Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
So I was searching about the difference between redox reactions and Lewis Acid-Base reactions and came across classical example of latter,
$$\ce{NH3 + BH3 \rightarrow NH3BH3}$$ Now in this example what is the oxidation state of borane? Is it $+4$, if yes then can this example be considered a type of redox reaction also, since boron went from $+3$ oxidation state in the products to $+4$ oxidation state?
As per this source, the idea and definition of oxidation state is based on the following principle:
The oxidation number of an atom in a molecule is based on a formalism that forces a covalent compound to possess complete ionic character and may be defined as the charge that an atom would have if all bonds to it are broken such that the ligands retain a closed-shell configuration ; an exception, however, refers to homonuclear bonds, in which case the bond is broken homolytically and a single electron is transferred to each atom.
The oxidation number may thus simply be expressed as Oxidation number = charge on compound - charge on ligands
So,evidently, for the presented molecule $\ce{NH3BH3}$, we see that $\ce{NH3}$ is a ligand for the $\ce{BH3}$ moiety. So,you can heterolytically cleave the dative bond between $\ce{N}$ and $\ce{B}$ towards $\ce{N}$(as per the electronegativity trends),leaving no charge on the $\ce{B}$ as of now,as after this action,it's valence shell contains 3 electrons.Now,for finding oxidation state on $\ce{B}$,an interesting thing crops up due to the presence of the three $\ce{B-H}$ bonds. The article further enumerates:
In many cases, the charges assigned to simple monoatomic ligands do not vary from compound to compound, as illus- trated by $\ce{F-}$, $\ce{Cl-}$, and $\ce{O^2-}$. However, a notable exception is provided by hydrogen for which both $\ce{H+}$ and $\ce{H-}$ have per- missible closed-shell configurations ($\ce{1s^0}$ and $\ce{1s^2}$ , respectively). In this case, the charge assigned to hydrogen is determined by the relative electronegativity of the atom to which it is attached.
So again, due to sligthly higher electronegativity value of $\ce{H}$ than $\ce{B}$,the $\ce{H}$ atom becomes the ligand for the $\ce{B-H}$ bond. Hence,all the $\ce{B-H}$ heterolytically cleave towards $\ce{H}$,each cleavage leading to a +1 charge on $\ce{B}$ and -1 on $\ce{H}$. In totality,$\ce{B}$ ends up with +3 oxidation state, as it has lost all three electrons from it's valence shell.
Note: I highly recommend everybody to read that paper cited above. It's really insightful into the difference between valence,oxidation number and coordination number,which are often used interchangeably
Don't count bonds. Count electrons. Here all the bonds to boron are polarized away from that atom as boron is less electronegative than both hydrogen and nitrogen. Since the boron also has no valence-shell lone pairs we count zero valence electrons dominated by the boron, versus the neutral atom having three. That drop from three valence electron to zero means an oxidation state of $+3$.
To reach $+4$ the boron would have to engage another electron in bonding to a more electronegative element, but that electron would have to come from the $1s$ core and it ain't happening.
