Some reactions of the halide ions

a. Complete the equations below, including state symbols.

i. `AgNO_3(aq) + NaCl(aq) -> …`

ii. `AgNO_3(aq) + NaBr(aq) -> …`

iii. `AgNO_3(aq) + NaI(aq) …`

b. What would you observe in each reaction in part a?

c. What would you observe in each case if dilute ammonia solution were subsequently added?

d. What would you observe in each case if concentrated ammonia solution were subsequently added?

a.

i. `AgNO_3(aq) + NaCl(aq) → AgCl(s) + NaNO_3(aq)`

ii. `AgNO_3(aq) + NaBr(aq) → AgBr(s) + NaNO_3(aq)`

iii. `AgNO_3(aq) + NaI(aq) → AgI(s) + NaNO_3(aq)`

b. In each reaction in part a, it is easily noticed that the precipitation with different colours is occurred during these reactions

For the first one, that is white. The second one is cream and the last one is pale yellow

c. Silver chloride forms complex ions with dilute ammonia

d. Silver bromide and silver iodide form complex ions with concentrated ammonia

Write an equation, including state symbols, to show the reaction between silver nitrate solution, AgNO₃ (aq), and sodium iodide solution, NaI (aq).

The equation for the reaction between silver nitrate solution, AgNO₃ (aq), and sodium iodide solution, NaI (aq)

`AgNO_3 (aq) + NaI (aq) -> AgI (s) + NaNO_3 (aq)`

Sodium iodide reacts with concentrated sulfuric acid via the following equation. 

`H_2SO_4 (l) + NaI (s) → HI (g) + NaHSO_4 (s)`

The hydrogen iodide gas, HI (g), formed can react in a number of ways

A. The hydrogen iodide formed reacts with concentrated `H_2SO_4` to form a yellow solid, a smell of bad egg and a purple vapour. 

Name the three products responsible for these observations.

B. Write the equation which forms the yellow solid and purple vapour from HI and `H_2SO_4`. Include state symbols in your answer

A.

A yellow solid is sulfur

Smell of bad egg is hydrogen sulfide

A purple vapour is iodine

B.

The equation:

`6HI (g) + H_2SO_4 (l) -> 3I_2 (g) + S(s) + 4H_2O (l)`

When sodium bromide reacts with concentrated sulfuric acid, bromine and sulfur dioxide are formed.

Write the half equation for the formation of bromine from bromide ions.

The full equation is 

`NaBr (s) + H_2SO_4 (aq) -> NaHSO_4 (s) + HBr (g)`

`2HBr (g) + H_2SO_4 (aq) -> Br_2 (g) + SO_2 (g) + 2H_2O (l)`

The half equation is

`2Br^-) -> Br_2 + 2e^-`

a. This question is about the redox reactions of halide ions. Chlorine can displace iodine from potassium iodide.

i. Write the balanced equation for this reaction.

ii. Write the ionic equation for this reaction.

b. Using oxidation numbers, explain how iodide ions act as a reducing agent in this reaction.

a.

i. The balanced equation for this reaction

`Cl_2 + 2KI -> 2KCl + I_2`

ii. The ionic equation for this reaction

`Cl_2 + 2I^-) -> 2Cl^-) + I_2`

b.

Because Iodide ions cause the oxidation number of Cl to decrease from 0 to -1, as such, iodide ions serves as a reducing agent which will decrease the oxidation number of another atoms.

A. Chlorine also displaces bromine from potassium bromide but not as readily as it can displace iodine from potassium iodide. 

Explain how you can use these results to determine which is the stronger reducing agent, bromide ions or iodide ions.

B. Explain why the halide ions show this trend in reducing power.

A. To determine which is the stronger reducing agent, we will look at which one donates electrons more easily. It’s easily noticed that iodide ions will donate electrons more easily than bromide ions, as such, it will be stronger reducing agent.

B. When we move down the Group, ionic radius will increase, thus, the outermost electrons get further away from the nucleus. That results in less attraction between the outer electrons and nucleus. Therefore, the halide ions lose electrons easily.

Which of these anions forms a white precipitate when tested with acidified silver nitrate solution?

A. `CO""_3^(2-`

B. `SO""_4^(2-`

C. `OH^-`

D. `Cl^-`

The answer is D

`Ag^+ (aq) + Cl^-) (aq) -> AgCl (s)`

A is incorrect because carbonate ions are will be tested with dilute acid and produce effervescence but no white precipitate

B is incorrect because sulfate ions are tested with barium chloride solution to produce a white precipitate of barium sulfate

C is incorrect because silver hydroxide is a dark-coloured precipitate

What are the correct observations for the reaction of solid sodium bromide and concentrated sulfuric acid?

A. White fumes

B. Reddish brown gas and yellow solid produced

C. Yellow solid produced

D. Reddish-brown gas and pungent odour

The answer is D

A is incorrect because white fumes are HCl gas generation

B and C are incorrect because a yellow solide of sulfur is produced from the reaction of solid sodium iodide and concentrated sulfuric acid

Why does the oxidising power of the halogens decrease as the group is descended?

A. Shielding increases down the group therefore it is more difficult to accept an extra electron

B. Atomic radius decreases down the group

C. Shielding increases down the group therefore it is easier to donate an electron

D. The halogens become more electronegative down the group

The answer is A

B is incorrect because atomic radius will increase

C is incorrect because it’s hard to donate an electron

D is incorrect because the halogens become less electronegative down the group

What is the complete list of all the products from the reaction of potassium bromide with concentrated sulfuric acid?

A. Potassium hydrogen sulfate, hydrogen bromide, bromine, water and sulfur dioxide

B. Potassium hydrogen sulfate, hydrogen bromide, bromine and water

C. Potassium hydrogen sulfate, hydrogen bromide and bromine

D. Potassium hydrogen sulfate and hydrogen bromide

The answer is A

A series of steps occur in this reaction

`KBr + H_2SO_4 -> KHSO_4 + HBr`

`2HBr + H_2SO_4 -> Br_2 + SO_2 + 2H_2O`