Atomic Struc Flashcards
Defn of atom
Atom is the smallest particle found in an element that can take part in a chemical rctn
Subatomic particle - location, relative mass, relative charge, symbol
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
Behaviour of subatomic particles in an electric field (3)
- Same speed
- diff direction of deflection
- diff angle of deflection
how to find angle of deflection and stuff (formula)
angle of deflection prop. to | q/m | when solving qn use the constant k
what is a nuclide
any species of given mass no. and atomic no. (eg hydrogen - 1) - elemental name & mass no
What is nucleon no.
mass no
what is mass no.
nucleon no.
What is atomic no.
proton no.
sig of atomic no.
determines the identity of an atom
isotopes & their properties
same no. of protons/e- but diff neutrons - same chem properties n diff/masses/phy properties
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 =
Nucleon no. of Y = 235+1-90-2 = 144
Proton no. of Y = 92-36 = 56
Hence Y = (144 56) Ba
Principle quantum shell, All the subshells, no. of orbitals, types of orbitals - name, shape, direction, size
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
the greater the value of n … (4)
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
energy lvl diagram & the exception
4s<3d
3 basic rules to write e- configuration
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
why are paired e- stable
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
2 anomalous e- configuration and why are they more stable ( general reason and indiv reason)
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
excited state
one or more e- absorb energy and are promoted to a higher energy lvl
are 4s e- lost before or after 3d e-? and why
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
isoelectronic species
species with the same total no. of e-
e- configuration from the periodic table
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
Defn of atomic radius
Half the shortest inter-nuclear dist found in the struc of the element
Variation in atomic radii across period
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
Variation in atomic radii down grp
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
Cationic radius vs atom
rcation
Anionic radius vs atom
ratom
Ionic radii of isoelectronic species across period
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
Why would there be a sharp increase in ionic radius from Al3+ to P3-
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+
Trend of 1st ionisation energies across the period
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
Irregularity in ionisation energies across a period (2)
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
Trend of 1st e- energies down a grp
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
Trend in successive ionisation energies of an element
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-
How to deduce grp no. from successive ionisation data
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
Defn of electronegativity
Relative measure of its ability to attract bonding e-
Trend in electronegativity across period
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
Trend in electronegativity down grp
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
Why is the 3rd IE of mg much higher than that of aluminium
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+
