NH3 reacts with BF3 to form a single species. Which of the following structural diagrams is the most — Bonding Chemistry Question

STEPS:

1. Analyze the Lewis structures of the individual reactants:
* Ammonia ($\text{NH}_3$): Nitrogen has five valence electrons and forms three single covalent bonds with hydrogen atoms. This leaves one lone pair of non-bonding electrons on the nitrogen atom. Because it can donate this electron pair, ammonia acts as a Lewis base (an electron-pair donor).
* Boron trifluoride ($\text{BF}_3$): Boron has three valence electrons and forms three single covalent bonds with fluorine atoms. In this molecule, boron has only six valence electrons in its outer shell, leaving it two electrons short of a stable octet. Because of this electron deficiency, boron has an empty valence orbital and acts as a strong Lewis acid (an electron-pair acceptor).
2. Determine the nature of the reaction:
* When $\text{NH}_3$ and $\text{BF}_3$ react, they undergo a Lewis acid-base reaction.
* The nitrogen atom in ammonia donates its lone pair of electrons directly into the empty orbital of the boron atom.
3. Analyze the chemical bond formed:
* This donation of an electron pair from nitrogen to boron forms a coordinate covalent bond (also called a dative bond).
* This bond connects the nitrogen and boron atoms directly to each other, creating a single stable species known as a Lewis acid-base adduct:
$$\text{H}_3\text{N}: + \text{BF}_3 \rightarrow \text{H}_3\text{N}\rightarrow\text{BF}_3$$
4. Verify the octets of the atoms in the product:
* In the resulting adduct, both nitrogen and boron successfully satisfy the octet rule:
* Nitrogen has four single covalent bonds (three to hydrogen atoms and one to boron), totaling 8 valence electrons.
* Boron has four single covalent bonds (three to fluorine atoms and one to nitrogen), totaling 8 valence electrons.
5. Identify the matching diagram:
* Examining the options, Diagram D is the only structure showing a direct central single bond linking the nitrogen atom of the $\text{NH}_3$ group to the boron atom of the $\text{BF}_3$ group. Therefore, the correct option is Option D.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This diagram shows a hydrogen atom forming two covalent bonds simultaneously to act as a bridging atom between nitrogen and fluorine ($\text{N–H–F}$). Because hydrogen only has a single $1s$ valence orbital, it can share a maximum of two electrons and cannot form more than one covalent bond.
• Option B is incorrect: Similar to Option A, this diagram shows a hydrogen atom forming two covalent bonds at the same time to bridge nitrogen and boron ($\text{N–H–B}$). This is chemically impossible due to the valence constraints of hydrogen.
• Option C is incorrect: This diagram shows a fluorine atom forming two single covalent bonds to act as a bridge between nitrogen and boron ($\text{N–F–B}$). Fluorine is extremely electronegative, has seven valence electrons, and highly prefers to form only a single covalent bond to complete its octet. It does not act as a central bridging atom in this reaction; the coordinate covalent bond must form directly between the electron donor (Nitrogen) and electron acceptor (Boron).