Interhalogen compounds — types, structure, and bonding of ClF₃, BrF₅, IF₇

Question

What are interhalogen compounds? Discuss the structure, geometry, and hybridization of ClF $_3$ , BrF $_5$ , and IF $_7$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Interhalogen compounds are formed between two different halogens. The general formula is $\text{AX}_n$ where A is the larger (less electronegative) halogen and X is the smaller (more electronegative) halogen. The value of $n$ can be 1, 3, 5, or 7.

The larger halogen is always the central atom because it can accommodate more bonds by expanding its octet using d-orbitals.

Type: AX $_3$
Total electron pairs around Cl: 3 bond pairs + 2 lone pairs = 5 pairs
Hybridization: $sp^3d$
Geometry (electron pair): Trigonal bipyramidal
Shape (molecular): T-shaped

The two lone pairs occupy equatorial positions (to minimise lone pair-lone pair repulsion), pushing the three F atoms into a T-shape. The axial F-Cl-F angle is less than 180° due to lone pair repulsion.

Type: AX $_5$
Total electron pairs around Br: 5 bond pairs + 1 lone pair = 6 pairs
Hybridization: $sp^3d^2$
Geometry (electron pair): Octahedral
Shape (molecular): Square pyramidal

The lone pair occupies one position of the octahedron, giving a square pyramidal shape with Br slightly below the plane of four F atoms.

Type: AX $_7$
Total electron pairs around I: 7 bond pairs + 0 lone pairs = 7 pairs
Hybridization: $sp^3d^3$
Geometry and shape: Pentagonal bipyramidal

This is the maximum number of halogen atoms that can bond to a central halogen. Only iodine (the largest halogen) can form AX $_7$ because it alone has enough space and available d-orbitals.

Why This Works

The structures follow from VSEPR theory. The central halogen uses its d-orbitals to expand beyond the octet. Larger halogens can accommodate more surrounding atoms because: (1) they have larger atomic radius (more space), and (2) they have available d-orbitals for bonding.

Fluorine, being the smallest and most electronegative halogen, is always the outer atom. It never acts as the central atom in interhalogens because it has no d-orbitals (Period 2 element).

Alternative Method — Quick Reference Table

Compound	Type	Hybridization	Shape	Lone pairs
ClF	AX	$sp^3$	Linear	3
ClF $_3$	AX $_3$	$sp^3d$	T-shaped	2
BrF $_5$	AX $_5$	$sp^3d^2$	Square pyramidal	1
IF $_7$	AX $_7$	$sp^3d^3$	Pentagonal bipyramidal	0

Pattern to remember: as $n$ increases by 2, the hybridization adds one more d-orbital. AX → $sp^3$ , AX $_3$ → $sp^3d$ , AX $_5$ → $sp^3d^2$ , AX $_7$ → $sp^3d^3$ . The number of lone pairs = (7 - $n$ )/2 for the central halogen.

Common Mistake

Students confuse the electron pair geometry with the molecular shape. ClF $_3$ has trigonal bipyramidal electron pair geometry but T-shaped molecular geometry. Examiners specifically ask for the molecular shape (not electron pair geometry). Always subtract the lone pairs from the full geometry to get the molecular shape.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quick Reference Table

Common Mistake

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