Chemistry · The p-Block Elements (Class 12)

Interhalogen Compounds

When two different halogens combine, they form a small but exceptionally instructive family of molecules called interhalogen compounds. Treated in the old-NCERT Group 17 supplement (§7.22) and in NIOS Chemistry (§20.7.4), this topic packs together periodic trends, bond strength, hydrolysis chemistry and VSEPR geometry into one set of formulae — XX′, XX′3, XX′5 and XX′7. For NEET it is a recurring source of match-the-shape and assertion–reason questions, so the structures of $\ce{BrF3}$, $\ce{IF5}$ and $\ce{IF7}$ are worth committing to memory.

What Interhalogen Compounds Are

The halogens of Group 17 — fluorine, chlorine, bromine and iodine — react not only with other elements but also with one another, producing a series of mixed binary compounds known as interhalogen compounds. Each molecule contains atoms of exactly two different halogens, written in the general form $\ce{XX'_n}$, where X is the larger, more electropositive halogen and X′ is the smaller, more electronegative halogen, and $n = 1, 3, 5$ or $7$.

The reason the larger halogen sits at the centre is geometric. A central atom can only accommodate as many surrounding atoms as its size permits. As the ratio of the radius of X to the radius of X′ increases, more X′ atoms can pack around X, so the number of atoms per molecule rises. This single principle explains the whole family: small differences in size give the 1:1 type $\ce{XX'}$, while the extreme case — large iodine surrounded by the smallest halogen, fluorine — gives $\ce{IF7}$, the only species with seven peripheral atoms.

Because the bonding is between two non-metals of similar but unequal electronegativity, every interhalogen is a covalent, diamagnetic molecule. Their behaviour is broadly intermediate between that of the two parent halogens, with one important twist in reactivity that NEET examiners love to probe.

The Four Types: XX′ to XX′7

Interhalogens are classified purely by stoichiometry into four types. The 1:1 type is known for every halogen pair; the higher types appear only where the size difference is large enough, and fluorine — being smallest — is the peripheral atom in almost all of the higher members.

TypeFormula examplesRepresentative speciesPhysical state / colour
XX′$\ce{ClF}$, $\ce{BrF}$, $\ce{ICl}$, $\ce{IBr}$, $\ce{BrCl}$$\ce{ClF}$colourless gas
XX′$\ce{ICl}$ruby-red solid (α-form); brown-red (β-form)
XX′3$\ce{ClF3}$, $\ce{BrF3}$, $\ce{IF3}$, $\ce{ICl3}$$\ce{BrF3}$yellow-green liquid
XX′5$\ce{IF5}$, $\ce{BrF5}$, $\ce{ClF5}$$\ce{IF5}$colourless gas (solid below 77 K)
XX′7$\ce{IF7}$ only$\ce{IF7}$colourless gas

Note that fluorine never sits at the centre of an interhalogen molecule. As the most electronegative and smallest halogen, it cannot expand its valence shell or act as the larger, electropositive partner — so it is always the peripheral X′ atom, never X. This is also why iodine, the largest halogen, is the central atom in the highest-coordination species $\ce{IF5}$ and $\ce{IF7}$.

Preparation

Two general routes give interhalogen compounds: direct combination of the two halogens, or the action of a halogen on a lower interhalogen. Which product forms depends on the proportions of reactants and the temperature, so controlling conditions is the whole game.

ConditionReactionType
Equal volumes, 437–473 K$\ce{Cl2 + F2 -> 2ClF}$XX′
Excess $\ce{F2}$, 573 K$\ce{Cl2 + 3F2 -> 2ClF3}$XX′3
With $\ce{F2}$, ~293 K$\ce{I2 + 5F2 -> 2IF5}$XX′5
Excess $\ce{F2}$, 523–573 K$\ce{I2 + 7F2 -> 2IF7}$XX′7
Equimolar, 573 K$\ce{I2 + Cl2 -> 2ICl}$XX′
Iodine in excess $\ce{Cl2}$$\ce{I2 + 3Cl2 -> 2ICl3}$XX′3

The pattern is consistent: a larger excess of the smaller, more reactive halogen (almost always fluorine) drives the equilibrium toward the higher interhalogen. Lowering the fluorine ratio stops the reaction at the lower type. This dependence on stoichiometry is the key to selecting between $\ce{ClF}$ and $\ce{ClF3}$, or between $\ce{ICl}$ and $\ce{ICl3}$.

General Physical Properties

All interhalogen compounds are covalent and diamagnetic. They are volatile solids or liquids at 298 K, with the single exception of $\ce{ClF}$, which is a gas. Their physical properties — colour, melting and boiling points — are intermediate between those of the two constituent halogens, except that their melting and boiling points run a little higher than a naive average would predict, owing to the polarity of the $\ce{X-X'}$ bond.

Reading the trend

Why does the colour deepen down the table — $\ce{ClF}$ colourless, $\ce{ICl}$ ruby-red, $\ce{IBr}$ black?

As the constituent halogens get heavier and more polarisable, absorption shifts further into the visible region. The trend simply mirrors that of the parent halogens, consistent with the rule that interhalogen properties lie between those of their two halogens.

Why They Out-React the Halogens

The headline chemical fact — and the one most often tested — is that interhalogen compounds are, in general, more reactive than the halogens, the lone exception being fluorine. The cause is bond strength: the $\ce{X-X'}$ bond in an interhalogen is weaker than the $\ce{X-X}$ bond in the parent halogens, except for the unusually strong $\ce{F-F}$ bond. A weaker bond breaks more readily, so the molecule is a more aggressive reagent.

NEET Trap

"More reactive" needs the fluorine caveat

The statement is "interhalogens are more reactive than halogens except fluorine." A common assertion–reason item claims $\ce{ICl}$ is more reactive than $\ce{I2}$ because the $\ce{I-Cl}$ bond is weaker than the $\ce{I-I}$ bond — both statements are true and the reason explains the assertion. Do not blanket-state that every interhalogen beats every halogen.

Rule: weaker $\ce{X-X'}$ bond ⇒ higher reactivity; $\ce{F2}$ is the exception because $\ce{F-F}$ is exceptionally weak yet $\ce{F2}$ itself is extremely reactive.

Build the base first

The bond-strength logic here rests on Group 17 periodic trends. Revise them in Group 17: The Halogen Family.

Hydrolysis Pattern

Every interhalogen undergoes hydrolysis with water, and the products follow a clean rule. The halide ion comes from the smaller halogen (X′), while the oxoanion comes from the larger halogen (X), and the type of oxoanion rises step-by-step with the formula. The 1:1 case is the template:

$\ce{XX' + H2O -> HX' + HOX}$

TypeOxoanion of the larger halogen XExample anion
XX′hypohalite$\ce{OX^-}$
XX′3halite$\ce{XO2^-}$
XX′5halate$\ce{XO3^-}$
XX′7perhalate$\ce{XO4^-}$

The oxidation state of the larger halogen in the oxoanion climbs from +1 (hypohalite) to +7 (perhalate) exactly as the number of fluorine atoms in the formula climbs, which is a useful internal consistency check when reconstructing a hydrolysis product under exam pressure.

VSEPR Structures & Hybridisation

The geometries of the higher interhalogens are a direct application of VSEPR theory to the central atom. Count the central atom's seven valence electrons, subtract the bonding electrons, and the remainder gives the lone pairs that decide the shape. The four types map onto four distinct geometries.

TypeBond pairsLone pairsHybridisationMolecular shape
XX′13$sp^3$Linear
XX′332$sp^3d$Bent T-shape
XX′551$sp^3d^2$Square pyramidal
XX′770$sp^3d^3$Pentagonal bipyramidal

XX′3 — the bent T-shape of $\ce{BrF3}$

In $\ce{BrF3}$ the central Br has seven valence electrons. Three of these form electron-pair bonds to three fluorine atoms, leaving four electrons as two lone pairs. With three bond pairs and two lone pairs, the electron domains adopt a trigonal bipyramidal arrangement ($sp^3d$). The two lone pairs occupy the equatorial positions — minimising the strong lone-pair–lone-pair and lone-pair–bond-pair repulsions — and the two axial fluorines bend slightly toward the equatorial fluorine. The result is a slightly bent T-shape.

Figure 1 · BrF₃ geometry Br F F F 2 lone pairs (equatorial)

Two equatorial lone pairs push the axial Br–F bonds inward, distorting the ideal "T" into a slightly bent T-shape. The same logic applies to $\ce{ClF3}$.

XX′5 and XX′7 — $\ce{IF5}$ square pyramidal, $\ce{IF7}$ pentagonal bipyramidal

In $\ce{IF5}$ the central iodine forms five bonds and retains one lone pair (five bond pairs, one lone pair, $sp^3d^2$). The six electron domains sit at the vertices of an octahedron; the single lone pair takes one vertex, pushing the five fluorines into a square-pyramidal shape. In $\ce{IF7}$ the iodine forms seven bonds with no lone pairs ($sp^3d^3$), so the seven fluorines spread to the vertices of a pentagonal bipyramid.

Figure 2 · IF₅ vs IF₇ IF₅ · square pyramidal I F F F F F 1 lone pair IF₇ · pentagonal bipyramidal I F F F F F F F 2 axial (amber) + 5 equatorial

$\ce{IF5}$: five bond pairs and one lone pair give an octahedral arrangement that collapses to a square pyramid. $\ce{IF7}$: seven bond pairs, zero lone pairs, an undistorted pentagonal bipyramid with five equatorial and two axial $\ce{I-F}$ bonds.

NEET Trap

Electron geometry vs molecular shape

For $\ce{XX'3}$ the electron arrangement is trigonal bipyramidal, but the molecular shape is the bent T — the lone pairs are invisible in the name. Likewise $\ce{XX'5}$ has an octahedral electron arrangement but a square-pyramidal shape. NEET answer keys quote the molecular shape: T-shape, square pyramidal, pentagonal bipyramidal.

Map: $\ce{XX'}\to$ linear · $\ce{XX'3}\to$ T-shape · $\ce{XX'5}\to$ square pyramidal · $\ce{XX'7}\to$ pentagonal bipyramidal.

Uses as Fluorinating Agents

Interhalogen compounds serve as non-aqueous solvents and, more importantly, as very powerful fluorinating agents. The weak, polar $\ce{X-F}$ bonds release fluorine readily, making compounds such as $\ce{ClF3}$ and $\ce{BrF3}$ ideal for converting metals and oxides into fluorides.

The textbook industrial application is the production of uranium hexafluoride, the volatile compound used to enrich uranium-235:

$\ce{U(s) + 3ClF3(l) -> UF6(g) + 3ClF(g)}$

Here $\ce{ClF3}$ fluorinates elemental uranium straight to gaseous $\ce{UF6}$, which is then separated by diffusion or centrifugation. This single reaction is the most frequently cited use of interhalogens, so tie "fluorinating agent + uranium enrichment" together in your memory.

Quick Recap

Interhalogen compounds in one screen

  • Four types: $\ce{XX'}$, $\ce{XX'3}$, $\ce{XX'5}$, $\ce{XX'7}$; X is the larger/electropositive halogen, X′ the smaller/electronegative one. $\ce{IF7}$ is the only XX′7.
  • Made by direct halogen combination or halogen + lower interhalogen; excess fluorine drives toward higher types.
  • Covalent, diamagnetic; properties intermediate between the parent halogens; $\ce{ClF}$ alone is a gas.
  • More reactive than halogens (except $\ce{F2}$) because $\ce{X-X'}$ is weaker than $\ce{X-X}$ (except $\ce{F-F}$).
  • Hydrolysis → halide of the smaller halogen + hypohalite/halite/halate/perhalate of the larger halogen as the type rises.
  • Shapes: linear → bent T ($sp^3d$) → square pyramidal ($sp^3d^2$) → pentagonal bipyramidal ($sp^3d^3$).
  • Use: powerful fluorinating agents; $\ce{ClF3}/\ce{BrF3}$ make $\ce{UF6}$ for $\ce{^{235}U}$ enrichment.

NEET PYQ Snapshot — Interhalogen Compounds

Real NEET-pattern questions on interhalogen reactivity and geometry.

NEET PYQ · Assertion–Reason

Assertion (A): $\ce{ICl}$ is more reactive than $\ce{I2}$. Reason (R): The $\ce{I-Cl}$ bond is weaker than the $\ce{I-I}$ bond.

  • (1) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
  • (2) (A) is correct but (R) is not correct.
  • (3) (A) is not correct but (R) is correct.
  • (4) Both (A) and (R) are correct and (R) is the correct explanation of (A).
Answer: (4)

Interhalogens are generally more reactive than halogens (except fluorine) because the $\ce{X-X'}$ bond is weaker than the $\ce{X-X}$ bond — except the strong $\ce{F-F}$ bond. The weaker $\ce{I-Cl}$ bond compared with $\ce{I-I}$ makes $\ce{ICl}$ more reactive than $\ce{I2}$, so (R) correctly explains (A).

NEET PYQ · Match the Columns

Match the interhalogen compounds of Column I with the geometry in Column II and assign the correct code.
(a) XX′   (b) XX′3   (c) XX′5   (d) XX′7   |   (i) T-shape (ii) Pentagonal bipyramidal (iii) Linear (iv) Square pyramidal (v) Tetrahedral

  • (1) a–iv, b–iii, c–ii, d–i
  • (2) a–iii, b–iv, c–i, d–ii
  • (3) a–iii, b–i, c–iv, d–ii
  • (4) a–v, b–iv, c–iii, d–ii
Answer: (3)

$\ce{XX'}$ ($sp^3$) is linear (iii); $\ce{XX'3}$ ($sp^3d$) is T-shape (i); $\ce{XX'5}$ ($sp^3d^2$) is square pyramidal (iv); $\ce{XX'7}$ ($sp^3d^3$) is pentagonal bipyramidal (ii). The matching code is a–iii, b–i, c–iv, d–ii.

FAQs — Interhalogen Compounds

Quick answers to the questions NEET aspirants ask most about interhalogens.

What are interhalogen compounds and what are their four types?

Interhalogen compounds are mixed binary compounds formed when two different halogens combine. They occur in four composition types: XX′ (e.g. ClF, BrF, ICl, IBr), XX′3 (e.g. ClF3, BrF3), XX′5 (e.g. IF5, BrF5) and XX′7 (only IF7). Here X is the larger, more electropositive halogen and X′ the smaller, more electronegative one. As the radius ratio of X to X′ increases, the number of smaller atoms that can pack around X increases, which is why iodine and fluorine give IF7.

Why are interhalogen compounds more reactive than halogens?

In general interhalogen compounds are more reactive than the halogens (except fluorine) because the X–X′ bond is weaker than the X–X bond in the parent halogens, the only exception being the very strong F–F bond. The weaker, more polar X–X′ bond breaks easily, so a molecule such as ICl is more reactive than I2.

What is the shape of BrF3 by VSEPR theory?

In BrF3 the central Br atom has 7 valence electrons. Three of these form bonds to three fluorine atoms, leaving 3 bond pairs and 2 lone pairs, giving sp3d hybridisation and a trigonal bipyramidal electron arrangement. The two lone pairs occupy equatorial positions to minimise repulsions, and the two axial F atoms bend slightly towards the equatorial F. The resulting molecular shape is a slightly bent T-shape.

What happens when an interhalogen compound is hydrolysed?

All interhalogen compounds undergo hydrolysis to give a halide ion derived from the smaller halogen and an oxoanion derived from the larger halogen. The oxoanion depends on the type: XX′ gives a hypohalite, XX′3 a halite, XX′5 a halate and XX′7 a perhalate. For example, XX′ hydrolyses as XX′ + H2O → HX + HOX′.

Why is IF7 the interhalogen with the largest number of atoms?

The number of smaller halogen atoms that can surround the central atom rises as the radius ratio of the large halogen (X) to the small halogen (X′) increases. Iodine is the largest halogen and fluorine the smallest, so the I to F radius ratio is the maximum possible. This allows seven fluorine atoms to pack around iodine, giving IF7, which is why no XX′7 species other than IF7 exists.

What are interhalogen compounds used for?

Interhalogen compounds are used as non-aqueous solvents and as very powerful fluorinating agents. In particular ClF3 and BrF3 are used to produce UF6 during the enrichment of uranium-235, for example U(s) + 3ClF3(l) → UF6(g) + 3ClF(g).