---
title: "Boron trifluoride gas and ammonia gas react at room temperature to form a solid adduct according to the following equation:  \\[ \\text{BF}_3(g) + \\text{NH}_3(g) \\rightarrow \\text{F}_3\\text{B-NH}_3(s) \\]  Which of the following correctly predicts the change in hybridization and \\(\\text{F-B-F}\\) bond angle around the boron atom as the product forms, along with the correct justification?"
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url: "https://nerd-notes.com/ubq/123468/"
date_modified: "2026-09-28T11:59:51+00:00"
---

# Boron trifluoride gas and ammonia gas react at room temperature to form a solid adduct according to the following equation:

\[ \text{BF}_3(g) + \text{NH}_3(g) \rightarrow \text{F}_3\text{B-NH}_3(s) \]

Which of the following correctly predicts the change in hybridization and \(\text{F-B-F}\) bond angle around the boron atom as the product forms, along with the correct justification?

Boron trifluoride gas and ammonia gas react at room temperature to form a solid adduct according to the following equation:

\[ \text{BF}_3(g) + \text{NH}_3(g) \rightarrow \text{F}_3\text{B-NH}_3(s) \]

Which of the following correctly predicts the change in hybridization and \(\text{F-B-F}\) bond angle around the boron atom as the product forms, along with the correct justification?

- **A.** The hybridization of the boron atom changes from \(\text{sp}^3\) to \(\text{sp}^2\), and the \(\text{F-B-F}\) bond angle increases from approximately \(109.5^\circ\) to \(120^\circ\) because the formation of the \(\text{B-N}\) bond reduces electron-electron repulsions around the boron center.
- **B.** The hybridization of the boron atom changes from \(\text{sp}^2\) to \(\text{sp}^3\), and the \(\text{F-B-F}\) bond angle decreases from \(120^\circ\) to approximately \(109.5^\circ\) because the boron atom accepts an electron pair to form a fourth bonding domain.
- **C.** The hybridization of the boron atom remains \(\text{sp}^2\), and the \(\text{F-B-F}\) bond angle decreases from \(120^\circ\) to approximately \(90^\circ\) because the incoming lone pair from \(\text{NH}_3\) occupies an unhybridized \(2p\) orbital perpendicular to the molecular plane.
- **D.** The hybridization of the boron atom changes from \(\text{sp}^2\) to \(\text{sp}^3\), and the \(\text{F-B-F}\) bond angle increases from approximately \(107^\circ\) to \(120^\circ\) because the boron atom completes its valence octet and adopts a planar geometry.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123468/*