Group 7 Halogens: Trends, Reactions, and Halides

Flashcards covering Group 7 halogens and halides, including physical properties, periodic trends, displacement reactions, redox chemistry with sulfuric acid, silver nitrate tests, and hydrogen halides.

What are the appearance and physical states of fluorine, chlorine, bromine, and iodine at room temperature?

Fluorine ($F_2$) is a very pale yellow gas; chlorine ($Cl_2$) is a greenish, reactive gas; bromine ($Br_2$) is a red liquid that gives off dense brown/orange poisonous fumes; and iodine ($I_2$) is a shiny grey solid that sublimes to a purple gas.

Why do the melting points and boiling points of the halogens increase down the group?

As the molecules become larger down the group, they have more electrons and therefore have larger London forces between the molecules. More energy is required to overcome these stronger intermolecular forces, leading to higher melting and boiling points.

What are the melting and boiling points of fluorine, chlorine, bromine, and iodine?

Fluorine: melting point $-220\,^{\circ}\text{C}$, boiling point $-188\,^{\circ}\text{C}$; Chlorine: melting point $-101\,^{\circ}\text{C}$, boiling point $-35\,^{\circ}\text{C}$; Bromine: melting point $-7\,^{\circ}\text{C}$, boiling point $59\,^{\circ}\text{C}$; Iodine: melting point $+114\,^{\circ}\text{C}$, boiling point $184\,^{\circ}\text{C}$.

What is electronegativity, and why does it decrease down Group 7?

Electronegativity is the relative tendency of an atom in a molecule to attract electrons in a covalent bond to itself. It decreases down the group because atomic radii increase due to an increasing number of electron shells, making the nucleus less able to attract the bonding pair of electrons.

What is the trend in oxidising ability of the halogens down Group 7?

Oxidising strength decreases down the group. A halogen that is a stronger oxidising agent will displace a halogen of lower oxidising power from its compounds.

Which halides can chlorine displace, and which can bromine displace in aqueous solution?

Chlorine will displace both bromide ($Br^-$) and iodide ($I^-$) ions. Bromine will only displace iodide ($I^-$) ions.

What colours do aqueous solutions of free chlorine, bromine, and iodine show?

Chlorine is a very pale green solution (often colourless); bromine is a yellow solution; and iodine is a brown solution (sometimes with a black solid present).

What colours do free chlorine, bromine, and iodine appear when an organic solvent layer is added?

Chlorine produces a colourless organic layer, bromine produces a yellow organic layer, and iodine produces a purple organic layer.

What are the oxidation and reduction half-equations when chlorine oxidises aqueous bromide ions?

Oxidation half-equation: $2Br^-(aq) \rightarrow Br_2(aq) + 2e^-$. Reduction half-equation: $Cl_2(aq) + 2e^- \rightarrow 2Cl^-(aq)$.

How do chlorine, bromine, and iodine compare in their ability to oxidise $Fe^{2+}$ ions to $Fe^{3+}$?

Chlorine and bromine are strong enough oxidising agents to oxidise $Fe^{2+}$ to $Fe^{3+}$. Iodine is not a strong enough oxidising agent to do this; instead, the reaction reverses and $Fe^{3+}$ oxidises $I^-$ to $I_2$.

What is a disproportionation reaction?

A reaction where an element simultaneously oxidises and reduces.

What occurs when chlorine reacts with water, and what colour changes occur when universal indicator is added?

Chlorine disproportionates according to $Cl_2(aq) + H_2O(l) \rightleftharpoons HClO(aq) + HCl(aq)$, changing its oxidation number from $0$ to $-1$ in $HCl$ and from $0$ to $+1$ in $HClO$. Universal indicator first turns red due to the acidity of both products, and then turns colourless as $HClO$ bleaches the colour.

What reaction occurs when chlorine is bubbled through water in the presence of bright sunlight?

$2Cl_2 + 2H_2O \rightarrow 4H^+ + 4Cl^- + O_2$. The greenish colour of the chlorine water fades as $Cl_2$ reacts, and a colourless gas ($O_2$) is produced.

What reaction occurs between chlorine and cold dilute $NaOH$, and what is the use of the product mixture?

$Cl_2(aq) + 2NaOH(aq) \rightarrow NaCl(aq) + NaClO(aq) + H_2O(l)$. The colour fades to colourless. The mixture of $NaCl$ and $NaClO$ (sodium chlorate(I)) is used as bleach and to disinfect and kill bacteria.

What reaction occurs between chlorine and hot dilute $NaOH$?

$3Cl_2(aq) + 6NaOH(aq) \rightarrow 5NaCl(aq) + NaClO_3(aq) + 3H_2O(l)$. Chlorine disproportionates to a higher oxidation state ($+5$) in sodium chlorate(V).

What is the chemical equation and starch indicator colour change for the iodine-thiosulfate redox titration?

The reaction is $2S_2O_3^{2-}(aq) + I_2(aq) \rightarrow 2I^-(aq) + S_4O_6^{2-}(aq)$. Starch indicator is added near the end point when the iodine fades to pale yellow, changing the colour from blue/black to colourless.

Why is starch indicator not added until near the end point in an iodine-thiosulfate titration?

Because the blue complex formed between starch and high concentrations of iodine is insoluble and does not re-dissolve as more thiosulfate is added.

Why does the reducing power of the halide ions increase down Group 7?

As halide ions get larger, the pull from the nucleus on the outer electrons becomes smaller, making it easier for the ions to donate outer electrons.

What reactions and observations occur when solid $NaF$ or $NaCl$ reacts with concentrated sulfuric acid?

Only acid-base reactions occur because concentrated $H_2SO_4$ is not strong enough to oxidise fluoride or chloride ions. Reactions: $NaF(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HF(g)$ and $NaCl(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HCl(g)$. Observations: White steamy fumes of $HF$ or $HCl$ are evolved.

What products and observations occur when solid $NaBr$ reacts with concentrated sulfuric acid?

An initial acid-base step forms $HBr$: $NaBr(s) + H_2SO_4(l) \rightarrow NaHSO_4(s) + HBr(g)$, followed by a redox step: $2HBr + H_2SO_4 \rightarrow Br_2(g) + SO_2(g) + 2H_2O(l)$. Observations: White steamy fumes of $HBr$, orange fumes of $Br_2$, and a colourless, acidic gas ($SO_2$).

What reduction products and observations occur when solid $NaI$ reacts with concentrated sulfuric acid?

Iodide ions reduce sulfur from $+6$ in $H_2SO_4$ to $+4$ ($SO_2$), $0$ ($S$), and $-2$ ($H_2S$). Observations: White steamy fumes of $HI$, black solid and purple fumes of $I_2$, colourless acidic $SO_2$ gas, yellow solid of sulfur ($S$), and a gas with a bad egg smell ($H_2S$).

Why is nitric acid added before silver nitrate solution when testing for halide ions?

To react with any carbonates present to prevent the formation of an $Ag_2CO_3$ precipitate, which would mask the desired observations.

What precipitates are formed when acidified silver nitrate reacts with chloride, bromide, and iodide ions?

Chloride forms a white precipitate ($AgCl$); bromide forms a cream precipitate ($AgBr$); and iodide forms a pale yellow precipitate ($AgI$). Fluoride produces no precipitate.

How do aqueous ammonia and concentrated ammonia distinguish between $AgCl$, $AgBr$, and $AgI$?

$AgCl$ dissolves in dilute aqueous ammonia to form a colourless solution; $AgBr$ does not dissolve in dilute aqueous ammonia but dissolves in concentrated ammonia; and $AgI$ does not dissolve in either aqueous or concentrated ammonia.

Why is concentrated phosphoric acid ($H_3PO_4$) used instead of concentrated sulfuric acid to produce hydrogen halides?

Phosphoric acid is not an oxidising agent and therefore does not oxidise $HBr$ and $HI$. This ensures pure hydrogen halides are formed without extra redox reactions taking place.

What property of hydrogen halides is demonstrated by the apparatus shown in this diagram?

Their high solubility in water; the hydrogen halide gas dissolves so rapidly in water that water quickly rises up the inverted tube.

What is the test for the presence of hydrogen halides using ammonia gas?

Hydrogen halides react readily with ammonia gas to produce a white smoke of the ammonium halide, for example: $HCl(g) + NH_3(g) \rightarrow NH_4Cl(s)$.

Why does the thermal stability of hydrogen halides decrease down Group 7?

As the halogen atoms become bigger down the group, the $H\text{-Hal}$ bond length becomes longer, meaning the bonding pair of electrons is further from the halogen nucleus. This decreases the $H\text{-Hal}$ bond energy, making the hydrides less stable.