Chemical periodicity

Question 1

Nuclear charge and effective nuclear charge

Answer 1

nuclear charge : Attraction between protons and electrons are electrostatic

effective nuclear charge: also known as the net nuclear charge . this is actual magnitude of positive charge that experienced by an electron in the atom.

This refers to when core electron surrounding the nucleus shield the valence electron from full nuclear charge. (core electron provide substantial shielding)

calculation of ENC : proton - core electron

trend : across the period ENC WILL increase because core electron remain the same but proton number increase

Question 2

shielding effect

Answer 2

shielding effect : also known as electron-electron repulsion

this is when an electron is partially shielded from positive charge of the nucleus by the other electrons

all electrons shield one another to some context, the most effective shielders are core electron ( due its directly between nucleus and valence electron, and it’s closest to the neutron)

Question 3

Atomic radii and it’s trends

Answer 3

Atomic radii : distance between the nucleus of an atoms and it’s valence shell

determined by :

• ENC
• number of shells the electrons occupy

Trend :
period : across the period, the number of shell with electron stay the same but ENC increases hence the trend of radii across the period will decrease.

left to right , no. proton and electron increases in atom, electrons are added to valence shell hence the electrons added are not able to shield the increased nuclear charge well, this lead to the increase of ENC , this made the valence electron feel increasingly attracted to the nucleus and drawn closer so the atomic size will reduce going across any period

group: down the group atomic radii will increase

The ENC throughout the group remains similar with same number of valence electrons present but the number of energy level containing electrons increase. with the number of shell increases this means that they will be more shielding effect present from nucleus. hence the distance from nucleus to valence will also be greater thus explaining why the atomic size down the group is bigger

Question 4

Ionisation energy

Answer 4

it is the minimum amount of energy required to remove a valence electron from a gaseous atom or ion forming a gaseous cation of a higher oxidation number

• to remove electron,
  attractive force between electron and nucleus must be overcome
• the greater the attraction between electron and nucleus the harder it is to remove the electron and the greater the ionisation energy

note:
- core is harder to remove as compared to valence
- more energy is needed to remove electron from cation due to higher ENC

• first IE is first electron that was removed, second IE is the second electron that was removed

Question 5

ionisation energy trend

Answer 5

Across the period : (metallic to non-metallic)
ENC increases across the period while the number of shell with electrons remains constant, so atomic size decrease , the larger the ENC, the smaller the size of the atoms means that the outer electrons will be more tightly held

down the group :
due to atomic size increase, number of shells with electron also increases , ENC is similar down the group. With greater separation between electron and nucleus this means there’s a weaker attraction between them and it will be easier to remove an electron as we move down a group

Question 6

Electron affinity

Question 7

electron affinity trends

Answer 7

Across the period :
Due to across the period, no. of shell with electron remain constant but there’s a increasing valence electron, ENC will increase across the period and the atomic size will decrease. this means that the nucleus will be able to attract another electron more easily across the period and more energy will be release hence electron affinity will increase across the period

Down the group:
Down the group the atomic radii will increase as ENC stay the same but number of shell present w electron increases. this means that the electron added has to enter valance shell that is more further way from the nucleus. since the valence shell bus now further away from the nucleus this means that there will be not much energy released as it is harder to attract electron now. hence Electron affinity will decrease down the group

Question 8

Metallic character

Answer 8

Across the period table, metallic character will decrease and down the group the metallic character will increase

• due to increased number of shell, valence electron are less strongly attracted to nucleus and lose electron readily ( metals tends to lose electron readily)

Question 9

Deviation of first IE

Answer 9

BE and B

• the 2p subshell is higher in energy and further away from nucleus as compared to 2S subshell
• so B 2P electron is easily removed then Be 2s electron

Hence B have a lower IE than Be

N and O
-the electron to be removed from 2p orbital, for N it’s 2P subshell is occupied by 3 electron

• for O it’s 2p subshell is occupied by 4 electron of these electron, one of it is paired with one another .
• removing one of these paired electron eases electron electron repulsion and thus I has a lower IE as compared to N

Question 10

Exception to electron affinity trends

Answer 10

noble gasses . these elements are stable and do not have a tendency to accept additional electron