is ibr ionic or covalent?

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

IBr contains covalent bonds because both iodine and bromine are nonmetallic elements.

Consider a compound that forms between the elements iodine and bromine, iodine monobromide, IBr. Iodine and bromine both have 7 valence electrons and need one more electron to complete their valence shell:

\(\overset{⬝\,⬝}{\underset{⬝\,⬝}{:\mathrm{I}\,⋅}}\)   \(\overset{⬝\,⬝}{\underset{⬝\,⬝}{⋅\,\mathrm{Br}:}}\)

If either iodine or bromine were to given up valence electrons to form a cation, they would have to give up all seven valence electrons to reveal the next energy level below as the new valence shell. Giving up seven valence electrons is energetically impossible, as it would imply the formation of an ion with a 7+ charge. Since opposite charges attract, the removal of each successive electron becomes increasingly difficult due to the increasing attraction between the nucleus and electrons in the electron cloud. It is much easier for each atom to gain one electron than it is to lose seven.

As the case illustrates above, it isn't easy for an atom to lose a whole bunch of electrons. Likewise, it isn't easy for an atom to gain a whole bunch of electrons. When working with an element like magnesium, we noticed that magnesium lost its two valence electrons to form magnesium ions, Mg2+, rather than gaining six electrons to fill the valence shell to form ions with a 6– charge. Since like charges repel, the addition of each successive electron becomes increasingly difficult due to the increasing repulsions between electrons in the electron cloud. It is much easier for each atom to lose two electrons than it is to gain six.

If this is the case, how is a compound such as iodine monobromide, IBr, possible? A second type of chemical bond, called a covalent bond, involves the sharing of electrons to complete a valence shell rather than transferring electrons. A covalent bond typically forms between nonmetals and/or metalloids, since they have a valence shell that usually contains four or more valence electrons.

Let's illustrate how a covalent bond forms between iodine and bromine, with the understanding that each atom only needs one more electron to complete an octet in the valence shell. The iodine and bromine atoms each share one electron with the other.

Through sharing, the iodine atom now has access to eight valence electrons, as does the bromine atom. The portion where the circles overlap represent a shared pair of electrons, otherwise known as a covalent bond. Electrons that are not a part of a covalent bond are called lone pairs.

When two atoms share a single pair of electrons, the bond is called a single covalent bond, or simply, a single bond. When writing out a Lewis structure, a dash is used to represent a shared pair of electrons in place of two dots. Therefore, the Lewis structure for iodine monobromide, IBr, is shown in Figure \(\PageIndex{1}\):

Something might stand out when looking at the Lewis structure of IBr above – all of the electrons are paired up, whether they are bonding pairs (called covalent bonds) or nonbonding pairs (called lone pairs). In fact, this is usual standard for valence electrons in molecules – they are always paired, with a few exceptions that are not covered in this text. As we will learn in Section 11.7, the presence of lone pairs on the central atom(s) of a molecule contributes to its three-dimensional shape.

Let's look next at the structure of a water molecule. The chemical formula for water is H2O. Each hydrogen atom has 1 valence electron. Because the first energy level holds a maximum of two valence electrons, it only needs one more electron to complete its valence shell. This sometimes called the duet rule. An oxygen atom has 6 valence electrons and needs two additional electrons to complete its valence shell:

\({\color{0.8, 0.0, 0.0}\begin{array}{c}\mathrm H\,⋅\\\mathrm H\,⋅\end{array}}\;\;\;\;{\color{0.0, 0.0, 1.0}\overset{⬝\,⬝}{\underset⬝{⋅\,\mathrm O:}}}\)

Let's illustrate how covalent bonds form between hydrogen and oxygen:

Notice that the oxygen atom makes two single bonds, one to each of the two hydrogen atoms. This makes sense, since the oxygen atom needed two electrons to complete its valence shell. The Lewis structure for water is

There is nothing special about the orientation of the water molecule shown above. In fact, any of the orientations shown in Figure \(\PageIndex{2}\) may be drawn, as well as an infinite number of additional orientations in three-dimensional space, since molecules are in constant motion.

Figure \(\PageIndex{3}\) shows the bent three-dimensional shape of a water molecule that is due to the presence of lone pairs on the oxygen atom. The shapes of molecules are discussed in further detail in Section 11.7.