Wjec Chemistry Ch5- Topic 16 P Block

Front 1

What is the order of filling of the sub shells for p-block elements?

Back 1

1s2,2s2,2p6,3s2,3p6,4s2,3d10 etc

Front 2

Why do elements at the bottom of groups 3,4 and 5 form cations with different charges to the normal +3,+4 and +5 respectively?

Back 2

The 1st 2 elements do as expected but beyond that only the outer p electrons are lost. This is known as the inert pair effect, because the ns2 electrons don't take part in bonding. Therefore you get ions like Pb4+ and Pb2+.

Front 3

Why is CCl4 immiscible with water whereas SiCl4 dissolves?

Back 3

Because the electronic configuration of carbon is 1s2,2s2,2p2 whereas Si in 1s2,2s2,2p6,3s2,3p2. SiCl4 will dissolve in water because it has empty 3d orbitals which water can attack where C doesn't have any free orbitals that can be used.

Front 4

What does Amphoteric mean?

Back 4

It means that a compound can display both acidic and basic properties.

Front 5

Give the Acidic and Basic equations for Al2O3.

Back 5

Al2O3 + 6H+ -> 2Al3+ + 3H2O
Al2O3 + OH- + H2O -> 2Al(OH)4-

Front 6

Give the Acidic and basic equations for PbO.

Back 6

PbO + 2H+ -> Pb2+ + H2O
PbO + OH- + H2O -> Pb(OH)4,2-

Front 7

Why are compounds like AlCl3, BCl3 and BF3 are considered electron deficient even though all their electrons are paired?

Back 7

Because when e.g Al (with 3 electrons in it's outer shell) bonds with Cl (which donates 1 electron for bonding) the resulting compound has 6 electrons in it's outer shell rather than the desired octet.

Front 8

Explain why the AlCl3 compound forms a dimer and what is it's structure and bonding.

Back 8

AlCl3 forms a dimer so that it's outer shell is filled to a stable 8. It achieves this by one of the chlorines from each molecule 'sharing' two of it's electrons with the opposite aluminium in a dative covalent bond.

Front 9

How would a compound like BF3 fulfil it's octet?

Back 9

By bonding with a NH3 molecule that contains a lone pair that it shares with BF3 in a dative bond.

Front 10

What is a Lewis acid?

Back 10

A Lewis acid is a chemical that can accept a lone pair.

Front 11

What is a Lewis base?

Back 11

A Lewis base is a chemical that can donate a lone pair.

Front 12

What is AlCl3 acting as in the chlorination of benzene and why does it work?

Back 12

AlCl3 is acting as a halogen carrier by attracting one of the Cl2 atoms towards the positve Al. This leaves the other Cl with a slightly positive charge allowing it to attack the benzene ring.

Front 13

Why have room temp. ionic liquids been developed to replace volatile organic compounds?

Back 13

Because VOC's are highly flammable and toxic. Also ionic liquids are cleaner solvents and better for use in catalytic processes like the polymerisation of alkenes.

Front 14

What are other advantages of ionic liquids?

Back 14

1) they have a very small vapour pressure and are non-volatile so don't evaporate during use.
2) they are often easier to separate from product and catalyst than VOC's
3) They will also dissolve grease that may be in the machinery.

Front 15

Explain the structure of hexagonal and cubic BN.

Back 15

Hexagonal BN is comprised of covalently bonded B and N in a mesh of hexagonal shapes, there are then many layers of this mesh bonded by van der waals bonds (Same structure as graphite). Cubic BN forms the same structure as diamond with each atom forming 4 bonds.

Front 16

What properties does cubic BN posses due to it's structure?

Back 16

Cubic BN is the second hardest material known. It has high thermal conductivity and is chemically inert. It is used for cutting tools and abrasives due to it's hardness. Also due to it's high thermal conductivity it is used for mounting high power electronic components, to ensure heat dissapation. Wear resistant coatings have been developed and BN powders with high surface area have been used as supports for catalysts due to it's chemical inertness.

Front 17

What properties does hexagonal BN posses due to it's structure?

Back 17

Due to the weak van der waals bonds the layers can slide over each other, making it an excellent lubricant. It can also be pressed into complex shapes and can form insulating nanotubes for carbon nanotubes.

Front 18

What happens to the stability of the oxidation states as you descend group 4?

Back 18

For the 1st 3 elements the +4 state is more stable than the +2 state however for Sn (tin) both oxidation states are equally stable and for Pb the stability is reversed so the +2 state is more stable than the +4.

Front 19

Give an example of a group 4 compound as a reducing agent.

Back 19

CO reduces Fe2O3 and CuO with the following equations.
Fe2O3 + 3CO -> 2Fe + 3CO2
CuO + CO -> Cu + CO2

Front 20

Give an example of a group 4 oxidizing agent.

Back 20

PbO2 + 4HCl -> PbCl2 + Cl2 + H2O

Front 21

Describe the bonding and solubility of CO2, PbO and PbO2.

Back 21

Both the Pb compounds are insoluble in water but PbO bonds ionicly and PbO2 bonds covalently. CO2 is soluble and bonds covalently.

Front 22

Give equations in which CO2 displays it's acidic quality.

Back 22

CO2 + 2NaOH -> Na2CO3 + H2O
If CO2 is in excess
CO2 + NaOH -> NaHCO3 + H2O

Front 23

Describe the types of bonding in
the chlorides CCl4, SiCl4 and PbCl2 and their reactions with water.

Back 23

CCl4 and SiCl2 whereas PbCl2 in ionic. SiCl dissolves in water with the following reaction SiCl4 + 2H2O -> 4HCl + SiO2. PbCl2 dissolves sparingly in cold water and more readily in hot water. CCl4 is not soluble.

Front 24

Give the equation and what is observed when Pb2+ reacts with NaOH.

Back 24

Pb2+ + 2OH- -> Pb(OH)2
Produces a white precipitate.

Front 25

Give the equation and observations when Pb2+ reacts with an excess of NaOH.

Back 25

Pb(OH)2 + 2OH- -> Pb(OH)4,2-
White precipitate will re-dissolve

Front 26

Give the equation and observation for Pb2+ reacting with Cl-.

Back 26

Pb2+ + 2Cl- -> PbCl2
Produces a white precipitate.

Front 27

Give the equation and observation for when Pb2+ reacts with I-.

Back 27

Pb2+ + 2I- -> PbI2
Produces a yellow precipitate.

Front 28

What are the trends in oxidizing power of the halogens and how do they relate to their Eθ values?

Back 28

The halogens have weaker oxidizing power as you descend the group which correlates with their decreasing Eθ values. Conversly this means that they become stronger reducing agents down the group.

Front 29

Give the equation for the reaction of Cl2 and cold dilute NaOH.

Back 29

Cl2 + 2OH- -> Cl- + ClO- +H2O

Hint 29

This is an example of dosproportionation because chlorine is simultaneously oxidized and reduced. From 0 in Cl2 to -1 in Cl- and +1 in ClO-.

Front 30

Give the equation for when Cl2 reacts with warm, concentrated NaOH.

Back 30

3Cl2 + 6OH- -> 5Cl- + ClO3- + 3H2O

Hint 30

This is an example of disproportionation where Cl is being simultaneously oxidized and reduced from 0 in Cl2 to -1 in Cl- and +5 ClO3.

Front 31

Give one use of chlorate (I) ClO-.

Back 31

When chlorine gas is bubbled through water it produces the ClO- and the ClO- attacks and kills bacterium. It will also attack dye so is used in bleaching.

Front 32

Give a use of Chlorate (V) ClO3-.

Back 32

ClO3- is a very powerful oxidising agent so will attack any organic molecule it comes into contact with and therefore makes a very good weedkiller.

Front 33

Give the reaction for NaCl and conc. H2SO4.

Back 33

NaCl + H2SO4 -> NaHSO4 + HCl

Hint 33

This is a redox reaction with Sulphuric acid as the oxidizing agent but becuase Cl is also quite a strong oxidizing agent it cannot react any further than this. This is not the case for NaBr and NaI because they are reducing agents they can reduce S further.

Front 34

What occurs when you react NaBr with H2SO4? Give equation and observations.

Back 34

NaBr + H2SO4 -> NaHSO4 + HBr
H2SO4 + HBr -> Br2 + SO2 + 2H2O.
White misty fumes of HBr are seen along with the brown fumes of Br2

Hint 34

For WJEC the most important point to take is the varying oxidation states of S. Here it is +6 in H2SO4 and NaHSO4 but the HBr reduces it to +4 in SO2. Iodide is is an even stronger reducing agent so reduced S even further.

Front 35

Give the equation and observation for NaI reacting with H2SO4.

Back 35

NaI + H2SO4 -> NaHSO4 + HI
8HI + H2SO4 -> 4I2 + H2S + 4H2O (S solid is also produced which is a yellow ppt.) On warming purple fumes of iodine are observed along with the smell of rotten eggs (which is the H2S)

Hint 35

HI is the strongest reducing agent out of the three hydrogen halides so it reduces S the furthest. From +6 in H2SO4 to -2 in H2S.

Front 36

Give some uses of halogen containing compounds.

Back 36

PTFE (teflon)-non stick coatings
TCP antiseptic
PVC - Window frames, clothing
Bromine compounds - Dyes and medicine
Iodine compounds - Catalysts and pharmaceuticals.