Question

The following reaction, $BB{r}_3\left(g\right)+BC{l}_3\left(g\right)\to BB{r}_{2}Cl\left(g\right)+BC{l}_{2}Br\left(g\right)$ has a $\Delta H$close to zero. Sketch the lewis structures of the four compounds, and explain why$\Delta H$ is so small.

Short answer

The reaction only trades one atom while the rest of the molecule remains unchanged.

This is also the reason why the enthalpy change is practically 0.

The bonds stay the same, which means that the bond enthalpies stay the same.

The enthalpies for bond breaking and bond forming will cancel each other out since they are the same in value but different in sign.

Step-by-step solution

Given data

The reaction is $BB{r}_3\left(g\right)+BC{l}_3\left(g\right)\to BB{r}_{2}Cl\left(g\right)+BC{l}_{2}Br\left(g\right).$

Concept of Lewis structure

A Lewis structure is a graphic representation of the electron distribution around atoms. We use them to see the type of bond that will form in molecules. It is based on the octet rule meaning that each atom acts towards filling up it's valence shell, or in other words to reach the octet.

Some basic rules for drawing Lewis structures:

Count the number of valence e-each atom brings into the molecule.

Put electron pairs about each atom such that there are 8 electrons around each atom (octet rule), with the exception of H, which is only surrounded by 2 electrons (there are a few other exceptions but this is the most common one)

If there is not enough electrons to follow the octet rule, then the least electronegative atom is left short of electrons.

If there are too many electrons to follow the octet rule, then the extra electrons are placed on the central atom.

The reaction in Lewis structure

The reaction is shown below.

The reason for ΔH being small

The reaction only trades one atom while the rest of the molecule remains unchanged.

This is also the reason why the enthalpy change is practically 0.

The bonds stay the same, which means that the bond enthalpies stay the same.

The enthalpies for bond breaking and bond forming will cancel each other out since they are the same in value but different in sign.