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A2 CHEM UNIT3 > 3.3 P-block > Flashcards

3.3 P-block Flashcards

1
Q

What is meant by a P-block element?

A
  • electronic configuration is where your last electron is in a p orbital
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2
Q

What happens as you descend group 4?

A
  • become more ionic
  • decrease in MP
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3
Q

Define the inert pair effect

A
  • the resistance of S electrons to be involved in bonding due to a tendency of an S2 pair of electrons in an atom to stay paired, leading to a lower oxidation state
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4
Q

What happens to the inert pair effect as you descend groups?

A
  • it gets stronger
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5
Q

Define electron deficiency

A
  • where outer shells don’t have a full octet
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6
Q

Tin and carbon are stable in oxidation state +4 whilst lead is more stable as +2. Explain the difference instabilities of these oxidation states

A
  • as you go down group 4 the inert pair effect becomes stronger
  • the S electrons are more reluctant to be involved in bonding hence +2 is more stable than +4
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7
Q

How do electron deficient compounds fulfil their octet?

A
  • form acceptor-donor compounds
  • form dimers
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8
Q

What is meant by a dimer?

A
  • when 2 of the same molecule join together using a coordinate bond
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9
Q

What is meant by an acceptor donor compound?

A

ie BF3 . NH3
- ammonia has lone pair to donate into boron to fulfil octet
- coordinately bonded

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10
Q

What enables octet expansion?

A
  • empty d and f orbitals
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11
Q

Adding a lone pair reduces the bond angle by …?

A

2.5°

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12
Q

Explain why nitrogen can only form a chloride with 8 outer shell electrons, but phosphorus can form a chloride with a different number of outer shell electrons

A
  • phosphorous can form a chloride with a different number of outer shell electrons as P3 elements have access to vacant d and f orbitals
  • nitrogen cannot do this as it is P2 so does not have access to d and f orbitals
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13
Q

Reaction of sodium chloride with H2SO4

A

NaCl + H2SO4 —> NaHSO4 + HCl

  • HCl has low oxidising power
  • not redox
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14
Q

Reaction of sodium bromide with H2SO4

A

NaBr + H2SO4 —> HBr + NaHSO4

2HBr + H2SO4 —> Br2 + SO2 + 2H2O

  • HBr has strong oxidising power
  • redox reaction
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15
Q

Reaction of sodium iodide with H2SO4

A

NaI + H2SO4 —> HI + NaHSO4

2HI + H2SO4 —> I2 + SO2 + 2H2O

6HI + H2SO —> 3I2 + S + 4H2O

8HI + H2SO4 —> 4I2 + H2S + 4H2O

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16
Q

What happens to reducing power as you descend group 7?

A
  • reducing power increases
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17
Q

Products and observations of NaF with H2SO4

A

HF - misty fumes

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18
Q

Products and observations of NaCl with H2SO4

A

HCl - misty fumes

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19
Q

Products and observations of NaBr with H2SO4

A

HBr - misty fumes
Br2 - brown fumes
SO2 - colourless gas with choking odour

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20
Q

Products and observations of NaI with H2SO4

A

HI - misty fumes
I2 - purple fumes and black ppt
H2S - colourless gas with egg smell
SO2 - colourless gas with choking odour
S - yellow ppt

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21
Q

Define isoelectronic

A

Same number of outer electrons used in bonding

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22
Q

Uses of hexagonal boron nitride

A
  • chemically inert
  • used in reaction vessels and crucibles
23
Q

Similarities of boron nitride and graphite

A
  • layered structure
  • high melting point and boiling point
  • very strong
24
Q

Differences of boron nitride and graphite

A
  • layers in boron nitride line up - they are not offset
  • h-boron nitride does not conduct electricity because no delocalised electrons due to different electronegativities
25
Q

Explain the isoelectric relationship between boron nitride and carbon

A
  • adopt the same hexagonal structure due to the number of bonding outer electrons being equal in them
26
Q

Why can electrons in boron nitride not be delocalised?

A
  • difference in electronegativities holding onto electrons
27
Q

Describe and explain reaction of CCl4 in water

A
  • immiscible in water and cannot be hydrolysed
  • no vacant d-orbitals for reaction
28
Q

PbCl4 reaction

A
  • yellow fuming liquid
  • explodes at 105° before boiling
29
Q

Why is meant by amphoteric?

A
  • able to react as both an acid and base
30
Q

Al3+ ion

A
  • small and highly charged so will be extensively hydrated

[Al(H2O)6]3+

31
Q

Pb2+ with OH-

A

[Pb]2+ + OH- —> Pb(OH)2

Pb(OH)2 + 2OH- —> [Pb(OH)4]2-

32
Q

Define disproportionation

A

When the same atom is both oxidised and reduced in the same reaction, forming products containing the element in two different oxidation states

33
Q

Reaction of chlorine with cold NaOH

A

Cl2 + NaOH —> NaCl + NaClO + H2O

Cl2 + OH- —> Cl- + ClO- + H2O

34
Q

Chlorine with hot NaOH

A

3Cl2 + 6NaOH —> 5NaCl + NaClO3 + 3H2O

3Cl2 + 6OH- —> 5Cl- + ClO3- + 3H2O

35
Q

Describe the oxides of carbon

A
  • CO2 is most stable
  • CO is only stable compound to contain carbon in the +2 oxidation state
  • CO is a reducing agent as it easily becomes oxidised from +2 to +4
36
Q

Describe oxides of lead

A
  • PbO is the most stable oxide of lead
  • PbO2 acts as oxidising agent as it is easily reduced from +4 to +2
  • PbO is amphoteric
37
Q

Describe CCl4 and SiCl4

A
  • tetrachlorides
  • CCl4 does not react with water due to lack of available d orbitals meaning octet cannot be expanded
  • SiCl4 reacts very quickly with water in a hydrolysis reaction due to available 3d orbitals
38
Q

Reaction of KI with Pb2+

A
  • bright yellow ppt of lead iodide
39
Q

Shape of BF3

A

Trigonal planar

40
Q

Shape of NH3

A

Trigonal pyramid

41
Q

Role of lead oxide in the reaction with hydrochloric acid

A

Oxidising agent

42
Q

Use for cubic boron nitride

A

Cutting instruments

43
Q

Phosphorus is able to form two different chlorides, PCl3 and PCl5. Nitrogen is only able to
form one chloride, NCl3.
Explain this difference.

A

Phosphorus has vacant d and f orbitals so can expand its octet whereas nitrogen doesn’t so it can’t expand

44
Q

Explain why iron has more than one common oxidation state in its compounds.

A

Energy of 4s and 3d orbitals are all similar therefore ionisation energy to remove the electrons are too

45
Q

Explain why aluminium forms compounds that are electron deficient

A
  • aluminium has 3 outer shell electrons
  • forms 3 covalent bonds
  • giving 6 outer shell electrons
    —> draw Al2Cl6 dimer
46
Q

Aluminium chloride forms Al2Cl6 dimer using coord bonding. Explain why coord bonds form

A
  • Al is electron deficient
  • Cl donates a lone pair
47
Q

Properties of PbO and PbO2

A
  • lead forms positive ions and strong ionic bonds
  • amphoteric oxides
  • PbO2 contains lead in +4 but its most stable in +2 so is easily reduced / oxidising agent
48
Q

Equation for lead as an oxidising agent

A

PbO2 + 4HCl —> PbCl2 + Cl2 + 2H2O

49
Q

Equations to show amphoteric properties of PbO

A

PbO + 2NaOH —> Na2PbO2 + H2O

PbO + 2HNO3 —> Pb(NO3)2 + H2O

50
Q

Describe the properties of CO and CO2

A
  • weak forces between molecules so gaseous
  • acidic oxide

CO2 + 2NaOH —> Na2CO3 + H2O

51
Q

Equation showing CO as a reducing agent

A

CO + CuO —> Cu + CO2

52
Q

What happens to Pb2+ and Cr3+ when excess NaOH is added?

A

They dissolve as they are amphoteric

A2 CHEM UNIT3 (9 decks)

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