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chemistry ~ inorganic chmistry (3.2) > group 7 > Flashcards

group 7 Flashcards

1
Q

trend electronegativity

A

decreases

larger atoms so larger atomic radius so increased distance between nucleus and shared pair

increased shielding means e- density in cov bond is further from nucleus

decreased FoA, decreased ability

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2
Q

boiling pints

A

increases

larger molecules so vdw forces between molecules are stronger

more energy to overcome

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3
Q

trend atomic radius

A

increases

atoms are larger so there are more e- and the gore more completed energy levels

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4
Q

trend in 1st ionisation energy

A

decreases

bigger atoms
larger radius
increase shielding
more shells
weaker FoA between.. .
e is more easily lost

increased nuclear charge overcome by these effects

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5
Q

halogen molecules profiles

fluorine

A

F2
gas at room temp
yellow
toxic
v reactive

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6
Q

halogen molecules profiles

chlorine

A

Cl2
toxic
v reactive
gas at room temp
green

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7
Q

halogen molecules profiles

bromine

A

Br2
toxic
reactive
liquid at rtp
orange

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8
Q

halogen molecules profiles

iodine

A

I2
solid r rto
grey silver black

sublimes straight to gas, purple

uses: tests for starch
stains cells
antiseptic solution

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9
Q

reaction of chlorine with water (no sunlight)

importance of product

A

Cl2 + H2O (reversible) HCl + HClO

HClO is chlorate acid which dissociated to produce H+ and ClO- ions

ClO- is a powerful oxidising agent which kills bacteria BC if oxidises things in bacteria

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10
Q

reaction of chloride and water (sunlight)

A

Cl2 + H2O – 4Hcl + O2

not good bc no HClO
so need to replenish regularly in pools

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11
Q

reaction of chlorine and sodium hydroxide

A

Cl2 + NaOH – NaCl + NaClO + H2O

NaClO = sodium chlorate

conditions: cold, dilute, aq NaOH

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12
Q

what is the trend in oxidising ability of the halogens

explain

A

decreases

oxidising agent undergoes reduction so it gains e-.
because as go down there are more completed energy levels resulting in increased shielding and atomic radius,

eFoA is weaker between nucleus and e to gain so it’s harder to gain

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13
Q

outline a test you can use to prove the oxidising ability of halogens

A
  1. In a spotting tile put Cl2, BR2, and I2 in each row
  2. put H2O in one column (control)
  3. KCl in one column
  4. KBr in one column
  5. KI in one column
  6. the more powerful oxidising agent will displace the other halide in the metal halide
  7. Cl2 displaces Br and I
    - in Cl2 with KBr or KI, goes from colourless to orange// brown showing Br2//I2 formed
  8. Br2 displaces I not Cl
    - in BR2 with KCl there is no change, remains colourless
    - with KI goes from colourless to brown showing I2 formed
  9. I2 doesn’t replace Br and Cl, remains brown (?)
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14
Q

ionic equations for oxidising ability of halogens

A

Cl2 + 2Br - – 2Cl- + Br2

etc etc (symbols all aq)

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15
Q

why is fluorine not included in reaction (ox ability investigation reaction)

A

fully reacts with water

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16
Q

trend and explanation of reducing power of halide ions

A

increases as go down

outer e-being lost is further from nucleus due to ions getting bigger so more completed energy levels. increased shielding. so weaker eFoA so e- is more easily lost

reducing power = reducing agent, gets oxidised so losing e-

17
Q

how can we evidence reducing power

A

reaction of solid sodium halide with conced sulfuric acid (so symbol l)

1/2 spatula solid
5 drops acid

18
Q

what’s the general equation for the reaction of solid sodium halide with sulfuric acid

observation

which to halide salts don’t have further reactions

A

NaX (s) + H2SO4 (l) – HX (g) + NaHSO4 (s)

white misty fumes of hydrogen halide gas

chlorine (the gas is hydrogen chloride not hydrochloric acid) and fluorine don’t do anything more than this. they are not strong enough reducing agents

19
Q

(R1: NaX + H2SO4 – HX + NaHSO4)

general equation for HX gas formed in reaction 1, with H2SO4

which halides undergo this 2nd reaction

A

HX (g) + H2SO4 (l) – X2 (g) + SO2 (g) + 2H2O (l)
produces halogen molecule, sulfur dioxide and water

HBr and HI (bromide, iodide)

20
Q

(R1: NaX + H2SO4 – HX + NaHSO4)
(R2: HX (g) + H2SO4 (l) – X2 (g) + SO2 (g) + 2H2O (l))
—-
—-

observations with Br

A

1: white misty fumes of HBr gas
2:
- brown fumes from BR2 gas (reaction is exothermic so liquid Br vaporises)
- choking fumes from SO2 which is colourless

21
Q

(R1: NaX + H2SO4 – HX + NaHSO4)
(R2: HX (g) + H2SO4 (l) – X2 (g) + SO2 (g) + 2H2O (l))
—-
—-

observations with I

A

1: white misty fumes
2: v exothermic so solid iodine vaporises to purple fumes of I2 vapour ;
choking fumes from SO2

22
Q

(R1: NaX + H2SO4 – HX + NaHSO4)
(R2: HX (g) + H2SO4 (l) – X2 (g) + SO2 (g) + 2H2O (l))

there is a third equation involving the HX and the SO2 formed in R2.
which halogen undergoes this one

what’s the eq for it

observations

A

iodine

6HI + SO2 – H2S + 3I2 + 2H2O (g/g/l)

  • purple fumes of iodine vapour (will have alr been there from 2nd reaction, so more thick ig)
  • rotten egg smell from hydrogen sulfide
  • yellow solid observation due to sulfur
23
Q

6HI + SO2 – H2S + 3I2 + 2H2O (g/g/l)

there is a yellow solid observation due to sulfur. explain

A

ox state in SO2 = +4
ditto H2S = -2

therefore had to pass O
so was an element for a sec
sulfur solid is yellow

24
Q

equations for investigating reducing power of halogens is as shown

R1: NaX + H2SO4 – HX + NaHSO4

R2: HX (g) + H2SO4 (l) – X2 (g) + SO2 (g) + 2H2O (l)

3: 6HI + SO2 – H2S + 3I2 + 2H2O (g/g/l)

who is redox and who is not

A

1 not
2 3 are

25
Q

which of the reactions of solid sodium halide and sulfuric acid conced,

have the observation of purple fumes

ionic equations for this

A

R2/3 with NaI

2I- + H2SO4 + 2H+ —- I2 + SO2 + 2H2O

26
Q

which of the reactions of solid sodium halide and sulfuric acid conced,

have the observation of choking fumes

ionic equations for this

A

R2 and final of NaBr
R2[/3(reactant)] of NaI

2I- + H2SO4 + 2H+ —- I2 + SO2 + 2H2O

27
Q

which of the reactions of solid sodium halide and sulfuric acid conced,

have the observation of rotten egg smell

ionic equations for this

A

r3 of NaI
actually the HI with the SO2

8I- + H2SO4 + 8H+ —- 4I2 + H2S + 4H2O

28
Q

which of the reactions of solid sodium halide and sulfuric acid conced,

have the observation of yellow solid

ionic equations for this

A

3rd reaction NaI
actually HI with SO2

6I- + H2SO4 + 6H+ —– 3I2 + S + 4H2O

29
Q

how to test for halide ions in solution

A
  1. acidify 2cm3 sample with 5 drops HNO3 j not HCl
    this removes impurities because ions present will bond to these ions

1.5 add water / make into a solution

  1. add 5 drops AgNO3 solution
  2. if halide ions present a silver halide precipitate forms
  • AgCl is white
  • AgBr is cream
  • AgI is yellow
30
Q

by using acidified silver nitrate you can distinguish halide ions in solution

how do you distinguish the silver halides formed

A
  1. add 1cm3 dilute ammonia to each ppt
  • silver chloride redissolves
    AgCl (s) + 2NH3 (aq) —- (Ag(NH3)2)+ (aq) + Cl- (aq)
  • there is no visible change for the others. ppt remains
  1. add 1cm3 conced ammonia (need gloves)
    - AgBr redissolves
    AgBr (s) + 2NH3 (aq) —- (Ag(NH3)2)+ (aq) + Br- (aq)
  2. AgI doesn’t redissolve
31
Q

why is fluorine not including in the acidified silver nitrate test

A

AgF is soluble, doesn’t give ppt

chemistry ~ inorganic chmistry (3.2) (3 decks)

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