Mateen Subhash

Iodine monobromide is an interhalogen compound formed when iodine reacts with bromine. It’s chemical formula is IBr. It is a dark red colored solid that gets melted in room temperature. It can be also called iodine bromide.

It’s molar mass is about 206.904g/mol. It’s melting point and boiling point is 420C and 1160C respectively. It gets completely miscible in water, ethanol, carbon disulphide, glacial acetic acid and ether. It is a source of I and can form charge transfer complexes when reacts with Lewis acids.

It is used in iodometric titrations, medical fields for heart imaging, as a powerful iodinating agent. Sometimes it is used for the generating electrophile for polyketide synthesis.

The Lewis structure of an atom can be used to represent the bonds and electrons in it in the most basic way. Any compound’s Lewis structure can be easily drawn through a few steps. The steps for drawing Lewis structure of iodine bromide is as follows

There is no resonance seen in iodine bromide.

The shape of iodine bromide is linear with a bond angle of 1800. It’s bond length is found to be 249 pm.

The equation used for calculating the formal charge of any compound is

Formal charge = Valence electrons – No. of non bonded electrons – No. of bonds

Formal charge of iodine = 7-6-1 =0

Formal charge of bromine =7-6-1 =0

So the formal charge of iodine bromide is zero.

The bond angle of iodine bromide is 1800.

The electrons which don’t involve in bond making is called lone pair of electrons or non bonded electrons. There is three lone pairs are found in both iodine and bromine. So in total 6 lone pairs or 12 non bonded electrons are seen in iodine bromide.

The term “valence electrons” refers to the total number of electrons that make up an atom’s outer shell. Iodine and bromine have seven electrons in their valence shells since they are both halogens.

The hybridisation is the process of formation of new set of orbitals with same energy from atomic orbitals of different energy to make more stable bond. The hybridisation of compound can be different. It can be sp3 ,sp2 , sp. The hybridisation of iodine bromide is sp3.

Iodine bromide is a polar compound. Since both are halogens there is slight electronegativity difference can be seen in between them. The electronegativity of iodine and bromine is 2.66 and 2.96 respectively.

Also bromine is more electronegative than iodine. So iodine bromide is found to be polar in nature.

The exchange of electrons between iodine and bromine results in the formation of iodine monobromide. Therefore, it is an entirely covalent compound.

The electronic and molecular geometry of iodine bromide is same. It is a linear shaped compound.

An atom’s charge after forming a bond is its oxidation state. There are certain rules for finding the oxidation state of an atom. Usually halogens are assigned with -1 .

Oxidation state of a compound is found in such a way that the sum of oxidation state of all the atoms present in a compound should be 0.Here in iodine bromide bromine has -1 and iodine has +1 charge. So in total it is 0.

Iodine bromide or iodine monobromide is a polar compound with dipole moment value is about 1.21 D.

Conclusion

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