Consider two molecules, \(\mathrm{BH}_{3}\) and \(\mathrm{NH}_{3} .\) Draw dot diagrams for both and indicate which one possesses and electron deficient atom.

Firstly, one needs to decide how many valence electrons each atom in the molecules BH3 and NH3 has. Boron (B) is in group 13 (3A) and has 3 valence electrons. Hydrogen (H) is in group 1 (1A) and has 1 valence electron. Nitrogen (N) is in group 15 (5A) and has 5 valence electrons.

Dot diagrams, or Lewis Structures, are drawn by locating the atoms, pairing up the valence electrons, and then distributing the remaining electrons. In \(\mathrm{BH}_{3}\), Boron (B) is the central atom with hydrogen atoms surrounding it. This is due to boron having less valence electrons. Here's how to draw it: - Place Boron (B) at the center and position the three Hydrogen (H) atoms around it. - Since Boron has 3 valence electrons and each Hydrogen has 1, pair them up. \[ \begin{array}{c} \text{H} \\ | \\ \text{B}--\text{H} \\ | \text{H} \end{array} \] For \(\mathrm{NH}_{3}\), Nitrogen (N) is the central atom surrounded by three Hydrogen atoms (H): - Place Nitrogen (N) at the center and three Hydrogen atoms around it. - Nitrogen has 5 valence electrons, each Hydrogen has 1. Start pairing up the electrons. Two electrons will remain unpaired on Nitrogen. \[ \begin{array}{c} \text{H} \\ | \text{N}--\text{H} \\ | \\ \text{H} \end{array} \] The two unpaired electrons are typically represented as a lone pair.

An electron-deficient atom is an atom that doesn't have a complete octet (8 electrons) in its valence shell. In \(\mathrm{BH}_{3}\), the central atom Boron (B) only has 6 electrons, so it's electron-deficient. In \(\mathrm{NH}_{3}\), the central atom Nitrogen (N) has a complete octet, so it's not electron-deficient. So, \(\mathrm{BH}_{3}\) possesses the electron deficient atom.