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Interhalogen compounds such as ClF₃, BrF₅ and IF₇ are short, high-scoring questions: give the shape, the hybridisation or the oxidation state. Students often memorise each one separately. One count does them all. This post explains why interhalogens come only in certain formulas, gives a single rule for their shapes, checks it against measured structures and works a mixed example.
Which exams and books cover interhalogens?
This is a case where the syllabus and the current NCERT books do not match.
The JEE (Advanced) 2026 brochure names them directly: under Group 17 it lists "preparation/manufacture, properties, and uses of chlorine, hydrogen chloride and interhalogen compounds".
The Syllabus for JEE (Main) 2026 and the NEET (UG) 2026 syllabus cover Group 13 to Group 18 elements only as a "general introduction": electronic configuration, trends and the unique behaviour of the first element. Neither names interhalogens.
The current NCERT Class 12 Chemistry Part I has five units, ending with the d- and f-block elements and coordination compounds, as its 2026-27 contents page shows. There is no chapter on groups 15 to 18, and the Class 11 book has no p-block unit either.
The shapes do survive. Table 4.7 of the Class 11 chapter Chemical Bonding and Molecular Structure lists ClF₃ as T-shape and BrF₅ as square pyramid, and VSEPR theory is on all three syllabi.
So for JEE Main and NEET, learn the shapes through VSEPR; for JEE Advanced, learn the compounds as well. Check the syllabus for your own year.
What is an interhalogen compound?
It is a compound of two different halogens. The central atom X is the larger, less electronegative one; the outer atoms X′ are smaller, most often fluorine. The formulas are XX′ₙ with n = 1, 3, 5 or 7, and the largest number, seven, is reached in IF₇.
Why only odd numbers? A halogen has seven valence electrons. Each bond to X′ uses one of them. The 7 − n electrons left over must pair up as lone pairs, which needs 7 − n to be even, so n is odd. An even n would leave an unpaired electron.
Why is the larger atom in the middle? Packing three, five or seven atoms around a centre needs a big central atom, and fluorine, small and the most electronegative, sits on the outside. Counting each more electronegative outer atom as −1, the central atom's oxidation state is simply +n.
One count for every interhalogen
For a central halogen with n bonds:
Lone pairs = (7 − n)/2
Electron domains = n + (7 − n)/2 = (n + 7)/2
That gives four cases:
n = 1 (ClF, ICl): 4 domains, 3 lone pairs. Shape: linear.
n = 3 (ClF₃, BrF₃): 5 domains, 2 lone pairs, sp³d. Both lone pairs go equatorial in the trigonal bipyramid, leaving a T-shape.
n = 5 (BrF₅, IF₅): 6 domains, 1 lone pair, sp³d². Shape: square pyramidal.
n = 7 (IF₇): 7 domains, no lone pairs, sp³d³. Shape: pentagonal bipyramidal.
Do the measured structures agree?
ClF₃. From its microwave spectrum, Smith (1953) found a planar, distorted T with one short Cl–F bond (1.598 Å), two long ones (1.698 Å), and 87°29′ (about 87.5°) between the two kinds of bond. The angle is under 90° because the two lone pairs push the outer fluorines back.
BrF₃. Magnuson (1957) found the same planar, distorted T, with bonds of 1.721 Å and 1.810 Å and an angle of 86°12.6′, about 86.2°.
IF₇. Reanalysing electron diffraction data, LaVilla and Bauer (1960) found the I–F distances roughly equal, averaging 1.825 Å, and a symmetry "almost D5h", the pentagonal bipyramid. They also found that the five fluorines around the middle are not exactly in one plane.
For an exam, write pentagonal bipyramidal for IF₇; the puckering is a refinement beyond the syllabus.
Worked example: four molecules in one pass
Give the oxidation state of the central atom, lone pairs, hybridisation, shape and polarity for IF₅, ClF, BrF₃ and IF₇.
IF₅: n = 5, so +5; (7 − 5)/2 = 1 lone pair; 6 domains, sp³d²; square pyramidal; polar, because the lone pair makes it lopsided.
ClF: n = 1, so +1; 3 lone pairs; linear; polar, because the two atoms differ in electronegativity.
BrF₃: n = 3, so +3; 2 lone pairs; 5 domains, sp³d; T-shaped; polar.
IF₇: n = 7, so +7; no lone pairs; 7 domains, sp³d³; pentagonal bipyramidal; non-polar in this ideal shape, because the bond dipoles cancel.
A check on IF₇: seven bonds put 14 electrons around iodine, far past an octet. That is possible for iodine, not for fluorine, which is why fluorine is never the central atom.
Which mistakes cost marks?
Calling ClF₃ trigonal planar by ignoring its two lone pairs.
Putting lone pairs axial in a trigonal bipyramid. They go equatorial.
Calling BrF₅ trigonal bipyramidal because it has five fluorines. Count the lone pair: six domains.
Writing an even n, such as ClF₂ or IF₄, as a neutral molecule.
Mixing up arrangement and shape. BrF₅ has an octahedral arrangement and a square pyramidal shape.
Our post on VSEPR and molecular shapes gives the general four-step method, and the one on p-block groups 15 to 17 covers halogen trends and oxoacids.
If the idea of electron domains itself is shaky, the Fermi Concept Library covers every concept in chemistry, each with its own lesson, so you can fix that first.
A practice routine
Day 1: Derive the (7 − n)/2 rule and write the four shapes from memory.
Day 2: Sketch ClF₃, BrF₃, BrF₅, IF₅ and IF₇ with lone pairs drawn in.
Day 3: Mix in five non-interhalogens, such as SF₄, XeF₄ and PCl₅, and repeat.
Day 4 (JEE Advanced): Add one preparation reaction from your textbook for ClF₃ and one for IF₇.
Today, run IF₃ and ClF₅ through the count and name each shape. You should get T-shaped and square pyramidal.
