Atomic structure ibdp

Question 1

ᴬ X

ᶻ

Answer 1

A= total number of protons and neutrons(mass number)
Z= proton number
(A-Z)= number of neutrons

Question 2

Isoelectronic?

Answer 2

Atoms/ions hv the same no. of e

Question 3

Isotonic?

Answer 3

Atoms/Ions tht hv the same no. of neutrons

Question 4

Isotopes properties

Answer 4

same chemical properties, diff physical properties, hv diff nucleon no., relative isotopic masses & no. of neutrons

Question 5

heavier neutron

Answer 5

isotope w more neutrons

Question 6

why are isotopes of some elements radioactive?

Answer 6

nuclei of these atoms r unstable & breakdown spontaneously, when they break down, they emit radiation which may be alpha particles, beta particles and gamma rays

Question 7

uses of isotopes

Answer 7

• biochemical tracers in nuclear medicine for diagnostics, treatment and research
• used as “chemical clocks” in geological and archaeological dating
• In nuclear medicine, PET(position emission tomography) scanners give 3D images of tracer conc in the body & r used to detect cancers

Question 8

What is the relative atomic mass of an element (Ar)? + instrument used to calculate it

Answer 8

its the weighted average atomic mass of its naturally occurring isotopes, modern mass spectrometer is used to determine the relative atomic mass of an element from its isotopic composition Each peak in the mass spectrum
indicates an isotope of the element and m/z values can be used to identify the isotopes. The height of the peak in the mass spectrum indicates the relative abundance of the isotopes.

Question 9

Mass spectrometer in detail:

Answer 9

Mass spectrometry is an analytical technique which allows the determination of relative isotopic masses, relative molecular masses and structural features of organic compound as well as identification of unknown compounds, e.g. in forensic science. It allows chemists to weigh atoms & molecules by measuring how ions formed from them are accelerated or deflected by electric and sometimes magnetic fields

Question 10

Mass spectrometer steps

Answer 10

1. Vaporisation
2. ionisation
3. acceleration
4. deflection–> if it has no charge it cannot b deflected and detected
5. Detection

Question 11

Electromagnetic spectrum

Answer 11

The arrangement of all the electromagnetic radiations in the increasing order of their wavelengths or decreasing order of their frequencies, rmivuxg, energy is inversely proportional to wavelength, and is directly proportinal to frequency

Question 12

c=

Answer 12

lv

Question 13

E=

Answer 13

hc/l

hv

Question 14

According to Bohr’s model

Answer 14

the energy of the electron in an atom is quantized–> quantity cannot vary continuously to have any arbitary value but can change only discontinously to have specific or discrete values

Question 15

lowest energy state

Answer 15

ground state, n=1, electrons r closest to nucleus

Question 16

hydrogen emission spectrum

Answer 16

an experiment tht can be conducted to give evidence of electron exhibiting discrete/different energy levels, its a line spectrum n the lines in the spectrum correspond to the photons of a particular wavelength(frequency). for each element, the emission spectrum is a characteristic and unique to the element & it can be used to identify the element, thus it provides crucial evidence for the existence of electrons in discrete energy levels which converges at higher energies/frequencies

Question 17

What is a electronic transition

Answer 17

The electron in the hydrogen atom may absorb energy and undergo an electronic transition to higher energy levels, e at higher energy level said to be in an excited state. When the electron returns to a lower energy lvl, energy is emitted in the form of light/photon

Question 18

orbitals having equal energy r termed

Answer 18

degenerate

Question 19

when is an electron in an atom stable

Answer 19

when the total attractive interactions are greater than the total repulsive interactions, most stable–. lowest orbital bcos thy r the closest to the nucleus and hence experience the strongest attractive force to the nucleus

Question 20

s orbitals

Answer 20

spherical in shape and thy r non-directional, for eveery quantam shell, only one s orbital and the size of the s orbital increases w the principal quantam shell

Question 21

p orbitals

Answer 21

dumb-bell shaped and they are directional, the 3 p orbitals hv diff directions in space, size also increases.w principal quantam shell

Question 22

aufbau principle

Answer 22

electrons r added progessively to the orbitals starting w the lowest energy, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s etc
exception: Cr and Cu bcos of syymetrical arrangement (thrs extra-stability in half-filled and fully-filled electronic configs)

Question 23

Pauli exclusion principle

Answer 23

Each orbital can hold a max of 2 es, paired es can only b stable when they spin in opp dorections so tht the magnetic attraction which results from their opp spins counterbalances the electrical repulsion

Question 24

Hund’s rule

Answer 24

When filling a sub-level, each orbital must b occupied singly b4 thy r occupied in pairs, can only pairup when each sublevel is alrdy half-filled

Question 25

electronic config of ions

Answer 25

ions r formed whn the atom loses or gains e, when forming cations–> es r lost frm the orbital w the highest e, anions r formed by adding es to vacant orbital of highest energy

Question 26

First IE

Answer 26

minimum energy req in removing one mole of valence e from one mole of gaseous atoms to form 1 mole of singly positively charge gaseous ions, endothermic as energy is req/absorbed to remove an e!

Question 27

factors affecting IE

Answer 27

1. nuclear charge: greater the nuclear charge, greater the ionisation energy
2. shielding effect by inner electrons, the greater the shielding effect, the lower the ie

Question 28

effective nuclear charge

Answer 28

diff btwn nuclear charge and shielding effect

Question 29

succesive ie

Answer 29

succesive ie of an atoms increase w the removal of ech e bcos of the increasing amt of energy req to remove succesive es from an increasingly +ve ion due to an increasing eletrostatic foa btwn the nuc n valence e

Question 30

trends in 1st ie

Answer 30

across a period, thr is an increase in eff nuclear charge bcos nuc charge increases but shielding effect remains relatively constant since the inner quantam shells of es remain the same, thus valence e r drawn closer to the nuc, so the efoa btwn the valence e and the nuc oncreases thus ie increase bcos more energy is req to remove a valence e frm the gasesous atom

down a grp, the no. of filled quantam shells increase & size of atoms increase thus valence e occuoy wnergy levels tht r increasingly further frm the nuc, this increased dist reduces the efoa btwn the protons in the nuc and the valence e, hence the efoa of the nuc for the valence e decreases n less energy is req to remove them, thus down the grp ie decreases

Question 31

deviation from trends in 1st ie

Answer 31

between mg:1s2 2s2 2p6 3s2 and al:1s2 2s2 2p6 3s2 3p1, less energy is req to remove a 3p e in the al atom than a 3s e in the mg atom as 3p>3s in energy levels thus 3p further frm nuc + 3p e exps shielding by the 3s , thus the 3p e exps a weaker efoa frm the nuc so less energy to remove it, thus a decrease in 1st ie even tho al is right of mg in the periodic table

between P: 1s2 2s2 2p6 3s2 3p3 and S: 1s2 2s2 2p6 3s3 3p4, bcos in the S atom, thr r 2 es occupying the same 3p orbital and this gives rise to inter-electronic repulsion, thus less energy is req to remove a paired 3p e from a sulfur atom compared to the enrgy req to remove an unpaired 3p e from a phosphorus atom(which is also half-filled), thus even tho S is right of P in periodic table, IE decreases rather than increasing!