Iodine Bromide Formula (Bonding Guide)
Iodine bromide has the formula IBr: one iodine atom bonded to one bromine atom. Its Lewis structure contains a single covalent bond and three lone pairs on each atom, for 14 valence electrons total. The bond is polar, with its dipole toward bromine. IBr is a linear, two-atom interhalogen molecule, not an ionic pair.
If you need the formula and bonding answer quickly, remember IBr and draw I—Br. The most common mistake is to see oxidation states of +1 and −1 and then draw separate ions. Those oxidation states are bookkeeping labels; they do not change the standard Lewis structure, which shows a shared bond and zero formal charges.
I use a short check to keep the ideas separate: count electrons, draw the bond, check charges, then describe polarity. The steps below show how to do that and explain why each result fits.
Iodine bromide formula and classification
Iodine bromide is a binary interhalogen compound with the chemical formula IBr. “Binary” means it contains two elements, while “interhalogen” means both elements are halogens. Each formula unit has one iodine atom and one bromine atom; the formula does not mean the atoms have ionic charges.
The symbols are written as I for iodine and Br for bromine. The 1-to-1 ratio needs no subscript, so the formula is IBr rather than I₁Br₁. It is also sometimes called iodine monobromide, but IBr is the standard concise formula used in chemistry.
The order of the symbols does not show which atom has a partial charge. Here, iodine comes first and bromine second, and bromine is more electronegative. That difference matters when describing the bond, but it does not turn the formula into a diagram of ions.
IBr is a molecular compound: its basic molecular unit contains two atoms joined by a covalent bond. It is not a polyatomic molecule, because it has only two atoms, and it is not correctly represented as an iodine ion next to a bromide ion when drawing its standard Lewis structure.
Quick identification
| Feature | Iodine bromide |
|---|---|
| Formula | IBr |
| Elements present | Iodine and bromine |
| Atom ratio | 1 iodine : 1 bromine |
| Compound class | Interhalogen |
| Molar mass | About 206.81 g/mol |
| Number of atoms per molecule | 2 |
The molar mass is the sum of the atomic masses for one iodine atom and one bromine atom. It is a useful calculation check, but it does not tell you whether the bond is ionic or covalent.
Valence electrons and the Lewis structure
A Lewis structure is a simple drawing of valence electrons, the outer electrons involved in bonding. For IBr, add iodine’s seven valence electrons to bromine’s seven. The total is 14 valence electrons, which must appear as one shared bond and six lone pairs.
Both iodine and bromine are in Group 17, the halogen group, so each has seven valence electrons. This gives the first step:
- Iodine: 7 valence electrons
- Bromine: 7 valence electrons
- Total: 7 + 7 = 14 valence electrons
Next, place one electron pair between the atoms to make a single bond: I—Br. A single bond uses two electrons. The remaining 12 electrons are arranged as three lone pairs on iodine and three lone pairs on bromine.
The completed structure can be read as I—Br, with three lone pairs around each symbol. Each atom counts the two shared bonding electrons toward its outer shell and has an octet: eight electrons around it. No extra bond is needed to complete either atom’s octet.
Electron-count check
| Part of structure | Electrons |
|---|---|
| One I—Br single bond | 2 |
| Three lone pairs on iodine | 6 |
| Three lone pairs on bromine | 6 |
| Total | 14 |
A helpful drawing habit is to count the pairs, not just the dots. There is one bonding pair and six lone pairs. If your drawing has 12 or 16 electrons, or gives either atom the wrong number of lone pairs, revisit the count before moving on.
Formal charges: why the Lewis structure is neutral
Formal charge is a way to check how electrons are assigned in a Lewis structure. It compares an atom’s usual valence-electron count with the electrons assigned to it in the drawing. For IBr, the single-bond structure gives both atoms a formal charge of zero.
Use this rule for each atom:
Formal charge = valence electrons − nonbonding electrons − half the bonding electrons
For iodine, the calculation is 7 − 6 − 1 = 0. For bromine, it is also 7 − 6 − 1 = 0. Each atom has six nonbonding electrons in its three lone pairs and shares the two bonding electrons, with half assigned to each atom for this calculation.
So the Lewis structure shows no formal charge on either atom. Write it as I—Br, not as I⁺—Br⁻. The latter suggests full charges on separate ions and does not represent the standard Lewis model for the molecule.
Formal charge and oxidation state answer different questions. Formal charge belongs to a specific Lewis structure and its electron assignment. Oxidation state uses a convention that assigns bonding electrons to the more electronegative atom. Keeping those tools distinct prevents a common mix-up.
Bond polarity, shape, and dipole direction
Bond polarity describes an uneven sharing of bonding electrons. Bromine is more electronegative than iodine, so the shared electrons are drawn more toward bromine. The molecule has a bond dipole pointing toward Br, often shown as Iδ+—Brδ−.
Electronegativity is a measure of how strongly an atom attracts shared electrons in a bond. Because bromine attracts the shared pair more strongly, iodine has a partial positive charge, written δ+, and bromine has a partial negative charge, written δ−. These symbols mean partial charge, not full ionic charge.
The bond is therefore polar covalent: the atoms share electrons, but not equally. “Covalent” describes the shared bond; “polar” describes the uneven sharing. Neither term means that the molecule consists of I⁺ and Br⁻ ions.
IBr has two atoms, so its molecular shape is linear. Any two-atom molecule lies along a straight line: there is only one bond connecting the atoms, not a central atom with several bonds that could form a bent shape.
When drawing the dipole, place the arrowhead toward bromine, the more electronegative atom. Some diagrams mark the iodine end with a small cross or plus sign. The exact notation may vary by class, so follow your course’s convention, but the direction remains toward Br.
Oxidation states are not formal charges
An oxidation state is a bookkeeping value assigned by a set of conventions. In IBr, the conventional assignments are iodine +1 and bromine −1, because bromine is more electronegative. These numbers help track electron assignment, but they are not the Lewis structure’s formal charges.
For oxidation-state bookkeeping, imagine assigning the bonding electrons to bromine. Bromine then has the −1 oxidation state, and iodine has +1. The values add to zero, which matches the neutral formula IBr.
That imagined assignment does not mean the molecule has fully separated ions. In the Lewis model, the bond remains covalent and the formal charges remain zero. This is why a chemistry answer can correctly report both oxidation states, I = +1 and Br = −1, while still drawing a neutral I—Br bond.
| Term | What it describes for IBr | Value or result |
|---|---|---|
| Formal charge | Electron assignment in the Lewis structure | I = 0, Br = 0 |
| Oxidation state | Bookkeeping assignment using electronegativity conventions | I = +1, Br = −1 |
| Partial charge | Uneven sharing in the polar bond | Iδ+, Brδ− |
| Bond type | How the atoms are bonded | Polar covalent |
A quick diagnostic: if a question asks for a Lewis structure, show the bond, lone pairs, and formal charges. If it asks for oxidation states, report +1 and −1. Do not use the oxidation-state numbers as a reason to replace the shared bond with ions.
Worked checks for common chemistry questions
A worked check applies the bonding rules to a typical classroom task. For IBr, the same facts answer several different questions: the formula gives the atom ratio, the electron count supports the Lewis drawing, and electronegativity identifies the dipole direction.
Question: How many valence electrons are in IBr?
Add the valence electrons from the two atoms: 7 from iodine plus 7 from bromine equals 14. Do not use the total number of electrons in the atoms; Lewis structures count valence electrons.
Question: How many lone pairs should the structure show?
Draw one single bond, which uses two electrons. Place the remaining 12 electrons as six pairs: three lone pairs on iodine and three on bromine.
Question: Does the molecule have a net formal charge?
No. Using the formal-charge calculation, each atom has a formal charge of zero. The molecule as drawn is neutral.
Question: Which end is partially negative?
The bromine end. Bromine is more electronegative than iodine, so the bond dipole points toward bromine.
Question: Is the molecule bent?
No. IBr contains just two atoms and one bond, so it is linear. The three lone pairs on each atom do not make this two-atom molecule bent.
When I explain this example, I find it useful to label each answer by the question it addresses: “Lewis structure,” “polarity,” or “oxidation state.” That small habit helps prevent correct facts from being combined in an incorrect way.
FAQ: iodine bromide formula and bonding
These short answers recap the formula, Lewis structure, polarity, and charge conventions. Use them as a final check, but follow your instructor’s notation if a course specifies a particular way to draw lone pairs or dipole arrows.
What is the chemical formula for iodine bromide?
The formula is IBr. It contains one iodine atom and one bromine atom.
How many valence electrons does IBr have?
It has 14 valence electrons: iodine contributes seven and bromine contributes seven.
What is the Lewis structure of IBr?
Draw a single bond, I—Br, and three lone pairs on each atom. Both atoms have formal charge zero.
Is iodine bromide ionic or covalent?
It has a polar covalent bond. The atoms share electrons unevenly; the standard Lewis structure is not a pair of separate ions.
Which atom is partially negative in IBr?
Bromine is partially negative because it is more electronegative than iodine. The bond dipole points toward Br.
What are the oxidation states in IBr?
The conventional oxidation states are iodine +1 and bromine −1. These are bookkeeping assignments, not formal charges.
What is the shape of IBr?
It is linear because it has only two atoms joined by one bond.
What is the molar mass of IBr?
The molar mass is about 206.81 g/mol, found by adding the atomic masses of one iodine atom and one bromine atom.
The reliable final picture is simple: 14 valence electrons, one shared I—Br bond, three lone pairs on each atom, zero formal charges, and a dipole toward bromine. Keep oxidation-state labels separate from that Lewis structure, and the main bonding questions follow consistently.
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
