Wavelength determines colour
• If constant energy, the same colour will always be emitted

Question 12

principal quantum number (n)

Answer 12

• main energy level
• number of energy levels/quantum shells
• can hold a maximum of 2n^2 electrons

Question 13

what is the main energy level divided into

Answer 13

• sublevels/subshells of successively higher energies
• s, p, d, f

Question 14

s atomic orbital

Answer 14

• spherical
• low energy
• most likely to find electrons

Question 15

p atomic orbitals

Answer 15

• 3: x, y, z
• dumbbell
• higher energy than s
• lobes are larger and longer as n increases

Question 16

ground state

Answer 16

most stable electronic configuration with the lowest amount of energy

Question 17

Aufbau’s principle

Answer 17

• fill the sub shells with lowest energy first (1s)

Question 18

exception to Aufbau’s principle

Answer 18

• 3d is higher in energy than 4s
• So 4s is filled first
• When filled, 4s is higher in energy

Question 19

Hund’s Rule

Answer 19

• Electrons fill all orbitals singularly first, then in pairs
• Electrons can spin clockwise or anticlockwise
• Those with the same spin repel (spin pair repulsion)
• So → occupy separate orbitals to minimise repulsion
• They then pair up with electrons with spin in the opposite direction

Question 20

Pauli exclusion principle

Answer 20

• Orbital must hold 2 electrons with the opposite spin
• Energy to jump to a higher energy orbital is greater than inter-electron repulsion
• So they pair up and occupy lower energy levels first

Question 21

transition metal rules

Answer 21

• Fill 4s before 3d
• Lose 4s before 3d

Question 22

exception to Aufbau’s principle: Chromium

Answer 22

• 1s2 2s2 2p6 3s2 3p6 3d5 4s1
• [Ar]3d5 4s1

Question 23

exception to Aufbau’s principle: Copper

Answer 23

• 1s2 2s2 2p6 3s2 3p6 3d10 4s1
• [Ar]3d10 4s1

Question 24

explanation for copper and chromium

Answer 24

it is more energetically favourable if a 3d sublevel is ½ full or full, therefore a 4s electron is used

Question 25

ions rules

Answer 25

• When removing electrons, we remove the highest energy electron first
  4s before 3d
• In an emission spectrum, the limit of convergence at higher frequencies corresponds to ionisation

Question 26

limit of convergence

Answer 26

where lines appear to meet, limit is the frequency at which spectral lines converge

Question 27

4 factors affecting ionisation energies

Answer 27

Size of nuclear charge
Atomic radii
Shielding effect of inner electrons
Spin-pair repulsion

Question 28

size of nuclear charge

Answer 28

Greater atomic number = greater nuclear charge
Greater attractive forces between nucleus and outer electron
More energy required to overcome attractive forces

Question 29

atomic radii

Answer 29

Electrons in shells further from nucleus are less attracted to nucleus
So lower ionisation energy

Question 30

shielding effect of inner electrons

Answer 30

Electrons in full inner shells repel electrons in outer shells
Prevents them to feel full nuclear charge
The greater the shielding the lower the ionisation energy

Question 31

spin-pair repulsion

Answer 31

Paired electrons in the same atomic orbital in a subshell repel each other (more than electrons in different atomic orbitals)
Easier to remove an electron, lower ionisation energy

Question 32

general trend in ionisation energies across a period

Answer 32

Increases
Increase in nuclear charge, same number of shells so same distance of outer electrons to nucleus
Reasonably constant shielding
Decreased atomic radius
Outer electron is held more tightly to nucleus → harder to remove

Question 33

trend in ionisation energies down a group

Answer 33

Decrease
Nuclear charge increases
Distance between nucleus and outer electron increases (atomic radii)
Shielding by inner electron increases
Effective nuclear charge decreases as shielding increases
Outer electron is held more loosely to nucleus → easier to remove

Question 34

explain the rapid decrease in IE between last element in a period and 1st element in next period

Answer 34

Increased atomic radii
Increased shielding by inner electrons
2 factors outweigh increased nuclear charge

Question 35

group 2 to 3 dip

Answer 35

drop in the I.E of group 3, there is an extra electron in the 2p orbital which is removed as the p orbital is higher energy than s (easier to remove)

Question 36

group 5 to 6 dip

Answer 36

drop in the I.E of group 6, extra electron paired up with one electron already in one of the 3p orbitals, there is a repulsive force between the 3 paired electrons, so less energy is needed to remove them

Question 37

successive ionisation energies of an element

Answer 37

• give information about its electron configuration
• increase because it is harder to remove electrons from an already positive ion
• as electrons are removed, attractive forces increase, due to decreased shielding and an increase in proton to electron ratio

Question 38

frequency and energy

Answer 38

E = hf
directly proportional

Question 39

frequency, energy, wavelength

Answer 39

f = c/λ
frequency & energy are inversely proportional to wavelength