Questions: Draw the Lewis electron dot structure for BeH2. What is the VSEPR shape of the particle? linear tetrahedral bent trigonal

Transcript text: Draw the Lewis electron dot structure for $\mathrm{BeH}_{2}$. What is the VSEPR shape of the particle? linear tetrahedral bent trigonal

Solution

Solution Steps

Step 1: Determine the Total Number of Valence Electrons

Beryllium (Be) is in group 2 of the periodic table and has 2 valence electrons. Each hydrogen (H) atom has 1 valence electron. Therefore, the total number of valence electrons in $\mathrm{BeH}_2$ is:

\[ 2 + 2 \times 1 = 4 \]

Step 2: Draw the Lewis Structure

In the Lewis structure, beryllium (Be) is the central atom, and the two hydrogen atoms are bonded to it. Each Be-H bond uses 2 electrons. Thus, the Lewis structure is:

\[ \text{H:Be:H} \]

Beryllium shares its 2 electrons with the 2 hydrogen atoms, forming two single bonds.

Step 3: Determine the VSEPR Shape

The VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict the shape of the molecule. Beryllium in $\mathrm{BeH}_2$ has no lone pairs and forms two bonds. According to VSEPR theory, the shape that minimizes repulsion between these two bonding pairs is linear.