2 Atomic Structure

Question 1

Define an atom.

Answer 1

The smallest particle found in an element that can take part in a chemical reaction.

Question 2

Describe electrons.

Answer 2

Electrons are negatively charged and are concentrated in certain regions of space around the nucleus called orbitals. They do not occupy fixed positions or move in orbits around the nucleus. Each electron occupies a discrete energy level.

Question 3

Describe the nucleus.

Answer 3

Contains positively charged protons and uncharged neutrons which are collectively known as nucleons and it contains the bulk of the mass of an atom.

Question 4

Describe the location, relative mass and relative charge of the 3 sub-atomic particles.

Answer 4

The 3 sub-atomic particles found in an atom are the proton, neutron and electron. The proton and neutron is located in the nucleus while the electron is located around the nucleus. The proton, neutron and electron have a relative mass of 1, 1 and 1/1840. respectively. The proton, neutron and electron have a relative charge of +1, 0 and -1. respectively.

Question 5

Describe the direction of deflection and the angle of deflection of the sub-atomic particles. What affects the direction of deflection and the angle of deflection?

Answer 5

The proton is deflected towards the negatively charged plate, the neutron is not deflected and the electron is deflected towards the positively charged plate with a greater angle of deflection than that of the proton. The direction of deflection depends on the charge of the sub-atomic particle and the angle of deflection is proportional to the magnitude of the q/m ratio of the particle. The larger the charge, the stronger the attraction towards the charged plate, the greater the angle of deflection. The larger the mass of the particle, the more difficult it is to cause it to deviate towards the charged plastic, the smaller the angle of deflection.

Question 6

Define a nuclide.

Answer 6

A nuclide is any species of a given mass number and atomic number. It is named by its elemental name followed by its mass number.

Question 7

Define isotopes. (Recall: Law of Conservation of Mass)

Answer 7

Isotope of an element are atoms that contain the same number of protons and electrons but a different number of neutrons. [Hence, they have the same chemical properties, but different masses hence different physical properties (density, m.p.).]

Question 8

Describe how electrons are arranged in atoms (electronic structure). Describe the characteristics of each component. What effects does a greater value of n have? Define an orbital. What information does the electronic structure provide us with?

Answer 8

Around the nucleus are electronic shells which contain subshells which contain orbitals that contain electrons.

The electrons are arranged in a series of shells/energy levels and each shell is described by a number known as the principal quantum number, n, comprising of one or more orbitals.

An orbital represents a region of space in which there is a high probability of finding an electron. Each orbital can accommodate 2 electrons, has a distinctive geometrical shape and has the same energy as the electron it contains. When the value of n increases from 1, 2, 3 or 4, the further the shell is from the nucleus, the higher the energy level of the shell, the weaker the electrostatic attraction between the nucleus and the electron and the larger the size of the orbital (orbital becomes more diffused).

It provides us with the number of electrons in an atom or ion, the distribution of electrons and the energies of the electrons.

Question 9

Name the types of subshell for each principal quantum number, the number of orbitals in each subshell and the types of orbitals in each subshell.

Answer 9

1s
2s2p
3s3p3d
4s4p4d4f

s - 1 orbital (s)
p - 3 orbitals (Px, Py and Pz)
d - 5 orbitals (Dxz, Dxy, Dyz, Dz^2, Dx^2-y^2)
f - 7 orbitals

Question 10

Describe the s orbital.

Answer 10

s orbitals have a spherical shape and are non-directional as the electron density is not concentrated in any particular direction. As n increases, the s orbital of different shells have the same shape but their size increases and the orbitals become more diffuse. (Electron density nearer the nucleus)

Question 11

Describe the p orbital and what happens when the value of n increases. Explain the term degenerate.

Answer 11

p orbitals have a dumbbell shape and are directional as the electron density is concentrated in certain directions along the x, y and z axes. As n increases, the P orbital of different shells have the same shape but their size increases and the orbitals become more diffuse. The 3 p orbitals in the same subshell have the same energy/ are degenerate.

Question 12

Describe the d orbital.

Answer 12

Dxy, Dxz and Dyz orbitals have a similar 4-lobed shape with their lobes pointing between the axes. The Dx^2-y^2 orbital has a 4-lobed shape but it has its lobes aligned along the x and y axes. The dz^2 orbital consists of a dumbbell surrounded by a small doughnut-shaped ring at its waist. The orbital is aligned along the z-axis.

Question 13

Rank the relative energies of the orbitals in the same shell.

Answer 13

From lowest, s, p, d, f

Question 14

Explain why the energy of orbitals decreases with increasing atomic number and why the s orbitals are affected to a greater degree.

Answer 14

As atomic number increases, the energies of all the orbitals decrease as there is increased nuclear charge for the same orbital and increased electrostatic F.O.A. between the electrons and the nucleus. Due to their proximity to the nucleus, the spherically symmetrical s orbitals are affected to a greater degree than the more angular p and d orbitals, so their energies decrease more rapidly than other types of orbitals.

Question 15

3D before 4S?

Answer 15

WRONG!!!!!!!

Question 16

State the formula used to calculate the number of subshells, number of orbitals and the maximum number of electrons that can be accommodated in an electronic shell.

Answer 16

Number of subshells = n
Orbitals = n^2
Maximum number of electrons that can be accommodated = 2n^2

Question 17

State 3 rules to writing electronic configurations using the s, p, d, f notation and the electron-in-boxes diagram.

Answer 17

(i) The Aufbau Principle
(ii) Hund’s Rule
(iii) The Pauli Exclusion Principle

AHP - A Harry Potter

Question 18

Explain the Aufbau Principle. Why do electrons fill the 4s orbital before the 3d orbital?

Answer 18

Electrons fill orbitals from the lowest energy orbital upwards. They occupy the 4s orbital before the 3d orbital because the 4s orbitals are at a lower energy level than the 3d orbitals.

Question 19

Explain Hund’s Rule.

Answer 19

Orbitals of a subshell (same energy) must be occupied singly by electrons of parallel spins before pairing can occur as this helps to minimise inter-electronic repulsion.

Question 20

Explain the Pauli Exclusion Principle.

Answer 20

Each orbital can hold a maximum of two electrons and they must be of opposite spins. (Paired electrons can only be stable when they spin in opposite directions so that the magnetic attraction which results from their opposite spins can counterbalance the electrical repulsion which results from their identical charges.)

Question 21

Name the 2 elements which have anomalous electronic configuration and explain their characteristic.

Answer 21

Cr and Cu both have only one electron in their 4s orbital while the other electron is in the 3d orbital. This is because the actual electronic configurations more stable than the expected electronic configuration. In Cr, the 3d and 4s orbitals are about equal in energy hence by having one electron each in the 3d and 4s orbitals, inter-electronic repulsion is minimised. In the actual electronic configuration of Cu, the fully-filled 3d subshell is unusually stable due to the symmetrical charge distribution around the metal centre (doughnut-shape).

Question 22

What is the ground state and excited state?

Answer 22

An atom is the ground state when the electrons are in the orbitals of the lowest available energy. This occurs at room temperature. When an atom is in the excited state, one or more electrons absorb energy and are promoted to a higher energy level. Such excited atoms are unstable and can emit energy to return to the ground state (Electrons in the 1s orbital are usually not promoted as the energy gap between n = 1 and n=2 is too large).

Question 23

With regard to the electronic configuration of ions, what happens when a cation or anion is formed?

Answer 23

During the formation of anions, electrons are added to the next available orbital in the atom. During the formation of cations, the electrons are first removed from the orbitals with the highest energy (as they are already of high energy levels and are easy to remove).

Question 24

Explain why electrons are removed from the 4s orbital before the 3d orbital in the formation of cations?

Answer 24

Although electrons occupy the 4s orbital first because the 4s orbital is at a lower energy level than the 3d orbitals. Once the electrons occupy the inner 3d orbitals, the electrons in the 3d orbital provide some shielding effect for the outermost 4s electrons and repel the 4s electrons to a slightly higher energy level. Hence, in the formation of cations, the 4s electrons are lost before the 3d orbitals (Always figure out the original electronic configuration before modifying it to give the ionic electronic configuration).

Question 25

What are isoelectronic species?

Answer 25

Species with the same total number of electrons.

Question 26

Recall how to come up with the electronic configuration of atoms from the periodic table using period, group number and principal quantum number

Answer 26

Horizontal rows in the periodic table are called periods and elements in the same period have the same number of electronic shells / n. The period number of an element is the number of electronic shells occupied with electrons in the atom.

The vertical columns in the periodic table are called groups and elements in the same group have the same number of valence electrons and hence similar outer electron configurations.

Group 1 and 2 are in the s block, Group 13 to 18 are in the p block and Group 3 to 12 are in the d block.

Valence electronic configurations of:
Group 1 elements - ns1
Group 2 elements - ns2
Group 13 elements - ns2np1
Group 18 elements - ns2np6
Group 3 to 12 elements -

Question 27

State and explain the main factors that affect the strength of the electrostatic attraction between the nucleus and the electrons? (Note that these factors can also be used to explain other trends and variations in first ionisation energies and electronegativity of an element across a period, down the group)

Answer 27

(i) Number of electronic shells: The electrons are arranged in a series of shells/energy levels and each shell is described by a number known as the principal quantum number n. When the number of shells increases, the principal quantum number n of the valence shell, the distance between the nucleus and the valence electron increases, the weaker the electrostatic attraction between the nucleus and the electron.
(ii) Nuclear charge Z: If the number of protons increases, Z increases and the electrostatic force of attraction between the nucleus and valence electrons increases.
(iii) Shielding (screening) effect by inner electrons: If the number of inner shell electrons increases, the shielding effect experienced by the valence electrons increases (are repelled) and the electrostatic force of attraction between the nucleus and the valence electron decreases (cannot experience the full effect of Z). *Electrons in the same shell provide very poor shielding effect for one another.

Question 28

Define effective nuclear charge and explain its effect on the electrostatic force of attraction between the nucleus and the electrons.

Answer 28

The effective nuclear charge is the resultant positive charge experienced by the valence electrons in a multi-electron atom taking into consideration the shielding effect of the inner electrons where Zeff = Z-S (Z is the proton number and S is the shielding effect which remains constant when comparing atomic properties across a period). If the effective nuclear charge increases, the electrostatic force of attraction between the nucleus and the electrons increases.

Question 29

Describe and explain the trend in the shielding ability of electrons in different orbitals.

Answer 29

From strongest, s, p, d, f
This is because d and f orbitals are rather diffuse and provide a poor shielding effect.

Question 30

Define atomic radius.

Answer 30

It is defined as half the shortest inter-nuclear distance found in the structure of the element.

Question 31

What is the observed atomic radius dependent on?

Answer 31

The observed radius is dependent on how the atom bonds or interacts with its neighbours.

Question 32

Define the three types of atomic radius and where they are usually found.

Answer 32

The metallic radius of an atom is half the inter-nuclear distance between the two neighbouring atoms in the metal. (Giant metallic lattice structure)

The covalent radius of an atom is half the inter-nuclear distance between the two covalently bonded atoms. (Simple/Giant molecular lattice structure)

The van der Waals’ radius of an atom is half the inter-nuclear distance between the two atoms which are not chemically bonded. (Monoatoms, Nobel gases)

Question 33

Describe the trend in atomic radii across a period.

Answer 33

Atomic radii decreases across a period as the number of electronic shells remain the same while the number of protons increases, hence nuclear charge increases. The number of electrons also increases but these electrons are added to the same outermost shell so the shielding effect remains approximately constant. As a result, effective nuclear charge increases and the electrostatic force of attraction between the nucleus and the electrons, resulting in a decrease in the size of the electron cloud and a decrease in atomic radii across a period.

Question 34

Describe the trend in atomic radii down the group. Does effective nuclear charge decrease?

Answer 34

Atomic radii increase down the group as the number of electronic shells increases and the distance between the valence electron and the nucleus increases. While the number of protons increases and the nuclear charge increases, the shielding effect increases. The electrostatic force of attraction between the nucleus and the electrons decreases, resulting in an increase in the size of the electron cloud and in atomic radii across a period. The effective nuclear charge does not decrease as both Z and S increase (unsure of extent).

Question 35

Compare the cationic radius with that of the parent atom.

Answer 35

The radius of a cation is always smaller than that of the parent atom. Both the parent atom and the cation have the same number of protons and hence have the same nuclear charge. However, the cation has one less electronic shell than its parent atom. Electrostatic attraction between the nucleus and the valence electrons increases, resulting in a decrease in the size of the electron cloud compared to the parent atom.

Question 36

Compare the anionic radius with that of the parent atom.

Answer 36

The radius of an anion is always greater than that of the parent atom. Both the parent atom and the cation have the same number of protons and hence have the same nuclear charge. However, the anion has one more electronic shell than its parent atom. Electron-electron repulsion increases and electrostatic attraction between the nucleus and the valence electrons decreases (electrons repelled further), resulting in an increase in the size of the electron cloud compared to the parent atom.

Question 37

Describe the trend in ionic radii of isoelectronic species across the period from Na+, Mg2+ to Al3+ / P3-, S2- to Cl-.

Answer 37

Ionic radii of isoelectronic species decrease across a period. The isoelectronic species experience the same shielding effect. However, the nuclear charge increases hence, effective nuclear charge increases. The electrostatic force of attraction between the nucleus and the electrons increases, resulting in a decrease in the size of the electron cloud and in ionic radii across a period.

Question 38

Explain why there is a sharp increase in the ionic radii from Al3+ to P3-.

Answer 38

From Al3+ to P3-, even though nuclear charge and shielding effect increases (we are unsure about effective nuclear charge), the number of electronic shells increases and the distance between the nucleus and the valence electrons increases. The electrostatic force of attraction between the nucleus and the electrons decreases, resulting in an increase in the size of the electron cloud and in ionic radii across a period.

Question 39

Define the first ionisation energy.

Answer 39

The first ionisation energy of an element is the energy required to remove one mole of electrons from one mole of gaseous M atoms to form singly charged gaseous cations (M+).

Question 40

Define the second ionisation energy.

Answer 40

The second ionisation energy of an element is the energy required to remove one mole of electrons from one mole singly charged gaseous cations (M+) to form doubly charged gaseous cations (M2+).

Question 41

Is ionisation an endothermic or exothermic process? Explain. How does ionisation energies affect the types of bonds being formed?

Answer 41

Ionisation is an endothermic process and ionisation energies are positive values since energy is absorbed during ionisation to overcome the attraction between the electron and nucleus. Ionisation energies affect the type of bond formed by the atom with other atoms. Elements with low ionisation energies will find it easy to lose an electron to form a cation, resulting in ionic bonds being formed.

Question 42

Describe the trend in first ionisation energies across a period. State and explain the two irregularities in the trend of the first ionisation energies.

Answer 42

The first ionisation energies of elements generally increase across a period. Across a period, the number of electronic shells remains the same while the number of protons increases, hence nuclear charge increases. The number of electrons increases but these electrons are added to the same outermost shell hence shielding effect remains approximately constant. The electrostatic force of attraction between the nucleus and the electrons increases, resulting in an increase in the energy required to remove the valence electron from an atom.

Group 2 and 13
3p electron to be removed from element A is of a higher energy level than the 3s electron to be removed from B. Hence, less energy is required to remove the 3p electron in A than the 3s electron in B. Hence, the first ionisation energy of element A is lower than B.

Group 15 and 16
3p electron to be removed from element A is a paired electron while that to be removed from element B is an unpaired electron. Due to inter-electronic repulsion between the paired electrons in the same orbital, less energy is required to remove the paired 3p electron from element A. Hence, the first ionisation energy of element A is lower than that of element B.

Question 43

Describe the trend in first ionisation energies down a group. Explain why there is a great decrease in ionisation energy from Ne to Na and Ar and K.

Answer 43

The first ionisation energies of elements generally decrease down a group. Down the group, the number of electronic shells increases, the distance between the nucleus and the valence electrons increases and shielding experienced by the valence electrons increases. Despite the increasing nuclear charge due to the increase in the number of protons in the elements, the electrostatic attraction between the nucleus and the valence electrons decreases, resulting in a decrease in the amount of energy required to remove the valence electron. Hence, the first ionisation energies of elements generally decrease down a group.

From Ne to Na and Ar and K, the electron removed from the group 1 element is from a larger electron shell compared to the preceding group 18 element. The distance between the nucleus and the valence electrons increases and shielding experienced by the valence electrons increases. Despite the increasing nuclear charge due to the increase in the number of protons in the elements, the electrostatic attraction between the nucleus and the valence electrons decreases, resulting in a decrease in the amount of energy required to remove the valence electron. Hence, the first ionisation energies of Na is much lower than Ne and that of K is much lower than Ar.

Question 44

Explain why the element at the lowest point of the nth IE trend is from group n for n = 1, 2 and group (10+n) for n > 2. What can be interpreted from the graph of variation of nth IE against atomic number?

Answer 44

For any nth IE element with n valence electrons, it still has an outer shell electron to lose. However, the preceding element with n-1 electrons loses an inner shell electron instead since it has already lost all its valence electrons. Thus, it has a much higher IE. The succeeding element will have a higher IE due to increasing effective nuclear charge. From the graph, the element with n valence electrons has a much lower nth IE than the preceding element. Hence, the element at the lowest point of the nth IE trend is from group n for n = 1, 2 and group (10+n) for n > 2.

Question 45

Describe the trend in successive ionisation energies of an element.

Answer 45

Successive ionisation energies of an element increases. Once, the first electron is removed from the neutral atom, each successive electron to be removed from an ion of increasing positive charge which attracts the electron more strongly.

Question 46

What information can a graph that shows the successive ionisation energies for an element provides us with?

Answer 46

The total number of electrons in an atom, the number of electronic shells (there is a large jump in ionisation energy - significantly more energy is required, when an electron is removed from a different shell, one that is inner and nearer to the nucleus and hence experiences a stronger electrostatic attraction to the nucleus e.g. 3 large jumps = 4 electronic shells).

Large jump - different shell
Moderate difference - removal of electrons from different subshells

Question 47

Define electronegativity. What is the significance of electronegativity?

Answer 47

The electronegativity of an atom in a molecule is the relative measure of its ability to attract bonding electrons to itself.

It can tell us the bond type in compounds. When two atoms of similar electronegativity are covalently bonded, a non-polar bond is formed. If the difference increases, the bond becomes more polar, becoming a polar covalent bond. With sufficient difference, the atoms will form an ionic bond instead.

Question 48

What is the electronegativity for noble gases?

Answer 48

Noble gases do not form compounds hence they are not studied for electronegativity.

Question 49

Describe the trend in electronegativity across a period.

Answer 49

Across a period, the number of electronic shells remains the same while the number of protons increases hence nuclear charge increase. The number of electrons also increases but these electrons are added to the same shell so the shielding effect remains approximately constant. Effective nuclear charge increase and electrostatic attraction between the nucleus and the valence electrons increases and hence electronegativity increases across a period.

Question 50

Describe the trend in electronegativity down the group.

Answer 50

Down a group, the number of electronic shells increases and the distance between the nucleus and the bonding* electrons increases hence shielding effect experienced by the bonding* electrons increases. Despite the increase in nuclear charge due to the increase in the number of protons, electrostatic attraction between the nucleus and the bonding* electrons decreases and hence electronegativity decreases across a period.

Question 51

Which elements are the least/most electronegative?

Answer 51

Metals are the least electronegative elements as reactive metals tend to give away electrons while non-metals are the most electronegative elements as they tend to take in electrons. They are usually located in the top right corner.

Question 52

Explain why the second ionisation energy of chromium is higher than that of manganese.

Answer 52

The second electron to be removed from manganese is a 4s-electron while the second electron to be removed from chromium is a 3d-electron which is closer to the nuclue and has lower energy. Hence, more energy is needed to remove it, causing its second ionisation energy to be higher.

Question 53

Explain why the fourth ionisation energy of Co is lower than that of iron.

Answer 53

The fourth electron to be removed from Co is removed from a 3d orbital containing a pair of electrons while the fourth electron to be removed from Fe is removed from a singly occupied 3d orbital. In this case, the removal of the fourth elecrton from Co is aided by inter-electronic repulsion so the fourth ionisation energy is lower although Co is a higher effective nuclear charge compared to Fe.

Question 54

Explain why Sr has a greater density than Ca.

Answer 54

Although the Sr atom is larger than the Ca atom, meaning that there will be fewer Sr atoms per unit volume compared to Ca, Sr has a higher molecular mass. Since density = mass/volume, the significantly larger molar mass of Sr causes its density to be larger than that of Ca.