Periodicity ibdp

Question 1

Period number corresponds to

Answer 1

principal quantum number, n of the highest occupied energy level in the elements of the period

Question 2

trends across period

Answer 2

Across a period, the chemical properties of the element gradually change from those of reactive metals to metalloids and non—metals (except for period 1).

Question 3

metalliods

Answer 3

They are poor conductors of electricity and, unlike metals, their conductivity increases with temperature. Some metalloids form amphoteric oxides.

Question 4

how to determine number of valence e

Answer 4

can b. found from the grp number of the s- & p-block elements

Question 5

different names for the elements

Answer 5

Main Group Elements: Group 1 (excluding H) and Group 2 and Groups 13 to 18.
Transition Elements: Groups 3 — 11
s—block Elements: Group I and 2 and He
p—block Elements: Group 13 to Group 18 (except He)
d—block Elements: Group 3 to 12 (include La-57 and Ac-89)
f—block Elements: Elements from Z=58 to Z=71 and from Z=90 to Z =103
Lanthanoids: Elements from Z = 57 to Z = 71 (many are radioactive)
Actinoids: Elements from Z = 89 to Z = 103.
the s-,p-,d-blocks are based on which subshell the valence e is occupying

Question 6

periodicity

Answer 6

repeating pattern of atomic, physical and chemical properties

Question 7

atomic radius

Answer 7

half of the distance between the nuclei of neighbouring atoms in the pure element, d/2

Question 8

atomic radius in noble gas is called

Answer 8

non-bonding atomic radius or the van der waal’s radius

Question 9

atomic radius in non-metals is called

Answer 9

boning atomic rasius aka covalent radius

Question 10

atomic radius in a metal is called

Answer 10

metallic radius

Question 11

factors affecting atomic radius

Answer 11

1. nuclear charge: nuclear charge is the total charge of the protons in the nuc(same a the no. of protons), thus the greater the no. of prootons in the nuc, the greater the efoa on the e, pulling them closer to the nuc, thus shorter atomic radius
2. shielding effect by inner e, e in the inner shells rpel the valence e, giving rise to the shielding effect–> usually given by the no. of es in the inner quantum shells. the core es in the inner non-valence energy levels of an atom reduce the positive nuclear charge experienced by a valence e, thus the greater the shielding effect, the lower efoa on the valence e thus the longer the atomic radius

Question 12

Effective nuclear charge

Answer 12

difference btwn nuclear charge & shielding effect. it gives a measure of the actual nuclear charge experience by valence electrons, hence how tightly the electrons r attracted to the nucleus

Question 13

Periodic trends in atomic radius–> across a period

Answer 13

• Nuclear charge ⬆️es but shielding effect remains relatively constant since the inner quantum shells of es remain the same
• effective nuclear charge increases
• this attracts the valence e closer to the nuc
• atomic radius ⬇️es

Question 14

Periodic trends in atomic radius–> down a grp

Answer 14

• in each new period the valence e enter a new energy lvl and r further away from the nuc
• the ⬆️sed dist reduces the efoa btwn the protons in the nuc & the val e
• thus atomic radius increases

Question 15

Periodic trends in ionic radius–> cations

Answer 15

The radii of cations are always smaller than their parent atoms.
- Number of protons, or nuclear charge, remains the same but there are fewer electrons in the cation.
- Hence, the valence electrons are more strongly attracted to the nucleus.
- Therefore, the ionic radii of positive ions are smaller than the corresponding atomic radii.
Across the period, the IONS contain the same number of electrons (isoelectronic), but an increasing number of protons. Consequently, the proton to electron ratio increases and the outermost electrons are drawn increasingly closer to the nucleus. Hence, the cationic radius decreases.

Question 16

Periodic trends in ionic radius–> anions

Answer 16

The radii of anions are always larger than their parent atoms.
- When one or more electrons are added to an atom, there is an increase in the repulsion between the electrons. Hence, the electron cloud increases in size.
- The number of protons, or the nuclear charge, in the ion remains the same as that of the atom. Hence, the ionic radii of negative ions are larger than the corresponding atomic radii.
- Anions have one more occupied quantum shell of electrons than the cations, hence, anionic radii are larger than cationic radii within the same period.
Across the period, the anionic radius decreases as the proton to electron ration increases

Question 17

First IE

Answer 17

minimum energy required in removing one mole of valence electrons from one mole of gaseous atoms to form 1 mole of singly positively charge gaseous ion.
- always +ve as energy is always req/absorbed to overcome the efoa btwn the e to b removed & the protons in the nuc
Na (g) Na+ (g) + e- –> 1st I.E. = +496 kJ mop

Question 18

Periodic trends in IE–> across a period

Answer 18

• There is an increase in effective nuclear charge, Zeff. This is because nuclear charge increases but shielding effect remains relatively constant since the inner quantum shells of electrons remain the same.
• The valence electrons are drawn closer to the nucleus, so the electrostatic attractions between the valence electrons and the nuclei increase.
• More energy is required to remove a valence electron from the gaseous atom, thus IE generally increases across a period

Question 19

Periodic trends in IE–> down a group

Answer 19

• The number of filled quantum shells increases and the size of the atoms increase.
• The valence electrons occupy energy levels that are increasingly further from the nucleus.
• This increased distance reduces the electrostatic attractions between the protons in the nuclei and the valence electrons.
• Hence the electrostatic forces of attraction of the nucleus for the valence electrons decreases and less energy is required to remove them so IE generally decreases down a grp

Question 20

Exceptions to periodic trends in IE

Answer 20

Periods 2 and 3:

• The group 13 element has a lower first ionisation energy than the group 2 element.
• The group 16 element has a lower first ionisation energy than the group 15 element.

Example 1:
Boron(grp 13) has lower first ionisation energy (+801 kJ mol-I) than that of beryllium(grp 2) (+900 kJ mol-I).
The electronic configuration of beryllium and boron atoms are,
Be 1s2 2s2
B 1s2 2s2 2p1
It is because less energy is required to remove a 2P electron in the boron atom than a 2s electron in the beryllium atom as the 2P sub-level is of higher energy than the 2s sub—level and the 2P electron is further from the nucleus. The 2P electron also experiences shielding by the 2s electrons. As a result, the 2P electron experiences a weaker electrostatic force of attraction from the nucleus.
Example 2:
Oxygen(grp16) has a lower first ionisation energy (+1314 kJ mol-I) than that Of nitrogen (+1402 kJ mol-I)(grp 15)
The electronic configs are,
N : 1s2 2s2 2p3
O: 1s2 2s2 2p4
It is because in an oxygen atom, there are two electrons occupying the same 2P orbital and this gives rise to inter- electronic repulsion. Thus, less energy is required to remove a paired 2P electron from an oxygen atom compared to the energy required to remove an unpaired 2P electron from a nitrogen atom.

Question 21

Electronegativity

Answer 21

relative attraction that an atom has for the shared pair of electrons in a covalent bond
(noble gases dh eletrongeativity!!)

Question 22

highly electronegative elements

Answer 22

fluorine(F), oxygen(O), chlorine(Cl) & Nitrogen(N)

Question 23

most electronegative element

Answer 23

fluorine, value of 4.0–> the atomic radius is vry small, thus the attraction btwn the -vely charged e & +vely charged e makes it very strong

Question 24

Periodic trends in electronegativity–> across a period

Answer 24

• The atoms get smaller, resulting in decreased distance between bonding electrons and the nuclei.
• Nuclear charge increases but shielding effect remains relatively constant since the inner quantum shells of electrons remain the same.
• Effective nuclear charge increases.
• As a result, electrostatic attraction between the bonding electrons and the nuclei increases.
• Therefore, electronegativity increases.

Question 25

Periodic trends in electronegativity–> down a group

Answer 25

• The atoms get larger, resulting in increased distance between bonding electrons and the nuclei.
• This increase in distance results in a decrease in electrostatic attraction between the bonding electrons and the nuclei of the atoms.
• Therefore, electronegativity decreases.

Question 26

Electron affinity

Answer 26

The first electron affinity, I st EA., is the enthalpy change when one mole of gaseous atoms acquires one mole of electrons to form one mole of singly negatively charged gaseous ions.
For example: S (g) + e- —> S- (g) —> 1st E.A = -200 kJ/mol
The first electron affinity is exothermic for most elements. Explanation:
• As the electron is added to the atom, there will be electrostatic attraction between the protons in the nucleus and the incoming electron.
• Energy is released when the nucleus attracts the electron which is in the outer shell (similar to bond forming).

Question 27

Periodic trends in EA–> down a group

Answer 27

• Atomic radius increases
• Decrease in the attraction between the nucleus and the added electron as the electron is brought to a shell which is further from the nucleus.
• Therefore, electron affinity becomes less exothermic.
  Going down group 17 the first electron affinity becomes less exothermic i.e. less negative (with the exception of fluorine). Group 17 elements have the most negative first E.A. values.

Question 28

Why is the first EA of fluorine less exo than chlorine?

Answer 28

bcos fluorine is a vry small atom, only hv 9 protons and 2 shells, thus when addig electrons, adding it to a vry corwded space filled w many es thus giving rise to inter-electronic repulsion btwn the incoming e & the es tht r alrdy present in the atom–> make the overall attraction lesser than expected thus 1st ea is lower

Question 29

Periodic trends in EA–> across a period

Answer 29

• Increase in effective nuclear charge
• Decrease in atomic radius
• Hence, the added electron will be more strongly attracted by the nucleus. Therefore, the first electron affinity becomes more exothermic.
  Nitrogen has a less exothermic first electron affinity than oxygen because of its electronic configuration. Nitrogen has three unpaired electrons in the 2P sublevel. When an electron is added to a nitrogen atom, the incoming electron “pairs up” with an electron in the p—orbital.
  This introduces an extra repulsion component that is not present in oxygen. Energy needs to be supplied to overcome the inter—electronic repulsion between the paired electrons(absorb energy–> energy enters– emdo) Hence, the process is less exothermic than expected.

The nuclear charges of fluorine and oxygen are larger than nitrogen. There is a stronger electrostatic force of attraction between the nucleus and the incoming electron. The energy released when the nucleus attracts an electron is larger than the energy taken in to overcome inter-electronic repulsion. Hence for fluorine and oxygen the first electron affinity values are exothermic.

Question 30

Second EA

Answer 30

The second electron affinity, 2nd EA., is the enthalpy change when one mole of singly negatively charged gaseous ions acquires one mole of electrons to form one mole of doubly negatively charged gaseous ions.

For example: M- (g) + e- —Y ME (g) 2nd E.A. = positive (kJ/mol)

It is always positive since the electron is now being added to a negative ion (anion). Energy needs to be supplied to overcome the repulsion between the two negatively charged species.

Question 31

normal melting point

Answer 31

the temperature at which a pure solid is in equilibrium with its pure liquid at one atmospheric pressure

Question 32

Melting

Answer 32

overcoming the forces of attraction between ions, atoms or molecules in different types of solids. If the forces of attraction between the particles are strong, the melting point is high; if the forces are weak, the melting point is low. Melting point depends on both the structure (packing) and the electrostatic forces of attraction between the particles.

Question 33

giant metallic structure(metals, grp1&2)–> bonding?

Answer 33

metallic bonds between cations & sea of delocalised e

Question 34

giant molecular structure(graphite, diamond)–> bonding?

Answer 34

covalent bonds thruout the entire 3-D network structure

Question 35

simple molecular structure(halogens)–> bonding?

Answer 35

instantaneous dipole-induced dipole forces btwn molecules

Question 36

monoatomic structure(noble gases)–> bonding?

Answer 36

instantaneous dipole-induced dipole forces btwn molecules

Question 37

Periodic trends in Melting Points

Answer 37

Across a period, the mp rises thru the metals and the metalloids and drops abruptly for the non-metals

Question 38

Periodic trends in mp for Na to Al

Answer 38

• Due to giant metallic structure
• Melting point increases from Na to Al because the metallic bond strength increases from Na to Al.
  The increase in the metallic bond strength is due to:
• decrease in metallic radius,
• increase in the number of electrons donated per atom to the mobile sea of electrons.
  This results in an increase in the charge density of cations across the period. As a result the electrostatic attraction between the cations in the metallic and the delocalised electrons in the metallic lattice increases.
  More energy is required to break the stronger metallic bond as from Na to Al.

Question 39

Periodic trends in mp for Si

Answer 39

• Higher melting point than sodium, magnesium and aluminium.
• Due to giant molecular structure/ giant network
• Its high melting point is due to energy required to overcome many strong covalent bonds in the entire structure.

Question 40

Periodic trends in mp for P to Ar

Answer 40

• Low melting point.
• Due to simple molecular structures,
• These molecules / atoms are attracted to each other by instantaneous dipole — induced dipole (London (dispersion)) forces. Thus, less thermal energy is required to overcome the forces of attraction between them.
• As the relative molecular mass increases the instantaneous dipole - induced dipole (London (dispersion)) forces between the molecules also increase as the electron clouds of the molecules increases in size and hence in polarisability.
• More energy is required to overcome the increasing strength of the instantaneous dipole — induced dipole (London (dispersion)) forces down the group.
  Therefore, the melting points decreases in the following order: S8> P4 > Cl2 > Ar

Question 41

Periodic trends in mp for noble gases

Answer 41

The elements in Group 18 were once called the inert gases (inert meaning unreactive). However, since a few compounds of these elements have been prepared, they are now termed the noble gases. The melting and boiling points of the noble gases are very low because the instantaneous dipole — induced dipole (London (dispersion)) forces are weak. Down the group, the melting points and boiling points increase because there are more electrons that are easily polarized.

Question 42

MP & BP down group 1

Answer 42

• Melting point decreases.
• The strength of the metallic bond decreases as the cations in the metallic lattice increase in size down the group.
• This results in a decrease in the charge density of cations upon descending the group. As a result the electrostatic attraction between the cations in the metallic lattice and the delocalised electrons in the metallic lattice decreases
• Less energy is required to break the weaker metallic bond as we go down group 1.

Question 43

MP & BP down group 17

Answer 43

The increase in electron cloud size of the molecules increases in size and increases polarizability, hence increasing the strength of intermolecular instantaneous dipole - induced dipole (London (dispersion)) forces
More energy is required to overcome the increasing strength of the instantaneous dipole - induced dipole (London (dispersion)) forces down the group.

Question 44

Group 1 alkali metals properties trends down the grp

Answer 44

down the grp:

• ⬆️sing atomic & ionic rad
• ⬇️sing first ie
• ⬇️sing electronegativity
• ⬆️sing reactivity

Question 45

grp 1 metals characteristics

Answer 45

• soft, malleable
• low mp relative to transition metals
• highly reactive metals
• good ras( by losing an e)

Question 46

grp 1 rctns

Answer 46

react w water to form an alkaline sln & h2 gas. Reactivity
increases down the group. Lithium floats and reacts quietly. Sodium immediately reacts with water and darts around on the surface. The heat generated from the reaction between potassium and water is sufficient to ignite the hydrogen gas; a lilac flame is observed.
2K (s) + 2H20(l) 2KOH (aq) + H2(g)
They all react with reactive non-metals (e.g. halogens) to form ionic compounds.
Sodium burns in a gas jar of chlorine gas to form white fumes Of sodium chloride.
Na (s) + Cl2(g) –>2NaCl (s)

Question 47

group 17 halogens trends down the grp

Answer 47

• ⬆️sing atomic & ionic rad
• ⬇️sing first ie
• ⬇️sing electronegativity
• ⬇️sing reactivity
• ⬇️sing oxidising strength–> tendency to gain e & undergo rctn ⬇️es as it moves down the grp cos valence e of I will be further away from nuc than valence e of F–> attractive force btwn valence e n nuc will b smaller as compared to tht of F thus the tendency for halogen to gain e ⬇️es as well, thus oxidising strength ⬇️es and reactivity ⬇️es

Question 48

grp 17 characteristics

Answer 48

• electronegative elements, diatomic molecules
• highly reactive non-metals
• good as

Question 49

grp 17 rctns

Answer 49

• react w alkali metals to form salts containg halide ion, these salts r ionic cmpnds, white/colourless n solutble in water giving colourless, neutral slns
• presence of halide ions in sln–> detected by adding agno3 sln, the ag ions react w halide ions to form a ppt of silver halide which can b distinguished by their colour. the silver halides when exposed to UV or sunlight, rapidly decomposes to silver(grey)
  Ag+ (aq) + X-(aq) –> AgX(s)
  X: Cl, Br, I(AgCl white, AgBr cream, AgI yellow)
  halogens form covalent cmpnds w non-metals
  the oxidising strength of halogens decreases down the grp as the radii of the atom increases & the efoa btwn the protons in the nuc n the e ⬇️es. This means Cl. has the strongest affinity for es and will remove es from bromide and iodide ions.hence, it displaces bromine and iodine from their salts.

Question 50

Oxidies of Period3 elements

Answer 50

An oxide is formed when the element combines with oxygen usually with heating. The bonding of the oxides changes from ionic to covalent across the period because the electronegativity difference between the element and oxygen decreases. The oxides of sodium, magnesium and aluminium exist as giant ionic lattice structures. They have high melting points as the ionic bonds present in the crystal lattice are strong. As MgO
has a more endothermic energy than Na20, it has a stronger ionic bond strength and hence, a higher melting point than Na20. As Al2O3 has some extent of covalent character, it has a lower melting point than MgO.
Si02 has a giant molecular structure. It also has a high melting point, as the numerous strong silicon-oxygen covalent bond must be broken as the oxide melts.
The oxides of phosphorous and sulfur have simple molecular structures. Their melting points are low relative to that of the earlier elements (exist as solids, liquids and gases at standard conditions) as the instantaneous dipole-induced dipole (London (dispersion)) forces holding the molecules together require little energy to overcome. P4O6 / P4O10 has a higher melting point than
SO2 / SO3 as it has a larger electron cloud. Hence, P4O6 / Paolo’s electron cloud is more polarisable. More energy is required to overcome the stronger instantaneous dipole- induced dipole forces.
Expansion of the octet structure is possible in phosphorous and sulfur because of the energetically accessible 3d orbitals available for accommodating the extra electrons. Thus, expansion of the octet is only possible for elements from the third period onwards because elements in the first and second period do not possess any energetically accessible d orbitals in their valence shell.

Question 51

Nature of the oxides

Answer 51

As the non-metallic character of elements increases across the period, so will the nature of the bonding in the oxides and their acid-base characters. The ionic oxides are basic and the covalent oxides are acidic. Aluminium oxide is ionic with covalent character, so its oxide is amphoteric. The Na2O and MgO are ionic, basic oxides. Na2O reacts to give an aqueous solution of sodium hydroxide.
Na2O (s) + H20 (l) –> 2Na+(aq) + 2OH- (aq) –> pH 13-14
MgO is not soluble in water, due to its high lattice energy but reacts with it to give magnesium hydroxide which is sparingly soluble. The magnesium hydroxide formed dissolves to a small extent to give hydroxide ions in water.
MgO(s) + H20 (l) –> Mg(OH)2 (s)
Mg(OH)2 (s)–>Mg2+ (aq) + 20H- (aq) —> pH 8-9
Al2O3, is an ionic compound with partial covalent character. It does not dissolve in water, due to its high lattice energy. However, it will react with both acids and strong bases to form salts, hence an amphoteric oxide.

Reaction With acids: Al2O3 (s) + 6HCl (aq)–> 2AlCl3 (aq) + 3H20 (l)
Reaction with bases: Al203 (s) + 2NaOH (aq) + 3H20 (l) –> 2NaAl(OH)4 (aq)

Silicon oxide does not dissolve in water. However, it is an acidic oxide as it forms sodium silicate(lV) with hot concentrated sodium hydroxide.
SiO2 (s) + 2NaOH (aq) –> Na2SiO3 (aq) + H2O(l)

The oxides of phosphorus and sulfur are simple covalent molecules and therefore acidic. They react with water to form acidic solutions (pH 1- 2).

P4O6(s) + 6H20 (l) –> 4H3P03 (aq) phosphorous acid or phosphoric(lll) acid

P4O10 (s) + 6H20 (l)–> 4H3P04 (aq) phosphoric acid or phosphoric(V) acid

SO2 (g) + H2O (l)–> H2SO3 (aq) sulfurous acid or sulfuric(lV) acid

SO3 (l) + H2O (l) —> H2SO4 (aq) sulfuric acid or sulfuric(Vl) acid

Nitrogen forms many oxides ranging from N2O, NOₓ to N2O5. Of importance to the impact on the environment is NO, nitrogen monoxide and NO2, nitrogen dioxide. NO is a neutral gas and it is virtually insoluble in water. NO is oxidised in the atmosphere to NO2, an acidic oxide which reacts with water to form nitric acid which is responsible for acid deposition
2NO2 (g) + H2O (l)–> HNO2 (aq) + HNO3 (aq)