7 - Periodicity

Question 1

left to right of the table elements are arranged in

Answer 1

increasing atomic number
- each successive element has one extra proton

Question 2

groups

Answer 2

vertical columns, where each element has the same number of electrons in the outer shell and similar properties

Question 3

end across a period of electron configuration

Answer 3

• across period 2, the 2s sub shells fill with 2 e- then the 2p sub shells with 6e-
  -across Ps, it is the same patter on filling for 3s and 3p

Question 4

what are blocks

Answer 4

it corresponds to their highest energy sub-shell

Question 5

first IE

Answer 5

energy required to remove one electron from each atom in one mole of gaseous atoms of an element to form one mole of gaseous 1+ ions

Question 6

factors affecting IE

Answer 6

Atomic radius- the greater the distance between the nucleus and e-, the less energy needed to remove the e-

nuclear charge - the more protons the greater the attraction between the nucleus and outer e-

electron shielding - the more shells the more shielding, therefore less attraction between nucleus and outer e-

Question 7

second IE
- compare to first IE of a group 7 element

Answer 7

energy required to remove one electron from each ion in one mole of gaseous 1+ ion of an element to form one mole of gaseous 2+ ions

• it is greater as the outer electron is pulled closer to the nucleus as there is more protons than electrons

Question 8

trends in IE down a group

Answer 8

• atomic radius increases
• more inner shells so shielding increases
• nuclear attraction on outer electrons decreases
• first IE energy decreases down a group

Question 9

trend in IE across a period

Answer 9

• nuclear charge increases
• same shells so similar shielding
• nuclear attraction increases
• atomic radius decreases
• first IE increases

Question 10

comparing IEs of beryllium and boron

Answer 10

• there is a fall in 1st IE from beryllium to boron as it marks the start of filling of the 2p sub shell
  -the 2p sub shell in boron has a higher energy than the 2s of beryllium.
  so the 2p electron in boron is easier to remove than the 2s in beryllium. so the 1st IE of boron is less than beryllium

Question 11

comparing the 1st IE of nitrogen and oxygen

Answer 11

• the fall in 1st IE from nitrogen to oxygen marks the start of the pairing in the p orbitals of the 2p sub she;;
• in nitrogen and oxygen the highest energy e- is in the 2p sub shell
• in O the paired e- in one of the orbitals repel one another, making it easier to remove an e- form O compared to N
• so the 1st IE of oxygen is less than N

Question 12

what metals at RTP isn’t solid

Answer 12

mercury

Question 13

metallic bonding

Answer 13

o the electrostatic attraction between positively charged metal ions and delocalized electrons within a metal structure
-> each atom donates its valence e- to a shared pool of e- which are delocalised throughout the structure. positive ions are in a fixed position

Question 14

properties of metals

Answer 14

• strong metallic bonds
• high electrical conductivity
  ->Delocalized electrons are free to move , they carry electrical charge along the metal lattice

High Tm and Tb
- large amounts of energy is needed to overcome the strong electrostatic attraction between metal ions and e-

are not soluble
- any interaction with a polar solvent would cause a reaction

Question 15

giant covalent structures

Answer 15

characterized by a vast network of covalent bonds extending throughout the entire substance. These structures are typically composed of atoms with strong covalent bonds that form a continuous lattice structure, resulting in high melting and boiling points, as well as exceptional hardness and electrical conductivity properties.
B,S and C

Question 16

giant covalent structure of carbon and silicon

Answer 16

carbon (in its diamond form) and silicon use their 4 valence e- to form covalent bonds with other C or S
- this results in a tetrahedral structure - 109.5

Question 17

properties of giant covalent structures

Answer 17

High Tm and Tb
-> due to the strong covalent bonds, lots of energy is reuired to overcome them

Insoluble
-> covalent bonds are too strong to be broken by the weak interactions with the solvents

Conductivity
- mostly cant conduct as in diamond and silicon all 4 e- are involved in a bond and are not delocalised
-> in graphene and graphite they can conduct

Question 18

graphene

Answer 18

Graphene is a single layer of graphite, composed of carbon atoms arranged in a hexagonal lattice and linked by strong covalent bonds
-120
- only three of the 4 electrons in carbon is used to form covalent bonds. so the left over ones are released into a pool of delocalised electrons shared by all atoms of the structure

Answer 19

only three of the 4 electrons in carbon is used to form covalent bonds. so the left over ones are released into a pool of delocalised electrons shared by all atoms of the structure
- 120
- parallel layers of hexagonally arranged carbon atoms, they are bonded by week London forces

Question 20

periodic trend in melting points

Answer 20

across period 2 and 3
- melting point increases from group 1 to 14
- there is a sharp decrease in Tm between group 14 and 15
- the melting points are comparatively low from group 15 to 18

the sharp decrease indicates the change from giant to simple molecular structures