Chapter 2 Electrons in atoms

Question 1

Behaviour of Subatomic Particles in an Electric Field

Answer 1

• Protons, neutrons and electrons behave differently when they move at the same velocity in an electric field
• When a beam of electrons is fired past the electrically charged plates, the electrons are deflected very easily away from the negative plate towards the positive plate
• A beam of protons is deflected away from the positive plate towards the negative plate
• A beam of neutrons is not deflected at all

Question 2

Behaviour of Subatomic Particles in an Electric Field according to weight

Answer 2

• proton deflects less due to its weight compared to electrons
• neutral particles are not deflected at all

Question 3

Atomic radius

Answer 3

• The atomic radius of an element is a measure of the size of an atom
• It is half the distance between the two nuclei of two covalently bonded atoms of the same type

Question 4

Atomic radii show predictable patterns across the Periodic Table

Answer 4

• They generally decrease across each Period
• They generally increase down each Group

Question 5

electron shell theory

Answer 5

• Atomic radii decrease as you move across a Period as the atomic number increases (increased positive nuclear charge) but at the same time extra electrons are added to the same principal quantum shell
• The larger the nuclear charge, the greater the pull of the nuclei on the electrons which results in smaller atoms
• Atomic radii increase moving down a Group as there is an increased number of shells going down the Group
• The electrons in the inner shells repel the electrons in the outermost shells, shielding them from the positive nuclear charge
• This weakens the pull of the nuclei on the electrons resulting in larger atoms

Question 6

Ionic radius

Answer 6

• The ionic radius of an element is a measure of the size of an ion
• Ionic radii show predictable patterns
• These trends can also be explained by the electron shell theory

Question 7

These trends can also be explained by the electron shell theory (ionic radius)

Answer 7

• Ions with negative charges are formed by atoms accepting extra electrons while the nuclear charge remains the same
• The outermost electrons are further away from the positively charged nucleus and are therefore held only weakly to the nucleus which increases the ionic radius
• The greater the negative charge, the larger the ionic radius
• Positively charged ions are formed by atoms losing electrons
• The nuclear charge remains the same but there are now fewer electrons which undergo a greater electrostatic force of attraction to the nucleus which decreases the ionic radius
• The greater the positive charger, the smaller the ionic radius

Question 8

Ionic radii show predictable patterns

Answer 8

• Ionic radii increase with increasing negative charge
• Ionic radii decrease with increasing positive charge

Question 9

Electron Shells: Basics

Answer 9

• The arrangement of electrons in an atom is called the electronic configuration
• Electrons are arranged around the nucleus in principal energy levels or principal quantum shells
• Principal quantum numbers (n) are used to number the energy levels or quantum shells
• The lower the principal quantum number, the closer the shell is to the nucleus
• The higher the principal quantum number, the lesser the energy of the shell
• Each principal quantum number has a fixed number of electrons it can hold
  n = 1 : up to 2 electrons
  n = 2 : up to 8 electrons
  n = 3 : up to 18 electrons
  n = 4 : up to 32 electrons

Question 10

Subshells

Answer 10

• The principal quantum shells are split into subshells which are given the letters s, p and d
• Elements with more than 57 electrons also have an f shell
• The energy of the electrons in the subshells increases in the order s < p < d

Question 11

Orbitals

Answer 11

The subshells contain one or more atomic orbitals
Orbitals exist at specific energy levels and electrons can only be found at these specific levels, not in between
–Each atomic orbital can be occupied by a maximum of two electrons

Question 12

This means that the number of orbitals in each subshell is as follows:

Answer 12

• s : one orbital (1 x 2 = total of 2 electrons)
• p : three orbitals ( 3 x 2 = total of 6 electrons)
• d : five orbitals (5 x 2 = total of 10 electrons)
• f : seven orbitals (7 x 2 = total of 14 electrons)

Question 13

Ground state

Answer 13

• The ground state is the most stable electronic configuration of an atom which has the lowest amount of energy
• This is achieved by filling the subshells of energy with the lowest energy first (1s)
• The order of the subshells in terms of increasing energy does not follow a regular pattern at n= 3 and higher

Question 14

Electron Configurations: principal quantum number

Answer 14

• indicates the energy level of a particular shell but also indicates the energy of the electrons in that shell
• A 2p electron is in the second shell and therefore has an energy corresponding to n = 2
• Even though there is repulsion between negatively charged electrons (inter-electrons repulsion), they occupy the same region of space in orbitals
• This is because the energy required to jump to successive empty orbital is greater than the inter-electron repulsion
• For this reason, they pair up and occupy the lower energy levels first

Question 15

electron orbitials

Answer 15

• Each shell can be divided further into subshells, labelled s, p, d and f
• Each subshell can hold a specific number of orbitals:
• –s subshell : 1 orbital
• –p subshell : 3 orbitals labelled px, py and pz
• –d subshell : 5 orbitals
• –f subshell : 7 orbitals

-Each orbital can hold a maximum number of

Question 16

Subshells & Energy

Answer 16

-The principal quantum shells increase in energy with increasing principal quantum number

• The subshells increase in energy as follows: s ＜ p ＜ d ＜ f
• —The only exception to these rules is the 3d orbital which has slightly higher energy than the 4s orbital, so the 3d orbital is filled before the 4s orbital

-All the orbitals in the same subshell have the same energy and are said to be degenerate
Eg. px, py and pz are all equal in energy

Question 17

p orbitals

Answer 17

• The p orbitals are dumbbell-shaped
• Every shell has three p orbitals except for the first one (n = 1)
• The p orbitals occupy the x, y and z-axis and point at right angles to each other so are oriented perpendicular to one another
• The lobes of the p orbitals become larger and longer with increasing shell number

Question 18

s orbitals

Answer 18

• The s orbitals are spherical in shape
• The size of the s orbitals increases with increasing shell number
• Eg. the s orbital of the third quantum shell (n = 3) is bigger than the s orbital of the first quantum shell (n = 1)

Question 19

Electron Configuration

Answer 19

• The electron configuration gives information about the number of electrons in each shell, subshell and orbital of an atom
• The subshells are filled in order of increasing energy

Question 20

Electron Configurations: Explained

Answer 20

• Electrons can be imagined as small spinning charges which rotate around their own axis in either a clockwise or anticlockwise direction
• The spin of the electron is represented by its direction

Question 21

Electron Configurations: SPIN-PAIR REPULSION

Answer 21

• Electrons will therefore occupy separate orbitals in the same subshell to minimize this repulsion and have their spin in the same direction
• Eg. if there are three electrons in a p subshell, one electron will go into each px, py and pz orbital

Question 22

Electron Configurations: what happens to electrons when they are paired

Answer 22

• Electrons are only paired when there are no more empty orbitals available within a subshell in which case the spins are the opposite spins to minimize repulsion
• Eg. if there are four electrons in a p subshell, one p orbital contains 2 electrons with opposite spin and two orbitals contain one electron only

Question 23

Free Radicals

Answer 23

• A free radical is a species with one or more unpaired electron
• The unpaired electron in the free radical is shown as a dot
• Eg. a chlorine free radical has the electron configuration 1s^2 2s^2 2p^6 3s^2 3p^5
• Two of the three p orbitals have paired electrons whereas one of them has an unpaired electron

Question 24

Determining Electronic Configurations

Answer 24

• Writing out the electronic configuration tells us how the electrons in an atom or ion are arranged in their shells, subshells and orbitals
• This can be done using the full electron configuration or the shorthand version
• Ions are formed when atoms lose or gain electrons
• The Periodic Table is split up into four main blocks depending on their electronic configuration

Question 25

shorthand version

Answer 25

• The full electron configuration describes the arrangement of all electrons from the 1s subshell up
• The shorthand electron configuration includes using the symbol of the nearest preceding noble gas to account for however many electrons are in that noble gas

Question 26

Ions are formed

Answer 26

when atoms lose or gain electrons

• Negative ions are formed by adding electrons to the outer subshell
• Positive ions are formed by removing electrons from the outer subshell
• The transition metals fill the 4s subshell before the 3d subshell but lose electrons from the 4s first and not from the 3d subshell (the 4s subshell is lower in energy)

Question 27

The Periodic Table is split up into four main blocks depending on their electronic configuration

Answer 27

• s block elements (valence electron(s) in s orbital)
• p block elements (valence electron(s) in p orbital)
• d block elements (valence electron(s) in d orbital)
• f block elements (valence electron(s) in f orbital)

Question 28

Exceptions to electron configurations

Answer 28

• Chromium and copper have the following electron configurations:
• Cr is [Ar] 3d^5 4s^1 not [Ar] 3d^4 4s^2
• Cu is [Ar] 3d^10 4s^1 not [Ar] 3d^9 4s^2
• This is because the [Ar] 3d^5 4s^1 and [Ar] 3d^10 4s^1 configurations are energetically stable

Question 29

First Ionization Energy

Answer 29

• The ionisation energy (IE) of an element is the amount of energy required to remove one mole of electrons from one mole of atoms of an element in the gaseous state to form one mole of gaseous ions
• Ionisation energies are measured under standard conditions which are 298 K and 1 atm
• The units of IE are kilojoules per mole (kJ mol-1)
• The first ionisation energy (IE1) is the energy required to remove the first electron from an atom of an element

Question 30

Ionisation Energies: Equations

Answer 30

• The second ionisation energy (IE2) is the energy required to remove the second electron from each ion in a mole of gaseous +1 ions
• The third ionisation energy (IE3) is the energy required to remove the third electron from each ion in a mole of gaseous +2 ions
• The electrons from an atom can be continued to be removed until only the nucleus is left
• This sequence of ionisation energies is called successive ionisation energies

Question 31

Ionisation Energies: Trends

Answer 31

• Ionisation energies show periodicity
• As could be expected from their electronic configuration, the group I metals show low IE whereas the noble gases have very high IE’s
• The first ionisation energy increases across a period and decreases down a group caused by four factors that influence the ionisation energy

Question 32

four factors that influence the ionisation energy

Answer 32

• Size of the nuclear charge
• Distance of outer electrons from the nucleus
• Shielding effect of inner electrons
• Spin-pair repulsion

Question 33

factors that influence the ionisation energy: Size of the nuclear charge

Answer 33

the nuclear charge increases with increasing atomic number, which means that there are greater attractive forces between the nucleus and electrons, so more energy is required to overcome these attractive forces when removing an electron

Question 34

factors that influence the ionisation energy: Distance of outer electrons from the nucleus

Answer 34

electrons in shells that are further away from the nucleus are less attracted to the nucleus so the further the outer electron shell is from the nucleus, the lower the ionisation energy

Question 35

factors that influence the ionisation energy: Shielding effect of inner electrons

Answer 35

the shielding effect is when the electrons in full inner shells repel electrons in outer shells preventing them to feel the full nuclear charge so the greater the shielding of outer electrons by inner electron shells, the lower the ionisation energy

Question 36

factors that influence the ionisation energy: Spin-pair repulsion

Answer 36

electrons in the same atomic orbital in a subshell repel each other more than electrons in different atomic orbitals which makes it easier to remove an electron (which is why the first ionization energy is always the lowest)

Question 37

Ionisation energy down a group due to factors

Answer 37

• Across a period the nuclear charge increases
• The distance between the nucleus and outer electron increases
• The shielding by inner shell electrons increases

Question 38

Successive ionisation energies of an element

Answer 38

• The successive ionisation energies of an element increase as removing an electron from a positive ion is more difficult than from a neutral atom
• As more electrons are removed the attractive forces increase due to decreasing shielding and an increase in the proton to electron ratio
• The increase in ionisation energy, however, is not constant and is dependent on the atom’s electronic configuration

Question 39

spin-par repulsion when removing electrons

Answer 39

• The first electron removed has a low IE1 as it is easily removed from the atom due to the spin-pair repulsion of the electrons in the 4s orbital
• The second electron is more difficult to remove than the first electron as there is no spin-pair repulsion
• The third electron is much more difficult to remove than the second one corresponding to the fact that the third electron is in a principal quantum shell which is closer to the nucleus (3p)
• Removal of the fourth electron is less difficult as the orbital is no longer full and there is less spin-pair repulsion

Question 40

Ionisation energy across a period: There is a rapid decrease in ionisation energy between the last element in one period and the first element in the next period caused by:

Answer 40

• The increased distance between the nucleus and the outer electrons
• The increased shielding by inner electrons
• These two factors outweigh the increased nuclear charge

Question 41

Ionisation energy across a period: There is a slight decrease in IE1 between beryllium and boron

Answer 41

• as the fifth electron in boron is in the 2p subshell which is further away from the nucleus than the 2s subshell of beryllium
• Beryllium has a first ionisation energy of 900 kJ mol-1 as its electron configuration is 1s^2 2s^2
• Boron has a first ionisation energy of 800 kJ mol-1 as its electron configuration is 1s^2 2s^2 2px^1

Question 42

Ionisation energy across a period: There is a slight decrease in IE1 between nitrogen and oxygen and phosphorus why?

Answer 42

due to spin-pair repulsion in the 2px orbital of oxygen

• Nitrogen has a first ionisation energy of 1400 kJ mol-1 as its electron configuration is 1s^2 2s^2 2px^1 2py^1 2pz^1
• Oxygen has a first ionisation energy of 1310 kJ mol-1 as its electron configuration is 1s^2 2s^2 2px^2 2py^1 2pz^1

Question 43

Ionisation Energies: Explained

Answer 43

• Energy is required to remove an outer shell electron as this involves breaking the attractive forces between the electron and the positively charged nucleus
• There are several factors which affect the magnitude of the ionisation energy:

Question 44

factors which affect the magnitude of the ionisation energy: nuclear charge

Answer 44

• Positive nuclear charge increases with increasing number of protons
• The greater the positive charge, the greater the attractive forces between the outer electron(s) and the nucleus
• More energy is required to overcome these forces so ionisation energy increases with increasing nuclear charge

Question 45

factors which affect the magnitude of the ionisation energy: shielding

Answer 45

• Electrons repel each other and electrons occupying the inner shells repel electrons located in shells further outside the nucleus and prevent them from feeling the full effect of the nuclear charge
• The greater the shielding effect is, the weaker the attractive forces between the positive nucleus and the negatively charged electrons
• Less energy is required to overcome the weakened attractive forces so ionisation energy decreases with increasing shielding effects

Question 46

factors which affect the magnitude of the ionisation energy: atomic/ionic radius

Answer 46

• The larger the radius, the greater the distance between the nucleus and the outer shell electron(s)
• Increasing distance weakens the strength of the attractive forces
• Larger atoms/ions also result in greater shielding due to the presence of more inner electrons
• Less energy is required to remove the outer shell electron(s) so ionisation energy decreases with increasing atomic/ionic radius

Question 47

factors which affect the magnitude of the ionisation energy: Spin-pair repulsion

Answer 47

• Spin pair repulsion occurs when the electron being removed is spin paired with another electron in the same orbital
• The proximity of the like charges of electrons in the orbital results in repulsion
• Less energy is required to remove one of the electrons so ionisation energy decreases when there is spin-pair repulsion

Question 48

Sodium

Answer 48

• For sodium, there is a huge jump from the first to the second ionisation energy, indicating that it is much easier to remove the first electron than the second
• Therefore, the first electron to be removed must be the last electron in the valence shell thus Na belongs to group I
• The large jump corresponds to moving from the 3s to the full 2p subshell
  Na 1s^2 2s^2 2p^6 3s^1

Question 49

Magnesium

Answer 49

• There is a huge increase from the second to the third ionisation energy, indicating that it is far easier to remove the first two electrons than the third
• Therefore the valence shell must contain only two electrons indicating that magnesium belongs to group II
• The large jump corresponds to moving from the 3s to the full 2p subshell
  Mg 1s^2 2s^2 2p^6 3s^2

Question 50

Aluminium

Answer 50

• There is a huge increase from the third to the fourth ionisation energy, indicating that it is far easier to remove the first three electrons than the fourth
• The 3p electron and 3s electrons are relatively easy to remove compared with the 2p electrons which are located closer to the nucleus and experience greater nuclear charge
• This is due to weakened shielding effects through the loss of three electrons
  The large jump corresponds to moving from the third shell to the second shell
  Al 1s^2 2s^2 2p^6 3s^2 3p1

Question 51

Ionisation Energies: Electronic Configuration

Answer 51

• Successive ionisation data can be used to:
  
  • Predict or confirm the simple electronic configuration of elements
  • Confirm the number of electrons in the outer shell of an element
  • Deduce the Group an element belongs to in the Periodic Table

By analyzing where the large jumps appear and the number of electrons removed when these large jumps occur, the electron configuration of an atom can be determined (more The double = jump in level