第17族

The halogens 卤素 are the Group 族 17 elements. They exist as diatomic molecules ($\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$). Going down the group:

The volatility 挥发性 (how easily a substance turns to vapour) decreases down the group: chlorine is a gas, but iodine is a solid. This is because the molecules get larger and have more electrons, so the instantaneous dipole 瞬时偶极 and induced dipole 诱导偶极 forces between them get stronger. Stronger forces are harder to break, so the boiling point rises and volatility falls.

The bond energy 键能 (bond strength) of the $\text{X}\text{–}\text{X}$ molecules generally falls from $\text{Cl}_2$ to $\text{I}_2$, because the shared electrons are further from the nuclei in the larger atoms.

Halogens as oxidising agents

Each halogen reacts by gaining one electron to form a $1-$ ion, so it acts as an oxidising agent 氧化剂. This power decreases down the group, because the larger atoms attract an extra electron less strongly. A more reactive halogen can push out a less reactive one from its salt:

Reactions with hydrogen

Each halogen reacts with hydrogen to form a hydrogen halide 卤化氢:

The reaction gets less vigorous down the group: chlorine reacts explosively in light, bromine needs heat, and iodine reacts slowly and only partly.

Thermal stability of the hydrogen halides

The thermal stability 热稳定性 of the hydrogen halides decreases down the group. The H–X bond gets weaker as the halogen atom gets larger, so HI breaks apart on gentle heating while HCl is very stable.

Halide ions as reducing agents

A halide ion 卤离子 (such as $\text{Cl}^-$) can give away an electron, acting as a reducing agent 还原剂. This power increases down the group, because a larger ion holds its outer electron less tightly.

Reaction with aqueous silver ions

Add aqueous silver nitrate, then aqueous ammonia 氨, to identify the halide from the colour of the silver halide precipitate 沉淀:

Reaction with concentrated sulfuric acid

All the halides first give the hydrogen halide. The lower halides are then oxidised by the acid, because they are stronger reducing agents:

• chloride gives only $\text{HCl}$ (no redox).
• bromide also gives some brown $\text{Br}_2$ and $\text{SO}_2$.
• iodide gives $\text{I}_2$ and the smelly gases $\text{H}_2\text{S}$ and $\text{SO}_2$, because $\text{I}^-$ is the strongest reducing agent.

With sodium hydroxide

With cold, dilute sodium hydroxide, chlorine reacts to form chloride and chlorate(I):

With hot, concentrated sodium hydroxide, it forms chloride and chlorate(V):

In both, the oxidation number 氧化数 of chlorine goes both up and down (from $0$), so both are disproportionation 歧化 reactions.

Chlorine in water purification

A little chlorine is added to water for water purification 水净化. It reacts with water:

The active species $\text{HOCl}$ and $\text{ClO}^-$ kill bacteria 细菌, making the water safe to drink.

Worked example. Solid $\text{NaCl}$, $\text{NaBr}$ and $\text{NaI}$ are each warmed with concentrated $\text{H}_2\text{SO}_4$. Predict the products. Reducing power increases down the group, so how far each halide reduces the sulfuric acid differs. $\text{Cl}^{-}$ is too weak to reduce it at all, so you get only steamy $\text{HCl}$ - an acid-base reaction. $\text{Br}^{-}$ reduces it a little: $\text{HBr}$ plus brown $\text{Br}_2$ and $\text{SO}_2$. $\text{I}^{-}$ is the strongest reducing agent: $\text{HI}$ plus $\text{I}_2$, and it drives the sulfur all the way down to $\text{H}_2\text{S}$, with its bad-egg smell. Every halide gives the hydrogen halide first; the extra products appear only where the halide is a strong enough reducing agent to attack the sulfur.

Common mistakes.

• Saying the halogens' boiling points rise because the covalent bonds get stronger. Boiling separates molecules: the instantaneous dipole–induced dipole forces between them get stronger with more electrons. From chlorine to iodine, the X–X bond gets weaker; fluorine is an exception because its F–F bond is unusually weak.
• Mixing up the two trends. Halogens (oxidising agents) get weaker down the group; halide ions (reducing agents) get stronger.
• Testing for halide ions without acidifying. Add dilute $\text{HNO}_3$ first to remove carbonate and other ions that also give precipitates; do not use HCl, which adds chloride.
• Calling the reaction of chlorine with cold NaOH a displacement. Chlorine (oxidation number 0) goes to $-1$ in $\text{NaCl}$ and $+1$ in $\text{NaClO}$ at the same time: disproportionation.
• Giving chlorine the oxidation number $+3$ in $\text{NaClO}_3$. Oxygen is $-2$ three times and sodium is $+1$, so chlorine is $+5$: chlorate(V).
• Saying iodide and concentrated sulfuric acid give only HI. $\text{I}^-$ is a strong reducing agent, so it reduces the acid to $\text{SO}_2$ and $\text{H}_2\text{S}$ and is oxidised to $\text{I}_2$.

• Halogens get less reactive down the group (harder to gain an electron); oxidising power decreases.
• Displacement: a more reactive halogen displaces a less reactive halide — state the colour change.
• Test halide ions with $\text{AgNO}_3$: white ($\text{Cl}^-$), cream ($\text{Br}^-$), yellow ($\text{I}^-$), then confirm with dilute/concentrated ammonia.
• Chlorine with water and with cold $\text{NaOH}$ are disproportionation — show the oxidation-number changes.