# Can you give examples for a halide ion reacting with interhalogen compound to give polyhalides?

Interhalogen compounds are formed when two halogens $$(\ce{F},$$ $$\ce{Cl},$$ $$\ce{Br},$$ $$\ce{I})$$ react together. They are represented by $$\ce{XX'_n},$$ where $$\ce{X}$$ is halogen of larger size and $$\ce{X'}$$ of smaller size. The total number of halogens in $$\ce{XX'_n}$$ depends on the radii of $$\ce{X}$$ and $$\ce{X'}.$$ The correct statement(s) regarding $$\ce{XX'_n}$$ is/are

❌ 1. $$\ce{FCl3}$$ is one of the stable interhalogen compounds.

✅ 2. Iodine can form an interhalogen compound containing maximum numbers of halogen atoms.

✅ 3. Interhalogen compounds are covalent in nature and more reactive than halogens, in general.

✅ 4. Halide ions reacts with interhalogen compounds to give polyhalides.

Regarding the 4th option: will it be like the following:

$$\ce{X'^- + XX'_n -> XX'_{n + 1}}?$$

Something like this? ($$\ce{X}$$ and $$\ce{X'}$$ are halogen compounds), so the polyhalide ion here would be $$\ce{X'_{n + 1}}.$$

What would be you views on this?

• $\ce{Cl- + ICl3}$ will do. – Ivan Neretin Aug 20 '19 at 7:49

TLDR: There is no exact generalized reaction. The reaction depends on type of interhalogen compound(value of n) and the nature of metal halides(ionic or covalent)

1. $$\ce{XX^{'}}$$ form

Let us take iodine monochloride($$\ce{ICl}$$). If it is reacted with ionic chloride like $$\ce{KCl}$$, it will form $$\ce{ICl2-}$$ ion.

$$\ce{MCl + ICl -> M+ICl2-, (M=K, Rb)}$$

If a different counter-ion is taken like bromide, it will form $$\ce{ClBrI-}$$ ion

$$\ce{ICl + KBr -> K+[ClBrI]-}$$

But if we add pure diatomic halogen, it will form a additive product

$$\ce{Cl2(excess) + ICl -> ICl3}$$

Covalent halides like $$\ce{AlCl3}$$ however gives different product:

$$\ce{ICl + AlCl3 -> I+ + AlCl4-}$$

1. $$\ce{XX^{'}3}$$ form

Let us take iodine trichloride. It will form $$\ce{ICl4-}$$ ions on reacting with ionic chloride:

$$\ce{MCl + ICl3 -> M+ICl4-}$$

If we take any other counter ion like fluorine, it will form $$\ce{MICl3F}$$ which is again kind of an additive product.

1. $$\ce{XX^{'}5}$$ form

Taking chlorine pentafluoride as examples, reacting with covalent fluorides like $$\ce{AsF5}$$ and $$\ce{SbF5}$$ will give adducts: $$\ce{ClF5.AsF5}$$ and $$\ce{ClF5.SbF5}$$

Iodine pentafluoride gives iodine heptafluoride on reacting with fluorine at 250 C

So, there is exact no generalized reaction. It all depend upon type of inter-halogen compound, reaction conditions and nature of metal halide. The close generalization I can make is that interhalogen compound of form $$\ce{XX^{'}_n}$$ if reacted with ionic metal halide will form metal polyhalide of form $$\ce{MXX^{'}_{n+1}}$$ or $$\ce{MXX^{'}_{n}X^{''}}$$(if a different counter-ion is taken). Additive product is formed when inter-halogen compound is reacted with pure diatomic halogens provided the second halogen is big enough to occupy electrons like bromine or iodine. In some case, covalent halides forms adducts with interhalogen compounds.

Thus the generalized reactions are(of course, there are certain exceptions):

$$\ce{X^{'-} + XX^{'}_n -> XX^{'-}_{n+1} (ionic)}$$

$$\ce{X^{'}_2 + XX^{'}_n -> XX^{'}_{n+2}}$$

Reference

All interhalogen compounds are prepared by direct combination of elements, and where more than one product is possible, the outcome of the reaction is controlled by temperature and relative proportions of the halogens. For example, reactions of $$\ce{F2}$$ with the later halogens at ambient temperature and pressure give $$\ce{ClF, BrF3}$$, or $$\ce{IF5}$$, but increased temperatures give $$\ce{ClF3, ClF5, BrF5,}$$ and $$\ce{IF7}$$. For the formation of $$\ce{IF3}$$, the reaction between $$\ce{I2}$$ and $$\ce{F2}$$ is carried out at $$\pu{228 K}$$. Same applies for halogen and interhalogen compounds hence these reactions are uncommon to find being pretty complex.

For this:
$$\ce{X'^- + XX'_n -> XX'_{n + 1}}$$
I can write reactions like:
$$\ce{I2 + IF3 -> 3IF}$$ at $$\pu{-78 °C}$$ in $$\ce{CCl3F}$$
$$\ce{Br2 + BrF3 -> 3BrF}$$
$$\ce{2I2 + IF5 -> 5IF}$$ (*not a point to mention that $$\ce{I-}$$ and $$\ce{Br-}$$ ions exist as $$\ce{I2}$$ and $$\ce{Br2}$$, respectively)

Keeping in mind that the generalization of above stated reactions seems incorrect.