Topic 1: Atomic Structure and the Periodic Table Flashcards
describe the structure of an atom
consists of a small dense central nucleus consisting of neutrons and protons, surrounded by orbiting electrons in electron shells
what is the relative mass and charge of a proton
+1 charge
1 mass
what is the relative mass and charge of a neutron
0 charge
1 mass
what is the relative mass and charge of an electron
-1 charge
1/1835
define relative molecular mass
the average mass of a molecule relative to 1/12 of the mass of a carbon -12 atom
what is the atomic number
the number of protons in a nucleus of an atom
what is the mass number
the sum of protons and neutrons in an atom
define the term isotope
atoms of the same element with the same number of protons but different number of neutrons
define relative isotopic mass
the average mass of an isotope relative to 1/12 the mass of a carbon - 12 atom
define relative atomic mass
average mass of an atom of an element relative to 1/12 of the mass of a carbon - 12 atom
what is the x and y axis on a mass spectra graph
x axis - m/z which is the relative isotopic mass
y axis - relative abundance
how can mass spectra’s determine the relative molecular mass of a molecule
( mass 1 x abundance 1) + ( mass 2 x abundance 2)
divide all by 100
how would you predict the mass spectra for diatomic molecules
using the example : Chlorine has two isotopes. 35Cl has an abundance of 75% and 37Cl has an abundance of 25%. Predict the mass spectrum of Cl2.
Step 1: Express each percentage as a decimal
75% = 0.75 and 25% = 0.25
Step 2: Create a table showing all different Cl2 molecules
35Cl-35Cl 0.75 × 0.75 = 0.5625
35Cl-37Cl 0.75 × 0.25 = 0.1875
37Cl-35Cl 0.25 × 0.75 = 0.1875
37Cl-37Cl 0.25 × 0.25 = 0.0625
Step 3: Combine abundances for identical molecules
For 35Cl - 37Cl and 37Cl - 35Cl: 0.1875 + 0.1875 = 0.375
Step 4: Calculate relative abundances and molecular masses by dividing by the smallest value
35Cl-35Cl relative abundances 0.5625 / 0.0625 = 9
molecular mass 35 + 35 = 70
35Cl-37Cl relative abundance 0.375 / 0.0625 = 6
molecular mass 35 + 37 = 72
37Cl-37Cl relative abundance 0.0625 / 0.0625=1
molecular mass 37 + 37 = 74
x axis - molecular mass
y axis - relative abundance
what is ionisation
the process of removing one or more electrons from an atom
requires an input of energy and so is endothermic
define first ionisation energy
the energy required to remove 1 mole of electrons from each atom in 1 moles of gaseous atoms, to form 1 mole of gaseous 1+ ions
define successive ionisation energy
the energy required to remove a second electron from each ion in 1 mole of gaseous 1+ ions to give gaseous 2+ ions
give an example of the first ionisation energy of magnesium
Mg(g) ➔ Mg+(g) + e−
give an example of the second ionisation energy of magnesium
Mg+(g) ➔ Mg2+(g) + e−
explain how the number of protons influence ionisation energy
- atoms with more protons have a stronger positive charge
- this creates a stronger electrostatic force between the nucleus and outer electrons.
- so more ionisation energy is needed to remove the electron
explain how electron shielding influences ionisation energy
- the more shielding the less attractive forces are felt by the outer electrons
- this means tat the outer electrons are easy to remove with more shielding
- so more shielding less ionisation energy needed
explain how the electron sub shell from which the electron is removed from influences the ionisation energy
- electrons which have half filled or fully filled sub shells are stable
- so it would require more energy to remove an electron from it
- if the the were two electrons in an orbit then there would be electron on electron repulsion so the ionisation energy would be less
why does is there an increase in first ionisation energy across a period
- from left to right the atomic radii decreases
- and the nuclear charge increases due to more protons
- this creates a stronger electrostatic force of attraction.
- so there is an increase in ionisation energy across a period
why is there a decrease in ionisation energy down a group
- down a group the atomic radii increases.
- so there is more shielding down the group
- so the electrostatic force of attraction between the outer electron and the nucleus reduces
- so there is less ionisation energy needed to remove the electron
how does atomics emission spectra provides existence of quantum shells
- when electrons are heated they get energy which causes them to jump to higher energy levels.
- this causes them to emit electromagnetic radiation at specific frequencies
- since each element has a distinct electron configuration the frequencies emitted is distinct for each element
explain how successive ionisation energies provide evidence for quantum shells and which element they belong to
- when there is a large jump in successive ionisation energies it suggest that the shell changes
- for Mg the second ionisation energy is large due to a shell change
what are the number of electrons that can fill the first four quantum shells
(1S)- the first shell has the s sub shell so can hold 2 electrons
(2S,2P)- the second shell has s and p sub shells so can hold 8 electrons (2+6)
(3S,3P,3D)- the third shell has s, p and d sub shells so can hold 18 electrons (2+6+10)
(4S,4P,4D,4F)- the fourth shell has s,p,d and f sub shells so can hold 32 electrons ( 2+6+10+14)
how many electrons fit in each sub shell
s - 2
p - 6
d - 10
f - 14
what is an orbital
a region within an atom that can hold up to two electrons
with opposite spins
what is the shape of a s - orbital
- a spherical shape
what is the shape of a p - orbital
- a dumbbell shape
describe how electrons fill up in subshells
- electrons fill up the lowest energy orbitals first eg 1S then 2S
- electrons first occupy orbitals single before pairing up
- two electrons in the same orbital means the electrons have opposite spins to minimise electron - electron repulsion
what is periodicity
- a trend in element properties with increasing atomic number
describe the trends of melting and boiling points of elements in period 2 and 3, across the period,
-Li, Be and B (period 2)
- Na, Mg, Al ( period 3)
- the Bp increases due to larger positive charge , causes a smaller ionic radius and there is more delocalised electrons so there is a higher electrostatic force of attraction, so the Bp increase
- C (period 2)
- Si ( period 3)
- the Bp continues increases as carbon have giant covalent structure, where each atom is bonded to 4 other atoms in a tetrahedral structure with strong bonds linking them together, so a lot of energy is needed to over come the bonds so the Bp increases
- N, O, F ( period 2)
- P, S, Cl ( period 3)
- the Bp drops low due to these elements having simple molecular structures with weak induced dipole - dipole forces, so there require very little energy to overcome, so the Bp is law
- Ne (period 2)
- Ar ( period 3)
-has an even lower Bp as they don’t form any bonds and only have very weak induced dipole, dipole forces so minimal energy is required to overcome these forces between atoms
explain why the atomic radius decreases across a period
- protons increase across the period, so the nuclear charge increases
- this results in a stronger electrostatics attraction between the nucleus and outer electrons drawing the the electrons closer to the nucleus
- this causes the atomic radius to decrease across the period
explain the trends between ionisation energy across a period
- generally increases across a period
- this is due to the nuclear charge increases so the atomic radius decreases
- and the shieling stays the same
- so there is more electrostatic forces of attraction so more energy is needed.
- However in period 3, the ionisation energy drops between group 2 and 3, and between group 5 and 6
- In group 3, the electron is removed from a p orbital rather than an s orbital like in group 2.
-p orbitals have slightly higher energy than s orbitals, so the outermost electron is on average further from the nucleus.
-As a result, less energy is required to remove the outermost p electron from the group 3 element compared to removing the outermost s electron from the group 2 element.
-The drop between groups 5 and 6
-This drop occurs because:
-In group 5, the electron is removed from a singly occupied orbital.
In group 6, the electron is removed from an orbital containing two electrons.
-The paired electrons in the group 6 element experience greater electron-electron repulsion.
-As a result, less energy is needed to remove one of these paired electrons in the group 6 element compared to the unpaired electron in the group 5 element.
