AQA AS Chemistry C10 The Halogens Kerboodle Answer

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 10.1 The Halogens AQA AS Chemistry C10 The Halogens Kerboodles:Page No.160

1.Ans- a) Solid         b)Greater                 c)Lesser   2.Ans-   

1. Astatine is solid as on moving down the group density increases.
2. The size of astatine is greater than other halogen atoms because it is the last element of the group and it has maximum number of shells.
3. Since on moving down size increases therefore electronegativity decreases this is why astatine has least electronegativity.

3.Ans-

457. As the boiling increases down the group so astatine is expected to have a boiling point greater than that of Iodine i.e greater than457.
458. The boiling point of astatine is the largest in the group as on moving down the group the strength of van der waals forces increases with increasing atomic masses.

10.2 The Chemical Reaction Of Halogens AQA AS Chemistry C10 The Halogens Kerboodles:Page No.162

4.Ans-

1. i)Will not react ii) will react
2. This is because a halogen can only replace the halogen ion from its compound if that ion lies below it.This is because on moving down the group reactivity decreases.
3. Cl₂ + 2NaI ->2NaCl + I₂

5.Ans-

1. Reducing power of fluoride ion is minimum amongst the halide ions
2. This is because fluorine has the smallest size and it is difficult for it to lose an electron in comparison to other halide ions.
3. 2NaF + H₂SO₄ ->Na₂SO₄ + 2HF
4. No this is not a redox reaction as the oxidation state of elements is same.

 10.3 Reaction Of Halide Ions AQA AS Chemistry C10 The Halogens Kerboodles:Page No. 164

6.Ans-

1. A cream colour precipitate is observed.
2. NaBr+AgNO₃->AgBr+NaNO₃
3. The precipitate formed will dissolve in concentrated ammonia solution.
4. An acid is added to sodium bromide to remove the carbonate and hydroxide impurities.
5. Neither hydrochloric acid nor sulfuric acid can be used because they will form a precipitate of silver sulfate and silver chloride which would invalidate the test.
6. This test cannot be used to find out if fluoride ions are present because it is soluble in water and does not form a precipitate.

7.Ans-

1. 2Br₂ + 2H₂O->4HBr + O₂
2. Br₂+2NaOH->NaBrO +NaBr + H₂O

  8.Ans- Chlorine is added to water to  purifyit.Chlorine on reaction with water forms HClO which is an oxidising agent and kills bacteria by oxidation.   9.Ans-

1. Hydrochloric acid and oxygen are formed.
2. 2Cl₂(g) + 2H₂O(l) -> 4HCl(aq) +O₂(g)

Cl goes from 0 to -1. O goes from -2 to 0.

1. Oxygen is oxidised.
2. Chlorine has been reduced.
3. Chlorine is oxidising agent.
4. Oxygen is reducing agent.

 

Practice Questions: Page No.168

1.Ans-

1. On moving down the group electronegativity of the elements decreases.This is because the size of atoms increases due to addition of shells.
2. i) The reducing ability of halide ions increases down the group.

ii)NaCl(s) + H₂SO₄ (l) -> NaHSO₄(s) + HCl(g) whereas Bromine being a strong reducing agent reduces the product futher. 2H⁺ + 2Br^– +H₂SO₄(l) -> SO₂(g) + 2H₂O (l) + Br₂(l) NaBr(s) + H₂SO₄(l) -> NaHSO₄ (s) + HBr(g)

1. It can be used to distinguish between bromide and chloride ions as in case of chloride ions a white precipitate is formed which dissolves in dilute ammonia but in case of bromide ions a creamy precipitate is formed which is insoluble in dilute ammonia and dissolves in conc ammonia solution.
2. Cl₂ (g) +2NaOH (aq) ->NaClO (aq) + NaCl(aq) +H₂O(l)

NaClO can be used to purify water and is used as bleaching agent. 2.Ans-

1. Reduction means gain of electron. A reducing agent is a substance that reduces the other elements and get oxidised itself.
2. i) NaHSO₄ oxidation state +6

H₂S oxidation state -2 S oxidation state 0

1. ii) A bad egg smell produced indicates the production of H₂S gas and a yellow solid produced indicates the presence of S.

iii) 8H⁺ + 8I^– +H₂SO₄ -> H₂S + 4H₂0 + 4I₂ 6HI + H₂SO₄ –> 3I₂ + S + 4H₂O

1. Cl₂(g) + H₂O(l) ^->_<-HClO(aq) + HCl(aq)

No,it is chlorine that is oxidised and reduced as it is a dispropotionation reaction and the oxidation state of H an O remains the same. 3.Ans-

1. The boiling points of the halogen increases from fluorine to iodine as on moving down from fluorine to iodine the strength of van der waal forces increases due to increasing mass.
2. i) A brownish gas(bromine) will be evolved.
3. ii) A creamy precipitate is formed which later dissolves in conc ammonia.

iii)Brown fumes of bromine is evolved.

1. KBr(s) + H₂SO₄(l) -> KHSO₄ (s) + HBr(g)

Br ions further react and forms- 2H⁺ + 2Br^– +H₂SO₄(l) -> SO₂(g) + 2H₂O (l) + Br₂(l)

4. Ans-
5. The electronegativity decreases on moving down the group from fluorine to iodine.
6. Cl₂ + 2KI -> 2KCl + I₂.A brownish blacksolid(Iodine) is formed.
7. Two sulfur containing reduced product formed are-Hydrogen sulfide and Sulfur.

8H⁺ + 8I^– +H₂SO₄ -> H₂S + 4H₂0 + 4I₂ 6HI + H₂SO₄ –> 3I₂ + S + 4H₂O

1. Cl₂ (g) +2NaOH (aq) ->NaClO (aq) + NaCl(aq) +H₂O(l)

Two chlorine containing products are sodium chloride and sodium chlorate. NaCl-oxidation state -1 NaClO-oxidation state +1  

5. Ans-
6. i)Iodine is the species responsible for brown colour.

Cl₂ + 2KI -> 2KCl + I₂(Redox reaction)

1. ii) Silver Chloride(AgCl) is responsible for white precipitate.

KCl + AgNO₃ -> KNO₃ + AgCl The precipitate dissolves on addition of dilute ammonia solution

1. i) KCl+H₂SO₄->K₂SO₄ + 2HCl.

The misty white fumes is due to HCl.

1. ii) 8I^– -> 4I₂ + 8e^–

8H⁺ +8e^– +H₂SO₄ -> H₂S + 4H₂O Sulfuric acid provides sulphur atoms which is reduced by Iodine.Yellow solid obtained is Sulfur. iii) According to Le Chatliers principle the equilibrium will move towards right on addition of sodium hydroxide because the NaOH will consume the hydrogenand bromide ions formed hence will decrease their concentration. Bromine can be used for killing microorganisms in swimming pool water even though it is toxic because Bromine reacts with water to form HBrO which is an oxidising agent and kills bacteria by oxidising them. Banner 2

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