Problem 95 - Chemistry

Question

Boron trifluoride reacts with dimethyl ether to form a compound with a coordinate covalent bond. Assign formal charges to the \(\mathrm{B}\) and \(\mathrm{O}\) atoms in both the reactants and products.

Step-by-Step Solution

Answer

In BF₃, B is 0 and in (CH₃)₂O, O is 0; in BF₃·O(CH₃)₂, B is -1 and O is +1.

Step 1: Review Formal Charge Formula

The formal charge (FC) of an atom can be calculated using the formula: \( FC = V - (L + \frac{N}{2}) \), where \( V \) is the number of valence electrons in the free atom, \( L \) is the number of lone pair electrons, and \( N \) is the number of electrons in bonds with other atoms.

Step 2: Assign Formal Charge to Boron in BF₃

Boron in boron trifluoride, \( \text{BF}_3 \), has 3 valence electrons, no lone pairs, and is bonded to three fluorine atoms (each sharing one electron). Thus, \( FC_{B} = 3 - (0 + \frac{6}{2}) = 3 - 3 = 0 \). The formal charge on boron is 0.

Step 3: Assign Formal Charge to Oxygen in (CH₃)₂O

Oxygen in dimethyl ether, \( \text{(CH}_3)_2\text{O} \), has 6 valence electrons, two lone pairs (4 electrons), and is bonded to two carbon atoms (each bond sharing one electron per carbon). Thus, \( FC_{O} = 6 - (4 + \frac{4}{2}) = 6 - 6 = 0 \). The formal charge on oxygen is 0.

Step 4: Consider the Formation of BF₃⋅O(CH₃)₂

When \( \text{BF}_3 \) reacts with \( \text{(CH}_3)_2\text{O} \) to form \( \text{BF}_3\cdot\text{O(CH}_3)_2 \), a coordinate covalent bond is formed where oxygen donates a lone pair.

Step 5: Assign Formal Charge to Boron in the Product

In \( \text{BF}_3\cdot\text{O(CH}_3)_2 \), boron is now bonded to four atoms: three fluorine and one oxygen. It has 3 valence electrons, no lone pairs, and participates in 4 bonds. Thus, \( FC_{B} = 3 - (0 + \frac{8}{2}) = 3 - 4 = -1 \). The formal charge on boron becomes -1.

Step 6: Assign Formal Charge to Oxygen in the Product

In the compound \( \text{BF}_3\cdot\text{O(CH}_3)_2 \), oxygen donates a lone pair to form the coordinate bond, remaining bonded to two carbon atoms as well. Thus, \( FC_{O} = 6 - (2 + \frac{6}{2}) = 6 - 5 = +1 \). The formal charge on oxygen becomes +1.

Key Concepts

Coordinate Covalent Bond

A coordinate covalent bond arises when one atom provides both electrons for a shared pair in a chemical bond.
Such bonds are also known as dative covalent bonds.

This happens when an atom with a lone pair of electrons bonds with another atom that lacks enough electrons to fulfill its valence shell.

• In a Lewis structure, this bond is often denoted by an arrow pointing from the electron donor to the electron acceptor.
• It does not differ in strength from other covalent bonds; it's essentially a matter of electron sharing origins.

The formation of a coordinate covalent bond plays a critical role in chemical reactions, as seen in the interaction between boron trifluoride and dimethyl ether. Here, oxygen from dimethyl ether donates a pair of electrons to boron trifluoride, forming a stable compound.
Understanding such bonds is crucial for predicting molecular behavior and reactivity.

Boron Trifluoride

Boron trifluoride, with the chemical formula \(\text{BF}_3\), is a compound formed from one boron atom and three fluorine atoms.
It is a planar molecule due to its trigonal planar geometry with \(120^\circ\) bond angles.

The boron atom has three valence electrons, which bond with one electron from each fluorine atom.

• This results in boron having just six electrons in its valence shell, making it electron-deficient.
• Such electron deficiency makes \(\text{BF}_3\) a Lewis acid as it is on the lookout for electrons to achieve a stable octet.

When it reacts with electron-rich molecules like dimethyl ether, \(\text{BF}_3\) easily forms a coordinate covalent bond, compensating for its electron shortfall.
This characteristic is often exploited in industrial chemical processes, where \(\text{BF}_3\) acts as a catalyst.

Dimethyl Ether

Dimethyl ether, \(\text{C}_2\text{H}_6\text{O}\), is a simple ether composed of two methyl groups attached to an oxygen atom.
It has a bent molecular geometry, similar to water, due to the lone pairs on oxygen causing a repulsion effect, leading to a calculated bond angle.

• This structure allows oxygen to easily form a coordinate covalent bond due to its lone pair of electrons.
• In chemistry, dimethyl ether is often utilized for its reactivity with substances that are electron-deficient.

It reacts with boron trifluoride by donating a pair of electrons from the oxygen to the boron, effectively creating a complex with altered electron distribution.
Consequently, in reaction systems, dimethyl ether serves both as a reactant and an influential component in catalytic processes and fuel formulations.