Atomic Struc

Question 1

Defn of atom

Answer 1

Atom is the smallest particle found in an element that can take part in a chemical rctn

Question 2

Subatomic particle - location, relative mass, relative charge, symbol

Answer 2

E- in orbitals around nucleus, 1/1840 , -1, (0 -1) e
Neutron - nucleus, 1, 0, (1 0) n
Proton -nuclues, 1, +1, (1 1) p

Question 3

Behaviour of subatomic particles in an electric field (3)

Answer 3

1. Same speed
2. diff direction of deflection
3. diff angle of deflection

Question 4

how to find angle of deflection and stuff (formula)

Answer 4

angle of deflection prop. to | q/m | when solving qn use the constant k

Question 5

what is a nuclide

Answer 5

any species of given mass no. and atomic no. (eg hydrogen - 1) - elemental name & mass no

Question 6

What is nucleon no.

Answer 6

mass no

Question 7

what is mass no.

Answer 7

nucleon no.

Question 8

What is atomic no.

Answer 8

proton no.

Question 9

sig of atomic no.

Answer 9

determines the identity of an atom

Question 10

isotopes & their properties

Answer 10

same no. of protons/e- but diff neutrons - same chem properties n diff/masses/phy properties

Question 11

How to solve
(235 92) U + (1 0) n -> Y + (90 36) Kr + 2 (1 0) n 
nucleon no. of Y = 
Proton no. of Y = 
Hence Y is =

Answer 11

Nucleon no. of Y = 235+1-90-2 = 144
Proton no. of Y = 92-36 = 56
Hence Y = (144 56) Ba

Question 12

Principle quantum shell, All the subshells, no. of orbitals, types of orbitals - name, shape, direction, size

Answer 12

Subshells: s, p, d, f
S: 1 - spherical shape, non directional
P: 3 -dumbbell shape, directional as the e- density is concentrated in certain directed along…axes
D: 5 -
d xz & d xy & d yz - 3 orbitals with similar 4-loped shape, orbitals have their lobes pointing b/w the axes
d x^2 - y^2 - 4 lobed shape, lobes aligned along the x & y axes
d z^2 - dumbbell shaped surrounded by a small doughnut shaped ring at its waist, orbital is aligned along the z axis

Question 13

the greater the value of n … (4)

Answer 13

further the shell is from the nucleus, higher the energy lvl of the shell
higher the energy lvl of the shell/e-
weaker the electrostatic attraction b/w nuceus and e-
larger size of orbital

Question 14

energy lvl diagram & the exception

Answer 14

4s<3d

Question 15

3 basic rules to write e- configuration

Answer 15

Aufbau principle - e- fill orbitals from the lowest energy orbital upwards
Hund’s rule - orbitals of a subshell must be occupied singly by an e- of parallel spins before pairing can occur
Pauli exclusion principle - each orbital can hold a max of 2 e- and they must be of opp. spins

Question 16

why are paired e- stable

Answer 16

when they spin in opp directions, the magnetic attraction which results from their opp spins can counterbalance the electrical repulsion which results from their identical charges

Question 17

2 anomalous e- configuration and why are they more stable ( general reason and indiv reason)

Answer 17

Cr & Cu - a ‘d’ subshell that is half-fulled or full is more stable -
for Cr: as 3d & 4s are abt equal in energy by the time Cr is reached, so by having 1 e- each in the 4d and 4s orbitals, inter-electronic repulsion is minimized
For Cu: the fully filled 3d subshell is unusually stable due to the symmetrical charge distribution around the metal center

Question 18

excited state

Answer 18

one or more e- absorb energy and are promoted to a higher energy lvl

Question 19

are 4s e- lost before or after 3d e-? and why

Answer 19

once e- occupy the inner 3d orbitals, they provide some shielding for the outermost 4s e- hence they repel the 4s e- to a slightly higher energy lvl

Question 20

isoelectronic species

Answer 20

species with the same total no. of e-

Question 21

e- configuration from the periodic table

Answer 21

s- block - ns^1 ns&2
d-block - (n-1)d1 ns^y - the 1 is starting from the first transition metal
p-block - ns^2np^1 -> ns^2np^6

Question 22

Defn of atomic radius

Answer 22

Half the shortest inter-nuclear dist found in the struc of the element

Question 23

Variation in atomic radii across period

Answer 23

DECREASE as
No.of e- shells remain the same
no. of protons, hence nuclear charge increases
e- increase but since they are added to the valence shell, the shielding effect remains approx const
effective nuclear charge increases
electrostatic attraction b/w nucleus and valence e- increases
decrease in size of e- cloud

Question 24

Variation in atomic radii down grp

Answer 24

INCREASE
although protons increase, nuclear charge increases
no. of e- shell increases, dist b/w nucleus and valence e- increase
hence, the electrostatic attraction b/w the nucleus and the valence e- decreased
increase in size of e- cloud

Question 25

Cationic radius vs atom

Answer 25

rcation

Question 26

Anionic radius vs atom

Answer 26

ratom

Question 27

Ionic radii of isoelectronic species across period

Answer 27

DECREASE
Na+, Mg2+, Al3+ are all isoe- species and hecne their valence e- exp the same shielding effect
nuclear charge increases from Na+ to Al3+
effective nuclear charge increases from Na+ to Al3+
electrostatic attraction b/w the nucleus and valence e- increase resulting in decreases in size of e- cloud

Question 28

Why would there be a sharp increase in ionic radius from Al3+ to P3-

Answer 28

P3- has 1 more e- shell than Al3+
Despite the increase in nuclear charge, the valence e- of P3- are less strongly attracted by the nucleus
Ionic radius of P3- is bigger than that of Al3+

Question 29

Trend of 1st ionisation energies across the period

Answer 29

INCREASE
no. of e- shell remain the same
no. of protons increase, nuclear charge increase
no. of e- increase, but they are added to the valence shell so shielding effect remains approx const
effective nuclear charge increases
electrostatic attraction b/w nucleus and valence e- increase
increase in energy required to remove the valence e- from an atom

Question 30

Irregularity in ionisation energies across a period (2)

Answer 30

Grp 2 & 13 ———— grp 13<2
Al vs Mg:
3p e- to be removed from Al is at a higher energy lvl than the 3s e- to be removed from Mg
Hence less energy is required to remove the 3p e- in Al than the 3s e- in Mg

Grp 15 & 16 ————- grp 16<15
S vs P:
The 3p e- to be removed from S is a paired e- while that to be removed from P is an unpaired e-
Due to inter-electronic repulsion b/w paired e- in the same orbital, less energy is required to remove the paired 3p e- from S

Question 31

Trend of 1st e- energies down a grp

Answer 31

DECREASE
no. of e- shells increase
dist b/w nucleus and valence e- increase
despite increase in nuclear charge,
electrostatic attraction b/w the nucleus and the valence e- deceases,
resulting in a decrease in the energy required to remove the valance e- from an atom

Question 32

Trend in successive ionisation energies of an element

Answer 32

INCREASE
once the 1st e- is removed from the neutral atom, each successive e- is removed from an ion of increasing +ve charge which attracts the e- more strongly, hence more energy is required for the removal of the e-

Question 33

How to deduce grp no. from successive ionisation data

Answer 33

A large jump in the 5th and 6th IE is observed
Sig more energy is required to remove the 6th e- as it is located in an inner electronic shell that is nearer to the nucleus and is hence more strongly attracted by the nucleus.
5e- in the valence shell
E is likely to be in grp 15

Question 34

Defn of electronegativity

Answer 34

Relative measure of its ability to attract bonding e-

Question 35

Trend in electronegativity across period

Answer 35

INCREASE
no. of e- shell remain the same
no. of protons increase, nuclear charge increase
no. of e- increase, but they are added to the valence shell so shielding effect remains approx const
effective nuclear charge increases
electrostatic attraction b/w nucleus and valence e- increase

Question 36

Trend in electronegativity down grp

Answer 36

DECREASES
no. of shells increase
dist b/w the nucleus and the bonding e- increase
despite increase in nuclear charge, electrostatic attraction b/w the nucleus and the bonding e- decreases

Question 37

Why is the 3rd IE of mg much higher than that of aluminium

Answer 37

Al2+ has one more e- shell than Mg2+, the 3s e- in Al2+ is further from the nucleus, has higher energy and shielded by more inner e- shells than the 2p e-
Hence the 3s e- in Al2+ is less strongly attracted by the nucleus and requires less energy to be removed compared to the 2p e- in Mg2+