Question

draw the lewis electron dot structure for beh2. what is the vsepr shape of the particle? tetrahedral linear trigonal bent

Explanation:

Step1: Determine valence electrons

Beryllium (\(Be\)) has \(2\) valence electrons (\(2s^{2}\)), and each hydrogen (\(H\)) has \(1\) valence electron. So total valence electrons \(=2 + 2\times1=4\).

Step2: Draw Lewis structure

\(Be\) is the central atom. \(Be\) forms two single bonds with two \(H\) atoms. The Lewis structure is \(H - Be - H\).

Step3: Apply VSEPR theory

The formula for VSEPR is \(AX_{n}E_{m}\), where \(A\) is the central atom, \(X\) is the number of bonded atoms, and \(E\) is the number of lone pairs. For \(BeH_{2}\), \(n = 2\) (two \(H\) atoms bonded to \(Be\)) and \(m=0\) (no lone pairs on \(Be\)). According to VSEPR, \(AX_{2}\) has a linear shape.

Answer:

linear