atomic structure Flashcards
state relationship between charge and mass –>angle of deflection
The larger the charge of the particle, the stronger the attraction of the particle towards the oppositely-charged plate, the larger the angle of deviation. The larger the mass of the particle, the more difficult it is to cause it to deviate towards the oppositely-charged plate, the smaller the angle of deviation.
state the rules for writing electronic configurations
the orbitals of a subshell must be occupied singly by electrons of parallel spins before pairing can occur, each orbital can hold a max of 2 electrons with opposite spins
state the reason behind the anomalous electronic configurations of Cr and Cu. (write out the electronic config)
Cr: because the 3d and 4s orbitals are about equal in energy by the time chromium is reached. Thus, by having one electron each in the 3d and 4s orbitals, inter-electronic repulsion is minimised.
Cu: the fully filled 3d subshell is unusually stable due to the symmetrical charge distribution about the metal centre.
define an atom in an excited state
whereby one or more electrons absorb energy and are promoted to a higher energy level.
rules for anions and cation
anions: electrons are added to the next available orbital (added to 4s before 3d, 5s before 4d etc.)
cations: electrons removed from the orbital with the highest energy (4s before 3d, 5s before 4d etc.)
why? intially 4s orbital is at a lower energy level than the 3d orbital but because once the 3d electrons occupy the inner 3d orbital, they provide some shielding to the outermost 4s electrons. The 3d electrons repel the 4s electrons to a higher energy level.
define isoelectronic species
species which contain the same total number of electrons.
state and explain trend of atomic radii across a period and down a group
atomic radii decreases across a period. This is because across a period:
-no. of electronic shells remain the same
-no. of protons increases, nuclear charge increases,
-no. of electrons increases but since they are added to the same valence shell, shielding effect remains approx constant.
-effective nuclear charge increases.
-electrostatic attraction between nucleus and valence electrons increases,
-electron cloud size decreases
-atomic radii decreases
atomic radii increases down the group because down the group:
-no. of electronic shells increases.
-distance between nucleus and valence electrons increases
-shielding effect on valence electrons increases due to increase in inner electrons
-despite increasing nuclear charge, the electrostatic attraction between the nucleus and valence electrons decreases,
-electron cloud size increases
-atomic radii increases.
state trend of ionic radii between cations and parent atoms and that between anions and parent atoms.
radius of a cation is always smaller than that of the parent atom.
- both have the same no. of protons and thus the same nuclear charge.
- however, the cation has one less electronic shell and thus electrostatic force between nucleus and valence electrons is stronger, smaller electron cloud size, smaller radius.
radius of an anion is always larger than that of the parent atom.
- both have the same no. of protons and thus the same nuclear charge.
- however the anion has more electrons than the parent atom and thus experiences stronger electron-electron repulsion, leading to weaker electrostatic attraction between the nucleus and the valence electrons. Thus, the electron cloud size is larger and the radius of the anion is larger.
state trend of ionic radii of isoelectronic species + explain the significant increase in ionic radius from Al3+ to P3-.
the ionic radii of isoelectronic ions decreases across a period. This is because across a period, the proton number increases and thus the nuclear charge increases. Since they are isoelectronic species, they have the same number of electrons and would experience the same shielding effect. Hence, effective nuclear charge increases and electrostatic attraction between the nucleus and valence electron increases, causing a decrease in the size of the electron cloud. Hence, ionic radii of isoelectronic ions decrease across a period.
There is a significant increase in ionic radius from Al3+ to P3- because from Al3+ to P3-, shielding effect and nuclear charge increases. Number of electronic shells and the distance between the nucleus and the valence electrons increases also increases. Hence, despite the increasing nuclear charge, the electrostatic attraction between the nucleus and valence electron decreases, meaning the electron cloud is larger.
define first ionisation energy and write the corresponding equation
First IE is the energy required to remove 1 mole of electrons from 1 mole of gaseous M atoms to form 1 mole of gaseous M+ ions.
state the general trend of first IE across the period and the anomalies.
first IE generally increases across a period. Across the period:
- no. of shells remain the same
- no. of protons increase and hence nuclear charge increases
- no. of electrons increases but since the electrons are added to the same valence shell, shielding effect remains approx constant.
- effective nuclear charge increases
- electrostatic attraction between the nucleus and valence electrons increases, resulting in an increase in the energy required to remove an electron from the atom.
anomalies: [just look at yellow booklet to see if the elements who don’t fit the trend and explain accordingly] could be due to 2 reasons:
1. e.g. The 3p electron to be removed from Al is at a higher energy level than the 3s electron to be removed from Mg. Thus, less energy is required to remove the 3p electron from Al which is less strongly attracted to the nucleus.. Hence, Al has a lower first IE than Mg.
2. e.g. 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. Hence, due to inter-electronic repulsion between the paired electrons in the same orbital, less energy is required to remove the 3p electron from S. Thus, S has a lower first IE than P.
state the general trend of first IEs down a group
Down a group:
- no. of electronic shells increase
- distance between the nucleus and valence electrons increases
- shielding effect increases
- despite the increasing nuclear charge, electrostatic attraction between the nucleus and valence electrons decreases,
- less energy required to remove one valence electron from the atom
Account for the great decrease in 1st IE between Ne and Na and between Ar and K.
[since it is not a period kind of trend, use group explanation]
- Na has one more electronic shell than Ne
- hence distance between the nucleus and valence electrons is larger for Na
- shielding effect is also large for Na which has more inner electrons
- despite the increasing nuclear charge in Na, electrostatic attraction between the nucleus and valence electrons decreases, meaning less energy is required to remove a valence electron, causing Na to have a significantly lower IE than Ne.
trend of successive ionisation energies
increase. Once the first electron is being removed from the neutral atom, successive electrons are being removed from an ion of increasing positive charge which attracts the electrons more strongly, meaning more energy is required to remove each subsequent electron.
define electronegativity
electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electron.
trend in electronegativity across a period and down a group
across a period:
- no. of shells remain the same
- no. of protons increases, nuclear charge increases
- no. of electrons increases but since they are added to the same valence shell, shielding effect remains relatively constant.
- electrostatic attraction between nucleus and bonding electrons increases, more electronegative
down a group:
- no of electronic shells increase
- distance between the nucleus and bonding electrons increases
- shielding effect experienced by bonding electrons increases
- despite the increasing nuclear charge,
- electrostatic attraction between the nucleus and bonding electrons decreases, less electronegative
explain why ionic radius of Mg2+ is less than that of Na+.
Both have the same number of electronic shells
Same number of electrons and thus experience the same shielding effect. However, Mg2+ has more protons than Na+ and thus has a higher nuclear charge. Thus, overall, Mg2+ has a higher effective nuclear charge. Electrostatic attraction between nucleus and valence electrons is stronger for Mg2+, causing the size of the electron cloud to be smaller and the ionic radius of Mg2+ to be less than that of Na+.
