atomic structure

Question 1

formula for angle of deflection

Answer 1

angle of deflection is proportional to |q/m|, charge over mass

Question 2

NESOE and what happens when principal quantum no. n increases

Answer 2

nucleus, electron shells, subshells, orbitals, electrons

when n increases, energy level increases and electrostatic attraction between electrons and nucleus decreases
the orbitals also become more diffuse

Question 3

definition of orbital

Answer 3

region of space where there is a high probability of finding an electron

Question 4

types of subshells

Answer 4

s, p, d, f

Question 5

types of orbitals

Answer 5

s
px, py, pz
dxz, dxy, dyz, dz^2, dx^2-y^2

Question 6

elaborate on types of orbitals (shape, axes, directional or not)

Answer 6

check notes

Question 7

formula for nth electron shell

Answer 7

• number of subshells = n
• number of orbitals = n^2
• max number of electrons that can be accommodated = 2n^2

Question 8

hunds rule and pauli exclusion principle

Answer 8

orbitals of a subshell must be occupied singly by electrons of parallel spins before pairing can occur

if there are 2 electrons in the same orbital, they must be of opposite spins

Question 9

exceptions for configuration

Answer 9

Cr 3d5 4s1

Cu fill 3d10 before 4s1

Question 10

ground state excited state definition

Answer 10

at ground state when it is overall at the lowest energy levels

at excited state when one or more electrons absorb energy and are promoted to a higher energy level (they are unstable and can emit energy to return to ground state)

Question 11

factors affecting electrostatic attraction

Answer 11

1. Number of electron shells
   - As number of electron shells increase
   - principle quantum number, n, of the outermost valence shell increases,
   - distance between the nucleus and valence electron increases,
   - electrostatic attraction between the nucleus and valence electron decreases
2. Nuclear charge, Z
   - If number of protons increases
   - nuclear charge increases
   - electrostatic attraction between the nucleus and valence electron increases
3. Shielding effect by inner electrons
   - if the number of inner electron shell electrons increases
   - shielding effect experienced by the valence electron increases
   - electrostatic attraction between the nucleus and valence electron decreases

Across a period: consider factor 2 and 3

Down the group: consider all factors (2 can be minor influencing)

Question 12

shielding effect

Answer 12

• electrons in inner electron shells repel the electrons in outer electron shells (prevent them from experiencing full effect of the actual nuclear charge)
• the greater the shielding of outer electrons by inner electron shell electrons, the weaker the attractive forces between the nucleus and the outer electrons
• electrons in same electron shell provide poorer shielding effect for one another
• for same shell n, shielding ability of electron decreases in order s>p>d>f (d and f provide very poor shielding effect)

Question 13

atomic radius

Answer 13

half the shortest inter nuclear distance found in the structure of the element

Question 14

atomic radii across period

Answer 14

decreases across the period

across the period,
- number of electron shells remain the same
- number of protons increases hence nuclear charge increases
- number of electrons also increases but these electrons are added to the outermost electron shell hence shielding effect remains approximately constant
- electrostatic attraction between nucleus and valence electrons increases hence electron cloud size decreases

hence atomic radii decreases across the period

Question 15

atomic radii down group

Answer 15

increases down the group

down the group,
- the number of electron shells increases
- the distance between the valence electron and nucleus increases
- shielding experienced by the valence electrons increases
- despite increasing nuclear charge,
- the electrostatic attraction between the nucleus and valence electrons decreases and electron cloud size increases

hence the atomic radii increases down the group

Question 16

radius of cation vs parent atom

radius of anion vs parent atom

Answer 16

radius of cation is always smaller than that of the parent atom
- they have the same number of protons and hence the same nuclear charge
- the cation has one less electron shell hence outermost electrons are shielded to a smaller extent
- electrostatic attraction between the nucleus and valence electron is stronger and electron cloud size decreases hence radii decreases

radius of anion always bigger than that of the parent atom
- they have the same number of protons and hence the same nuclear charge
- the anion has a greater number of electrons than the parent atom hence the outermost electrons experiences greater shielding (interelectronic repulsion)
- electrostatic attraction between the nucleus and valence electron is weaker and electron cloud size increases hence radii increases

Question 17

when to use outermost and when to use valence electrons

Answer 17

outermost is for ions
valence is for atoms

Question 18

graph for radii for isoelectronic species
+ explanation

Answer 18

look at notes for graph

Decrease from Na+ to Si4+ and P3- to Cl-
- Number of electron shells remain the same hence the outermost electrons experience the same shielding effect
- Number of protons increases, hence nuclear charge increases
- Electrostatic attraction between nucleus and outermost electrons increases, hence electron cloud size decreases and radii decreases

Sharp increase from Al3+ to P3-
- P3- has one more electron shell than Al3+ hence the distance between the nucleus and the outermost electrons increases
- the shielding effect experienced by the outermost electrons increases
- hence despite increasing nuclear charge, the electrostatic attraction between the nucleus and outermost electrons decreases, hence electron cloud size increases and radii increases

Question 19

definition of IE

Answer 19

1st IE
- amount of energy required to remove 1 mole of electrons from 1 mole of gaseous M atoms to form 1 mole of gaseous M+ ions.

2nd IE
- amount of energy required to remove 1 mole of electrons from 1 mole of gaseous M+ ions to form 1 mole of gaseous M2+ ions

Question 20

First IE across period trend + graph + exceptions

Answer 20

graph look at notes

Across period, increases
- number of electrons shells remain the same
- number of protons increases hence nuclear charge increases
- number of electrons increases but they are added to the same outermost electron shell hence shielding remains approx constant
- EA between nucleus and valence electrons increases hence energy required to remove 1 mole of electrons from the atom also increases

Exception 1: group 2 and 13
- the 3p electron to be removed from Al is at a higher energy level than the 3s electron to be removed from Mg
- hence less energy is needed to remove the 3p electron in Al than the 3s electron in Mg
- hence first IE of Al is less than that of Mg

Exception 2: Group 15 and 16
- the 3p electron to be removed from S is a paired electron while the 3p electron to be removed from P is an unpaired electron
- due to greater interelectronic repulsion between paired electrons in the same orbital, less energy is needed to remove the paired 3p electron from S
- hence first IE is S is lower than that of P

Question 21

1st IE down the group + same explanation for big drop between 1st IE of Ne and Na and Ar and K

Answer 21

it decreases down the group
- number of electron shells increase, distance between nucleus and valence electrons increases
- shielding experienced by the valence electrons also increases
- despite increasing nuclear charge, the electrostatic attraction between the nucleus and valence electrons decreases and energy required to remove a valence electron from the atom decreases

hence 1st IE decrease down the group

Question 22

successive IEs of the same element trend

Answer 22

increases
- no. of protons remain the same hence nuclear charge remains the same
- no. of electrons decreases and shielding experienced by the remaining outermost electrons decreases
- EA between the nucleus and the remaining electrons increases hence the amount of energy required to remove each subsequent electron increases

Question 23

identification of which group from IE

Answer 23

1. there is a significant increase between the 4th and 5th IE of __.
2. significantly more energy is needed to remove the 5th electron as it is in an inner electron shell that is nearer to the nucleus and experiences less shielding hence experiences a stronger electrostatic attraction to the nucleus

Question 24

electronegativity

Answer 24

electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electrons

Question 25

electronegativity trend across period and group

Answer 25

BETWEEN NUCLEUS AND BONDING ELECTRONS

across period, increase
- no. of electron shells the same
- no. of protons increase so nuclear charge increases
- no. of electrons increases but all added to the same outermost electron shell hence shielding effect remains approx constant
- EA between nucleus and bonding electrons increases, EN increases

down group, decrease
- no. of electron shells increase, distance between nucleus and valence electrons increase and shielding effect increases
- hence despite increasing nuclear charge, EA between nucleus and bonding electrons decreases and EN decreases