ClF, BrF₃, IF₅, IF₇. Structure using VSEPR.

Interhalogen Compounds — Types and Examples

Question

Classify interhalogen compounds into types with one example each. Using VSEPR theory, predict the shapes of BrF₃, IF₅, and IF₇.

(JEE Advanced 2023 — this exact classification + shape prediction appeared as a multi-correct question)

Solution — Step by Step

Interhalogens form when two different halogens bond together. The general formula is XYₙ where X is the larger (less electronegative) halogen and Y is always the smaller, more electronegative one — almost always fluorine for the higher members.

The larger halogen sits at the centre because it can accommodate more bond pairs using its d-orbitals.

There are exactly four types based on the number of Y atoms:

Type	Examples	Hybridisation of X
XY	ClF, BrF, BrCl, ICl, IBr	sp³ (like any diatomic)
XY₃	ClF₃, BrF₃, ICl₃	sp³d
XY₅	ClF₅, BrF₅, IF₅	sp³d²
XY₇	IF₇ (only known example)	sp³d³

Note: XY₇ has only one known compound — IF₇. No other halogen pair can manage 7 bonds.

Br has 7 valence electrons. Three are used to form bonds with F. That leaves $7 - 3 = 4$ non-bonding electrons = 2 lone pairs.

Total electron pairs around Br = 3 (bond) + 2 (lone) = 5 → trigonal bipyramidal electron geometry.

Now, in a trigonal bipyramid, lone pairs prefer equatorial positions (less repulsion). Two lone pairs occupy two equatorial spots, and the three F atoms take the remaining positions. Shape of molecule = T-shaped.

I has 7 valence electrons. Five are used in bonds with F. Remaining: $7 - 5 = 2$ electrons = 1 lone pair.

Total electron pairs = 5 (bond) + 1 (lone) = 6 → octahedral electron geometry.

The lone pair occupies one position of the octahedron. The five F atoms sit in the remaining positions. Shape = Square pyramidal.

\text{Bond angles: F-I-F} \approx 81.9° \text{ (less than 90° due to lone pair repulsion)}

I uses all 7 valence electrons in bonding — no lone pairs. Total electron pairs = 7.

Seven pairs with no lone pair distortion → Pentagonal bipyramidal. This is the only common inorganic compound with this shape.

IF₇ is special for two reasons: it’s the only XY₇ interhalogen, and it’s one of the few pentagonal bipyramidal molecules you’ll encounter at JEE/NEET level. The others (like PF₇) don’t exist stably. IF₇ appears in JEE Advanced precisely because it’s unusual.

Why This Works

The central atom in interhalogens must always be the larger halogen because it has d-orbitals available for expanded octets. Fluorine has no d-orbitals and can never be the central atom — it always forms exactly one bond. This is why every XYₙ compound with n ≥ 3 uses fluorine as Y.

The shapes follow directly from VSEPR: count all electron pairs (bonding + lone), arrange them to minimise repulsion, then assign lone pairs to positions of least repulsion (equatorial in trigonal bipyramid, any position in octahedron since they’re equivalent — but lone pairs prefer the spot that maximises distance from other lone pairs).

For board exams, memorise the final shapes. For JEE, understand why — the lone pair placement logic is what gets tested in multi-correct and paragraph questions.

Alternative Method — Hybridisation Route

Instead of VSEPR, you can directly determine hybridisation and look up the shape:

For BrF₃:

\text{Hybridisation} = \frac{1}{2}[V + M - C + A]

where V = valence electrons of central atom, M = monovalent atoms, C = cationic charge, A = anionic charge.

= \frac{1}{2}[7 + 3 - 0 + 0] = 5 \rightarrow sp^3d \rightarrow \text{T-shaped (with 2 lone pairs)}

For IF₅: $\frac{1}{2}[7 + 5] = 6 \rightarrow sp^3d^2 \rightarrow$ Square pyramidal

For IF₇: $\frac{1}{2}[7 + 7] = 7 \rightarrow sp^3d^3 \rightarrow$ Pentagonal bipyramidal

Both methods give the same answer. VSEPR is more fundamental; the formula is faster in exams.

Common Mistake

BrF₃ is NOT trigonal planar. A very common error: students see 3 fluorines and write trigonal planar (like BF₃). But BF₃ is sp² with no lone pairs. BrF₃ has two lone pairs — they push the fluorines into a T-shape, compressing the F-Br-F angles below 90°. If you write trigonal planar in JEE, you lose the mark even if you identified the hybridisation correctly.

Similarly, don’t write IF₅ as trigonal bipyramidal (that’s 5 pairs with no lone pair). IF₅ has one lone pair making it square pyramidal — the lone pair sits “on top” of the square base.

Summary of shapes:

Compound	Electron Pairs	Lone Pairs	Shape
BrF₃	5	2	T-shaped
IF₅	6	1	Square pyramidal
IF₇	7	0	Pentagonal bipyramidal

Question

Solution — Step by Step

Why This Works

Alternative Method — Hybridisation Route

Common Mistake

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